5 Sequences and Series 数列与级数

本页译自 OpenStax《Calculus Volume 2》第 5 章 Sequences and Series（数列与级数）：5.1–5.6 六节（数列／无穷级数／发散与积分判别法／比较判别法／交错级数／比值与根值判别法）+ Chapter Review / Key Terms / Key Equations / Key Concepts / Review Exercises 全译，段段对照。公式经本地 MathJax 渲染，自定义宏已注入。

5.1 Sequences 5.1 数列

5.1.1 Find the formula for the general term of a sequence.
5.1.2 Calculate the limit of a sequence if it exists.
5.1.3 Determine the convergence or divergence of a given sequence.

5.1.1 求数列通项的公式。
5.1.2 若数列的极限存在，计算该极限。
5.1.3 判定给定数列的收敛性或发散性。

In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.

本节引入数列，并给出数列收敛或发散的定义。我们将说明如何求收敛数列的极限，其方法通常是利用前面讨论过的函数极限的性质。本节最后介绍单调收敛定理，它是我们用来证明某些类型的数列收敛的工具。

Terminology of Sequences 数列的术语

To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form

为讨论这一新的主题，我们需要一些新的术语和定义。首先，无穷数列是形如下面这样的一个有序数列表：

$$a_{1},a_{2},a_{3}\text{,…},a_{n}\text{,…}\ \text{.}$$

Each of the numbers in the sequence is called a term. The symbol $n$ is called the index variable for the sequence. We use the notation

数列中的每一个数称为一项。符号 $n$ 称为该数列的下标变量。我们用记号

$$\left\{ a_{n} \right\}_{n = 1}^{\infty},\ \text{or simply}\ \left\{ a_{n} \right\},$$

to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number $a_{n}$ exists for each positive integer $n,$ we can also define a sequence as a function whose domain is the set of positive integers.

来表示这个数列。集合也使用类似的记号，但数列是有序的列表，而集合是无序的。由于对每个正整数 $n,$ 都存在一个确定的数 $a_{n}$，我们也可以把数列定义为一个定义域为正整数集的函数。

Let’s consider the infinite, ordered list

考虑下面这个无穷的有序列表：

$$2,4,8,16,32\text{,…}\ .$$

This is a sequence in which the first, second, and third terms are given by $a_{1} = 2,$ $a_{2} = 4,$ and $a_{3} = 8.$ You can probably see that the terms in this sequence have the following pattern:

这是一个数列，其第一、第二、第三项分别为 $a_{1} = 2,$ $a_{2} = 4,$ 与 $a_{3} = 8.$ 不难看出，该数列的各项具有如下规律：

$$a_{1} = 2^{1},\ a_{2} = 2^{2},\ a_{3} = 2^{3},\ a_{4} = 2^{4},\ \text{and}\ a_{5} = 2^{5}.$$

Assuming this pattern continues, we can write the $n\text{th}$ term in the sequence by the explicit formula $a_{n} = 2^{n}.$ Using this notation, we can write this sequence as

假定这一规律继续保持，我们可以用显式公式 $a_{n} = 2^{n}.$ 写出该数列的第 $n\text{th}$ 项。利用这一记号，该数列可写成

$$\left\{ 2^{n} \right\}_{n = 1}^{\infty}\ \text{or}\ \left\{ 2^{n} \right\}.$$

Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $n\text{th}$ term $a_{n}$ in terms of the previous term $a_{n - 1}.$ In particular, we can define this sequence as the sequence $\left\{ a_{n} \right\}$ where $a_{1} = 2$ and for all $n \geq 2,$ each term $a_{n}$ is defined by the recurrence relation$a_{n} = 2a_{n - 1}.$

另一方面，我们也可以换一种方式描述这个数列。由于每一项都是前一项的两倍，该数列可以递推地定义：把第 $n\text{th}$ 项 $a_{n}$ 用前一项 $a_{n - 1}.$ 表示。具体地说，我们可以把该数列定义为数列 $\left\{ a_{n} \right\}$，其中 $a_{1} = 2$，且对所有 $n \geq 2,$ 每一项 $a_{n}$ 由递推关系 $a_{n} = 2a_{n - 1}.$ 确定。

An infinite sequence$\left\{ a_{n} \right\}$ is an ordered list of numbers of the form

无穷数列 $\left\{ a_{n} \right\}$ 是形如下面这样的一个有序数列表：

$$a_{1},a_{2}\text{,…},a_{n}\text{,…}\ \text{.}$$

The subscript $n$ is called the index variable of the sequence. Each number $a_{n}$ is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case $a_{n} = f(n)$ for some function $f(n)$ defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.

下标 $n$ 称为该数列的下标变量。每个数 $a_{n}$ 是该数列的一项。有时数列由显式公式给出，此时 $a_{n} = f(n)$，其中 $f(n)$ 是定义在正整数上的某个函数。在另一些情形下，数列由递推关系给出。在递推关系中，数列的某一项（或若干项）被显式地给出，其后各项则用数列中较前的项来定义。

Note that the index does not have to start at $n = 1$ but could start with other integers. For example, a sequence given by the explicit formula $a_{n} = f(n)$ could start at $n = 0,$ in which case the sequence would be

注意，下标不必从 $n = 1$ 开始，也可以从其他整数开始。例如，由显式公式 $a_{n} = f(n)$ 给出的数列可以从 $n = 0,$ 开始，这时数列为

$$a_{0},a_{1},a_{2}\text{,…}\ \text{.}$$

Similarly, for a sequence defined by a recurrence relation, the term $a_{0}$ may be given explicitly, and the terms $a_{n}$ for $n \geq 1$ may be defined in terms of $a_{n - 1}.$ Since a sequence $\left\{ a_{n} \right\}$ has exactly one value for each positive integer $n,$ it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence $\left\{ a_{n} \right\}$ consists of all points $(n,a_{n})$ for all positive integers $n.$ Figure 5.2 shows the graph of $\left\{ 2^{n} \right\}.$

类似地，对于由递推关系定义的数列，可以显式给出项 $a_{0}$，而当 $n \geq 1$ 时各项 $a_{n}$ 用 $a_{n - 1}.$ 来定义。由于数列 $\left\{ a_{n} \right\}$ 对每个正整数 $n,$ 恰好取一个值，它可以描述为定义域是正整数集的函数。因此，讨论数列的图像是有意义的。数列 $\left\{ a_{n} \right\}$ 的图像由对所有正整数 $n.$ 的一切点 $(n,a_{n})$ 组成。图 5.2 给出了 $\left\{ 2^{n} \right\}.$ 的图像。

Figure 5.2 （图片占位）

图 5.2 （图片占位）

Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the *difference* between every pair of consecutive terms is the same. For example, consider the sequence

有两类数列经常出现，并有专门的名称：等差数列与等比数列（几何数列）。在等差数列中，任意相邻两项之间的*差*都相同。例如，考虑数列

$$3,7,11,15,19\text{,…}\ \text{.}$$

You can see that the difference between every consecutive pair of terms is $4.$ Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation

可以看到，任意相邻两项之差都是 $4.$ 假定这一规律继续保持，该数列就是等差数列。它可以用如下递推关系来描述：

$$\left\{ \begin{matrix} {a_{1} = 3} \\ {a_{n} = a_{n - 1} + 4\ \text{for}\ n \geq 2.} \end{matrix} \right.$$

Note that

注意到

$$\begin{array}{l} {a_{2} = 3 + 4} \\ {a_{3} = 3 + 4 + 4 = 3 + 2 \cdot 4} \\ {a_{4} = 3 + 4 + 4 + 4 = 3 + 3 \cdot 4.} \end{array}$$

Thus the sequence can also be described using the explicit formula

因此该数列也可以用显式公式来描述：

$$\begin{array}{cl} a_{n} & {= 3 + 4\left( {n - 1} \right)} \\ & {= 4n - 1.} \end{array}$$

In general, an arithmetic sequence is any sequence of the form $a_{n} = cn + b.$

一般地，等差数列就是任何形如 $a_{n} = cn + b.$ 的数列。

In a geometric sequence, the *ratio* of every pair of consecutive terms is the same. For example, consider the sequence

在等比数列中，任意相邻两项之*比*都相同。例如，考虑数列

$$2, - \frac{2}{3},\frac{2}{9}, - \frac{2}{27},\frac{2}{81}\text{,…}\ \text{.}$$

We see that the ratio of any term to the preceding term is $- \frac{1}{3}.$ Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as

可以看到，任一项与其前一项之比都是 $- \frac{1}{3}.$ 假定这一规律继续保持，该数列就是等比数列。它可以递推地定义为

$$\begin{array}{l} \\ {a_{1} = 2} \\ {a_{n} = - \frac{1}{3} \cdot a_{n - 1}\ \text{for}\ n \geq 2.} \end{array}$$

Alternatively, since

另一方面，由于

$$\begin{array}{l} \\ \\ {a_{2} = - \frac{1}{3} \cdot 2} \\ {a_{3} = \left( {- \frac{1}{3}} \right)\left( {- \frac{1}{3}} \right)(2) = \left( {- \frac{1}{3}} \right)^{2} \cdot 2} \\ {a_{4} = \left( {- \frac{1}{3}} \right)\left( {- \frac{1}{3}} \right)\left( {- \frac{1}{3}} \right)(2) = \left( {- \frac{1}{3}} \right)^{3} \cdot 2,} \end{array}$$

we see that the sequence can be described by using the explicit formula

我们看到，该数列可以用显式公式来描述：

$$a_{n} = 2\left( {- \frac{1}{3}} \right)^{n - 1}.$$

The sequence $\left\{ 2^{n} \right\}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2.$ In general, a geometric sequence is any sequence of the form $a_{n} = cr^{n}.$

我们前面讨论过的数列 $\left\{ 2^{n} \right\}$ 是等比数列，其中任一项与前一项之比为 $2.$ 一般地，等比数列就是任何形如 $a_{n} = cr^{n}.$ 的数列。

Finding Explicit Formulas 求显式公式

For each of the following sequences, find an explicit formula for the $n\text{th}$ term of the sequence.

对下面每个数列，求其第 $n\text{th}$ 项的显式公式。

1. $- \frac{1}{2},\frac{2}{3}, - \frac{3}{4},\frac{4}{5}, - \frac{5}{6}\text{,…}$

2. $\frac{3}{4},\frac{9}{7},\frac{27}{10},\frac{81}{13},\frac{243}{16}\text{,…}$

Solution 解答

1. First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the $n\text{th}$ term includes a factor of $(-1)^{n}.$ Next, consider the sequence of numerators $\left\{ 1,2,3\text{,…} \right\}$ and the sequence of denominators $\left\{ 2,3,4\text{,…} \right\}.$ We can see that both of these sequences are arithmetic sequences. The $n\text{th}$ term in the sequence of numerators is $n,$ and the $n\text{th}$ term in the sequence of denominators is $n + 1.$ Therefore, the sequence can be described by the explicit formula

1. 首先注意，该数列的符号在正负之间交错。数列中奇数项为负，偶数项为正。因此第 $n\text{th}$ 项含有因子 $(-1)^{n}.$ 其次，考察分子构成的数列 $\left\{ 1,2,3\text{,…} \right\}$ 与分母构成的数列 $\left\{ 2,3,4\text{,…} \right\}.$ 可以看到这两个数列都是等差数列。分子数列的第 $n\text{th}$ 项为 $n,$ 分母数列的第 $n\text{th}$ 项为 $n + 1.$ 因此，该数列可由显式公式描述：

$$a_{n} = \frac{(-1)^{n}n}{n + 1}.$$

2. The sequence of numerators $3,9,27,81,243\text{,…}$ is a geometric sequence. The numerator of the $n\text{th}$ term is $3^{n}$ The sequence of denominators $4,7,10,13,16\text{,…}$ is an arithmetic sequence. The denominator of the $n\text{th}$ term is $4 + 3\left( {n - 1} \right) = 3n + 1.$ Therefore, we can describe the sequence by the explicit formula $a_{n} = \frac{3^{n}}{3n + 1}.$

2. 分子构成的数列 $3,9,27,81,243\text{,…}$ 是等比数列。第 $n\text{th}$ 项的分子为 $3^{n}$。分母构成的数列 $4,7,10,13,16\text{,…}$ 是等差数列。第 $n\text{th}$ 项的分母为 $4 + 3\left( {n - 1} \right) = 3n + 1.$ 因此，该数列可由显式公式 $a_{n} = \frac{3^{n}}{3n + 1}.$ 描述。

Find an explicit formula for the $n\text{th}$ term of the sequence $\left\{ \frac{1}{5}, - \frac{1}{7},\frac{1}{9}, - \frac{1}{11}\text{,…} \right\}.$

求数列 $\left\{ \frac{1}{5}, - \frac{1}{7},\frac{1}{9}, - \frac{1}{11}\text{,…} \right\}.$ 第 $n\text{th}$ 项的显式公式。

Defined by Recurrence Relations 由递推关系定义的数列

For each of the following recursively defined sequences, find an explicit formula for the sequence.

对下面每个递推定义的数列，求该数列的显式公式。

1. $a_{1} = 2,$ $a_{n} = -3a_{n - 1}$ for $n \geq 2$

1. $a_{1} = 2,$ $a_{n} = -3a_{n - 1}$，当 $n \geq 2$

2. $a_{1} = \frac{1}{2},$ $a_{n} = a_{n - 1} + \left( \frac{1}{2} \right)^{n}$ for $n \geq 2$

2. $a_{1} = \frac{1}{2},$ $a_{n} = a_{n - 1} + \left( \frac{1}{2} \right)^{n}$，当 $n \geq 2$

Solution 解答

1. Writing out the first few terms, we have

1. 写出前几项，得

$$\begin{array}{l} {a_{1} = 2} \\ {a_{2} = -3a_{1} = -3(2)} \\ {a_{3} = -3a_{2} = (-3)^{2}2} \\ {a_{4} = -3a_{3} = (-3)^{3}2.} \end{array}$$

In general,

一般地，

$$a_{n} = 2(-3)^{n - 1}.$$

2. Write out the first few terms:

2. 写出前几项：

$$\begin{array}{l} \\ \\ {a_{1} = \frac{1}{2}} \\ {a_{2} = a_{1} + \left( \frac{1}{2} \right)^{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}} \\ {a_{3} = a_{2} + \left( \frac{1}{2} \right)^{3} = \frac{3}{4} + \frac{1}{8} = \frac{7}{8}} \\ {a_{4} = a_{3} + \left( \frac{1}{2} \right)^{4} = \frac{7}{8} + \frac{1}{16} = \frac{15}{16}.} \end{array}$$

From this pattern, we derive the explicit formula

由这一规律，我们导出显式公式

$$a_{n} = \frac{2^{n} - 1}{2^{n}} = 1 - \frac{1}{2^{n}}.$$

Find an explicit formula for the sequence defined recursively such that $a_{1} = -4$ and $a_{n} = a_{n - 1} + 6.$

求由 $a_{1} = -4$ 与 $a_{n} = a_{n - 1} + 6.$ 递推定义的数列的显式公式。

Limit of a Sequence 数列的极限

A fundamental question that arises regarding infinite sequences is the behavior of the terms as $n$ gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n\rightarrow\infty.$ For example, consider the following four sequences and their different behaviors as $n\rightarrow\infty$ (see Figure 5.3):

关于无穷数列，一个根本性的问题是当 $n$ 变大时数列各项的行为。由于数列是定义在正整数上的函数，讨论当 $n\rightarrow\infty$ 时各项的极限是有意义的。例如，考虑下面四个数列以及它们在 $n\rightarrow\infty$ 时的不同行为（见图 5.3）：

1. $\left\{ {1 + 3n} \right\} = \left\{ 4,7,10,13\text{,…} \right\}.$ The terms $1 + 3n$ become arbitrarily large as $n\rightarrow\infty.$ In this case, we say that $1 + 3n\rightarrow\infty$ as $n\rightarrow\infty.$

1. $\left\{ {1 + 3n} \right\} = \left\{ 4,7,10,13\text{,…} \right\}.$ 当 $n\rightarrow\infty$ 时，各项 $1 + 3n$ 变得任意大。在这种情况下，我们说当 $n\rightarrow\infty$ 时 $1 + 3n\rightarrow\infty$。

2. $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\} = \left\{ {\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16}\text{,…}} \right\}.$ The terms $1 - \left( \frac{1}{2} \right)^{n}\rightarrow 1$ as $n\rightarrow\infty.$

2. $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\} = \left\{ {\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16}\text{,…}} \right\}.$ 当 $n\rightarrow\infty$ 时，各项 $1 - \left( \frac{1}{2} \right)^{n}\rightarrow 1$。

3. $\left\{ (-1)^{n} \right\} = \left\{ \text{−}1,1,-1,1\text{,…} \right\}.$ The terms alternate but do not approach one single value as $n\rightarrow\infty.$

3. $\left\{ (-1)^{n} \right\} = \left\{ \text{−}1,1,-1,1\text{,…} \right\}.$ 各项正负交替，但在 $n\rightarrow\infty$ 时不趋于某个单一的值。

4. $\left\{ \frac{(-1)^{n}}{n} \right\} = \left\{ {-1,\frac{1}{2}, - \frac{1}{3},\frac{1}{4}\text{,…}} \right\}.$ The terms alternate for this sequence as well, but $\frac{(-1)^{n}}{n}\rightarrow 0$ as $n\rightarrow\infty.$

4. $\left\{ \frac{(-1)^{n}}{n} \right\} = \left\{ {-1,\frac{1}{2}, - \frac{1}{3},\frac{1}{4}\text{,…}} \right\}.$ 这个数列的各项同样正负交替，但当 $n\rightarrow\infty$ 时 $\frac{(-1)^{n}}{n}\rightarrow 0$。

From these examples, we see several possibilities for the behavior of the terms of a sequence as $n\rightarrow\infty.$ In two of the sequences, the terms approach a finite number as $n\rightarrow\infty.$ In the other two sequences, the terms do not. If the terms of a sequence approach a finite number $L$ as $n\rightarrow\infty,$ we say that the sequence is a convergent sequence and the real number $L$ is the limit of the sequence. We can give an informal definition here.

从这些例子中，我们看到当 $n\rightarrow\infty$ 时数列各项的行为有几种可能。在其中的两个数列中，各项当 $n\rightarrow\infty$ 时趋于一个有限数；在另外两个数列中则不然。如果数列的各项当 $n\rightarrow\infty$ 时趋于一个有限数 $L$，我们说该数列是收敛数列，而实数 $L$ 是数列的极限。这里我们可以给出一个非正式的定义。

Given a sequence $\left\{ a_{n} \right\},$ if the terms $a_{n}$ become arbitrarily close to a finite number $L$ as $n$ becomes sufficiently large, we say $\left\{ a_{n} \right\}$ is a convergent sequence and $L$ is the limit of the sequence. In this case, we write

给定数列 $\left\{ a_{n} \right\}$，如果当 $n$ 充分大时，各项 $a_{n}$ 与某个有限数 $L$ 任意接近，我们说 $\left\{ a_{n} \right\}$ 是收敛数列，而 $L$ 是数列的极限。在这种情况下，我们记作

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n} = L.$$

If a sequence $\left\{ a_{n} \right\}$ is not convergent, we say it is a divergent sequence.

如果数列 $\left\{ a_{n} \right\}$ 不收敛，我们说它是发散数列。

From Figure 5.3, we see that the terms in the sequence $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\}$ are becoming arbitrarily close to $1$ as $n$ becomes very large. We conclude that $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\}$ is a convergent sequence and its limit is $1.$ In contrast, from Figure 5.3, we see that the terms in the sequence $1 + 3n$ are not approaching a finite number as $n$ becomes larger. We say that $\left\{ 1 + 3n \right\}$ is a divergent sequence.

由图 5.3 可见，当 $n$ 变得非常大时，数列 $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\}$ 的各项与 $1$ 任意接近。我们由此得出 $\left\{ {1 - \left( \frac{1}{2} \right)^{n}} \right\}$ 是收敛数列，其极限为 $1$。相反，由图 5.3 可见，数列 $1 + 3n$ 的各项随着 $n$ 增大并不趋于有限数。我们说 $\left\{ 1 + 3n \right\}$ 是发散数列。

In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.

在数列极限的非正式定义中，我们使用了"任意接近"和"充分大"这两个说法。虽然这些措辞有助于说明收敛数列的含义，但它们多少有些含糊。为了更精确，我们现在给出数列极限的更正式的定义，并在图 5.4 中将这些思想用图形表示出来。

A sequence $\left\{ a_{n} \right\}$ converges to a real number $L$ if for all $\varepsilon > 0,$ there exists an integer $N$ such that $\left| {a_{n} - L} \right| < \varepsilon$ if $n \geq N.$ The number $L$ is the limit of the sequence and we write

数列 $\left\{ a_{n} \right\}$ 收敛于实数 $L$，当且仅当对任意 $\varepsilon > 0$，存在整数 $N$，使得当 $n \geq N$ 时 $\left| {a_{n} - L} \right| < \varepsilon$。数 $L$ 是数列的极限，我们记作

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n} = L\ or\ a_{n}\rightarrow L.$$

In this case, we say the sequence $\left\{ a_{n} \right\}$ is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.

在这种情况下，我们说数列 $\left\{ a_{n} \right\}$ 是收敛数列。如果一个数列不收敛，它就是发散数列，我们说极限不存在。

We remark that the convergence or divergence of a sequence $\left\{ a_{n} \right\}$ depends only on what happens to the terms $a_{n}$ as $n\rightarrow\infty.$ Therefore, if a finite number of terms $b_{1},b_{2}\text{,…},b_{N}$ are placed before $a_{1}$ to create a new sequence

我们指出，数列 $\left\{ a_{n} \right\}$ 的收敛或发散只取决于各项 $a_{n}$ 当 $n\rightarrow\infty$ 时的行为。因此，如果在 $a_{1}$ 之前放上有限多项 $b_{1},b_{2}\text{,…},b_{N}$，从而构造出一个新数列

$$b_{1},b_{2}\text{,…},b_{N},a_{1},a_{2}\text{,…},$$

this new sequence will converge if $\left\{ a_{n} \right\}$ converges and diverge if $\left\{ a_{n} \right\}$ diverges. Further, if the sequence $\left\{ a_{n} \right\}$ converges to $L,$ this new sequence will also converge to $L.$

那么当 $\left\{ a_{n} \right\}$ 收敛时这个新数列也收敛，当 $\left\{ a_{n} \right\}$ 发散时它也发散。进一步，如果数列 $\left\{ a_{n} \right\}$ 收敛于 $L$，那么这个新数列也收敛于 $L$。

As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences $\left\{ 1 + 3n \right\}$ and $\left\{ (-1)^{n} \right\}$ shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence $\left\{ (-1)^{n} \right\}$ diverges because the terms alternate between $1$ and $-1,$ but do not approach one value as $n\rightarrow\infty.$ On the other hand, the sequence $\left\{ 1 + 3n \right\}$ diverges because the terms $1 + 3n\rightarrow\infty$ as $n\rightarrow\infty.$ We say the sequence $\left\{ 1 + 3n \right\}$ diverges to infinity and write $\underset{n\rightarrow\infty}{\text{lim}}(1 + 3n) = \infty.$ It is important to recognize that this notation does not imply the limit of the sequence $\left\{ 1 + 3n \right\}$ exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence $\left\{ \text{−}5n + 2 \right\}$ diverges to negative infinity because $-5n + 2\rightarrow\text{−}\infty$ as $n\rightarrow\infty.$ We write this as $\underset{n\rightarrow\infty}{\text{lim}}\left( {-5n + 2} \right) = \rightarrow\text{−}\infty.$

如上所定义，如果一个数列不收敛，就称它为发散数列。例如，图 5.4 中所示数列 $\left\{ 1 + 3n \right\}$ 与 $\left\{ (-1)^{n} \right\}$ 都发散。然而，不同数列的发散方式可以不同。数列 $\left\{ (-1)^{n} \right\}$ 发散是因为各项在 $1$ 与 $-1$ 之间交替，但在 $n\rightarrow\infty$ 时不趋于某个值。另一方面，数列 $\left\{ 1 + 3n \right\}$ 发散是因为各项满足 $1 + 3n\rightarrow\infty$ 当 $n\rightarrow\infty$。我们说数列 $\left\{ 1 + 3n \right\}$ 发散到无穷大，并记作 $\underset{n\rightarrow\infty}{\text{lim}}(1 + 3n) = \infty$。重要的是认识到，这一记法并不意味着数列 $\left\{ 1 + 3n \right\}$ 的极限存在。事实上，该数列是发散的。把极限写成无穷大只是为了提供关于该数列为何发散的更多信息。一个数列也可以发散到负无穷大。例如，数列 $\left\{ \text{−}5n + 2 \right\}$ 发散到负无穷大，因为 $-5n + 2\rightarrow\text{−}\infty$ 当 $n\rightarrow\infty$。我们把它记作 $\underset{n\rightarrow\infty}{\text{lim}}\left( {-5n + 2} \right) = \rightarrow\text{−}\infty$。

Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence $\left\{ a_{n} \right\}$ and a related function $f$ defined on all positive real numbers such that $f(n) = a_{n}$ for all integers $n \geq 1.$ Since the domain of the sequence is a subset of the domain of $f,$ if $\underset{x\rightarrow\infty}{\text{lim}}f(x)$ exists, then the sequence converges and has the same limit. For example, consider the sequence $\left\{ \frac{1}{n} \right\}$ and the related function $f(x) = \frac{1}{x}.$ Since the function $f$ defined on all real numbers $x > 0$ satisfies $f(x) = \frac{1}{x}\rightarrow 0$ as $x\rightarrow\infty,$ the sequence $\left\{ \frac{1}{n} \right\}$ must satisfy $\frac{1}{n}\rightarrow 0$ as $n\rightarrow\infty.$

因为数列是定义域为正整数集的函数，我们可以利用函数极限的性质来判断数列是否收敛。例如，考虑数列 $\left\{ a_{n} \right\}$ 以及一个定义在全体正实数上的相关函数 $f$，使得对所有整数 $n \geq 1$ 有 $f(n) = a_{n}$。由于数列的定义域是 $f$ 定义域的子集，若 $\underset{x\rightarrow\infty}{\text{lim}}f(x)$ 存在，则数列收敛且具有相同的极限。例如，考虑数列 $\left\{ \frac{1}{n} \right\}$ 与相关函数 $f(x) = \frac{1}{x}$。由于定义在全体实数 $x > 0$ 上的函数 $f$ 满足当 $x\rightarrow\infty$ 时 $f(x) = \frac{1}{x}\rightarrow 0$，数列 $\left\{ \frac{1}{n} \right\}$ 必定满足当 $n\rightarrow\infty$ 时 $\frac{1}{n}\rightarrow 0$。

Limit of a Sequence Defined by a Function 由函数定义的数列的极限

Consider a sequence $\left\{ a_{n} \right\}$ such that $a_{n} = f(n)$ for all $n \geq 1.$ If there exists a real number $L$ such that

考虑数列 $\left\{ a_{n} \right\}$，对所有 $n \geq 1$ 有 $a_{n} = f(n)$。如果存在实数 $L$ 使得

$$\underset{x\rightarrow\infty}{\text{lim}}f(x) = L,$$

then $\left\{ a_{n} \right\}$ converges and

那么 $\left\{ a_{n} \right\}$ 收敛，且

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n} = L.$$

We can use this theorem to evaluate $\underset{n\rightarrow\infty}{\text{lim}}r^{n}$ for $0 \leq r \leq 1.$ For example, consider the sequence $\left\{ \left( {1\text{/}2} \right)^{n} \right\}$ and the related exponential function $f(x) = \left( {1\text{/}2} \right)^{x}.$ Since $\underset{x\rightarrow\infty}{\text{lim}}\left( {1\text{/}2} \right)^{x} = 0,$ we conclude that the sequence $\left\{ \left( {1\text{/}2} \right)^{n} \right\}$ converges and its limit is $0.$ Similarly, for any real number $r$ such that $0 \leq r < 1,$ $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = 0,$ and therefore the sequence $\left\{ r^{n} \right\}$ converges. On the other hand, if $r = 1,$ then $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = 1,$ and therefore the limit of the sequence $\left\{ 1^{n} \right\}$ is $1.$ If $r > 1,$ $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = \infty,$ and therefore we cannot apply this theorem. However, in this case, just as the function $r^{x}$ grows without bound as $n\rightarrow\infty,$ the terms $r^{n}$ in the sequence become arbitrarily large as $n\rightarrow\infty,$ and we conclude that the sequence $\left\{ r^{n} \right\}$ diverges to infinity if $r > 1.$

我们可以用这个定理来计算 $0 \leq r \leq 1$ 时的 $\underset{n\rightarrow\infty}{\text{lim}}r^{n}$。例如，考虑数列 $\left\{ \left( {1\text{/}2} \right)^{n} \right\}$ 与相关指数函数 $f(x) = \left( {1\text{/}2} \right)^{x}$。由于 $\underset{x\rightarrow\infty}{\text{lim}}\left( {1\text{/}2} \right)^{x} = 0$，我们得出结论：数列 $\left\{ \left( {1\text{/}2} \right)^{n} \right\}$ 收敛，其极限为 $0$。类似地，对任意满足 $0 \leq r < 1$ 的实数 $r$，有 $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = 0$，因此数列 $\left\{ r^{n} \right\}$ 收敛。另一方面，若 $r = 1$，则 $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = 1$，从而数列 $\left\{ 1^{n} \right\}$ 的极限为 $1$。若 $r > 1$，则 $\underset{x\rightarrow\infty}{\text{lim}}r^{x} = \infty$，因此我们不能应用这个定理。不过在这种情况下，正如函数 $r^{x}$ 当 $n\rightarrow\infty$ 时无界增长，数列中的各项 $r^{n}$ 当 $n\rightarrow\infty$ 时也变得任意大，因此我们得出结论：当 $r > 1$ 时，数列 $\left\{ r^{n} \right\}$ 发散到无穷大。

We summarize these results regarding the geometric sequence $\left\{ r^{n} \right\}\text{:}$

我们把关于几何数列 $\left\{ r^{n} \right\}$ 的这些结果总结如下：

$$\begin{array}{l} {r^{n}\rightarrow 0\ \textit{if}\ 0 < r < 1} \\ {r^{n}\rightarrow 1\ \textit{if}\ r = 1} \\ {r^{n}\rightarrow\infty\ \textit{if}\ r > 1.} \end{array}$$

Later in this section we consider the case when $r < 0.$

在本节稍后，我们将考虑 $r < 0$ 的情形。

We now consider slightly more complicated sequences. For example, consider the sequence $\left\{ {(2\text{/}3)}^{n} + {(1\text{/}4)}^{n} \right\}.$ The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences $\left\{ {(2\text{/}3)}^{n} \right\}$ and $\left\{ {(1\text{/}4)}^{n} \right\}.$ As we describe in the following algebraic limit laws, since $\left\{ {(2\text{/}3)}^{n} \right\}$ and $\left. \left\{ 1\text{/}4) \right.^{n} \right\}$ both converge to $0,$ the sequence $\left\{ {(2\text{/}3)}^{n} + {(1\text{/}4)}^{n} \right\}$ converges to $0 + 0 = 0.$ Just as we were able to evaluate a limit involving an algebraic combination of functions $f$ and $g$ by looking at the limits of $f$ and $g$ (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $a_{n}$ and $b_{n}$ by evaluating the limits of $\left\{ a_{n} \right\}$ and $\left\{ b_{n} \right\}.$

我们现在考虑略微复杂一些的数列。例如，考虑数列 $\left\{ {(2\text{/}3)}^{n} + {(1\text{/}4)}^{n} \right\}$。这个数列的各项比我们之前讨论过的其他数列更复杂，但幸运的是，该数列的极限由两个数列 $\left\{ {(2\text{/}3)}^{n} \right\}$ 与 $\left\{ {(1\text{/}4)}^{n} \right\}$ 的极限所决定。正如我们在下面的代数极限法则中所描述的，由于 $\left\{ {(2\text{/}3)}^{n} \right\}$ 与 $\left. \left\{ 1\text{/}4) \right.^{n} \right\}$ 都收敛于 $0$，数列 $\left\{ {(2\text{/}3)}^{n} + {(1\text{/}4)}^{n} \right\}$ 收敛于 $0 + 0 = 0$。正如我们可以通过考察函数 $f$ 与 $g$ 的极限来求涉及 $f$ 与 $g$ 的代数组合的极限（见"极限导论"），我们也能够通过求 $\left\{ a_{n} \right\}$ 与 $\left\{ b_{n} \right\}$ 的极限来求通项为 $a_{n}$ 与 $b_{n}$ 的代数组合的数列的极限。

Algebraic Limit Laws 代数极限法则

Given sequences $\left\{ a_{n} \right\}$ and $\left\{ b_{n} \right\}$ and any real number $c,$ if there exist constants $A$ and $B$ such that $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = A$ and $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = B,$ then

给定数列 $\left\{ a_{n} \right\}$ 与 $\left\{ b_{n} \right\}$ 以及任意实数 $c$，如果存在常数 $A$ 与 $B$ 使得 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = A$ 且 $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = B$，那么

1. $\underset{n\rightarrow\infty}{\text{lim}}c = c$

2. $\underset{n\rightarrow\infty}{\text{lim}}ca_{n} = c\underset{n\rightarrow\infty}{\text{lim}}a_{n} = cA$

3. $\underset{n\rightarrow\infty}{\text{lim}}\left( {a_{n} \pm b_{n}} \right) = \underset{n\rightarrow\infty}{\text{lim}}a_{n} \pm \underset{n\rightarrow\infty}{\text{lim}}b_{n} = A \pm B$

4. $\underset{\text{n}\rightarrow\infty}{\text{lim}}\left( {a_{n} \cdot b_{n}} \right) = \left( \underset{n\rightarrow\infty}{\text{lim}}a_{n} \right) \cdot \left( \underset{n\rightarrow\infty}{\text{lim}}b_{n} \right) = A \cdot B$

5. $\underset{n\rightarrow\infty}{\text{lim}}\left( \frac{a_{n}}{b_{n}} \right) = \frac{\underset{n\rightarrow\infty}{\text{lim}}a_{n}}{\underset{n\rightarrow\infty}{\text{lim}}b_{n}} = \frac{A}{B},$ provided $B \neq 0$ and each $b_{n} \neq 0.$

Proof 证明

We prove part iii.

我们证明第 (iii) 部分。

Let $\epsilon > 0.$ Since $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = A,$ there exists a constant positive integer $N_{1}$ such that $\left| a_{n} - A \middle| < \frac{\varepsilon}{2} \right.$ for all $n \geq N_{1}.$ Since $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = B,$ there exists a constant $N_{2}$ such that $\left| b_{n} - B \middle| < \varepsilon\text{/}2 \right.$ for all $n \geq N_{2}.$ Let $N$ be the larger of $N_{1}$ and $N_{2}.$ Therefore, for all $n \geq N,$

令 $\epsilon > 0$。由于 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = A$，存在常数正整数 $N_{1}$，使得对所有 $n \geq N_{1}$ 有 $\left| a_{n} - A \middle| < \frac{\varepsilon}{2} \right.$。由于 $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = B$，存在常数 $N_{2}$，使得对所有 $n \geq N_{2}$ 有 $\left| b_{n} - B \middle| < \varepsilon\text{/}2 \right.$。令 $N$ 为 $N_{1}$ 与 $N_{2}$ 中较大的一个。于是，对所有 $n \geq N$，

$|(a_{n} + b_{n})\text{−}(A + B)| \leq |a_{n} - A| + |b_{n} - B| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$

□

The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence $\left\{ \frac{1}{n^{2}} \right\}.$ As shown earlier, $\underset{n\rightarrow\infty}{\text{lim}}1\text{/}n = 0.$ Similarly, for any positive integer $k,$ we can conclude that

代数极限法则使我们能够求许多数列的极限。例如，考虑数列 $\left\{ \frac{1}{n^{2}} \right\}$。如前所示，$\underset{n\rightarrow\infty}{\text{lim}}1\text{/}n = 0$。类似地，对任意正整数 $k$，我们可以得出结论：

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{k}} = 0.$$

In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.

在下面的例子中，我们利用这一事实以及极限法则来求其他数列的极限。

Determining Convergence and Finding Limits 判定收敛性与求极限

For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.

对下面每个数列，判定它是否收敛。如果收敛，求出它的极限。

1. $\left\{ {5 - \frac{3}{n^{2}}} \right\}$

2. $\left\{ \frac{3n^{4} - 7n^{2} + 5}{6 - 4n^{4}} \right\}$

3. $\left\{ \frac{2^{n}}{n^{2}} \right\}$

4. $\left\{ \left( {1 + \frac{4}{n}} \right)^{n} \right\}$

Solution 解答

1. We know that $1\text{/}n\rightarrow 0.$ Using this fact, we conclude that

1. 我们知道 $1\text{/}n\rightarrow 0$。利用这一事实，我们得出结论：

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{2}} = \underset{n\rightarrow\infty}{\text{lim}}\left( \frac{1}{n} \right).\underset{n\rightarrow\infty}{\text{lim}}\left( \frac{1}{n} \right) = 0.$$

Therefore,

因此，

$$\underset{n\rightarrow\infty}{\text{lim}}\left( {5 - \frac{3}{n^{2}}} \right) = \underset{n\rightarrow\infty}{\text{lim}}5 - 3\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{2}} = 5 - 3 \cdot 0 = 5.$$

The sequence converges and its limit is $5.$

该数列收敛，其极限为 $5$。

2. By factoring $n^{4}$ out of the numerator and denominator and using the limit laws above, we have

2. 将 $n^{4}$ 从分子与分母中提出，并运用上面的极限法则，我们得到

$$\begin{array}{cl} {\underset{n\rightarrow\infty}{\text{lim}}\frac{3n^{4} - 7n^{2} + 5}{6 - 4n^{4}}} & {= \underset{n\rightarrow\infty}{\text{lim}}\frac{3 - \frac{7}{n^{2}} + \frac{5}{n^{4}}}{\frac{6}{n^{4}} - 4}} \\ & {= \frac{\underset{n\rightarrow\infty}{\text{lim}}\left( {3 - \frac{7}{n^{2}} + \frac{5}{n^{4}}} \right)}{\underset{n\rightarrow\infty}{\text{lim}}\left( {\frac{6}{n^{4}} - 4} \right)}} \\ & {= \frac{\left( {\underset{n\rightarrow\infty}{\text{lim}}(3)\text{−}\underset{n\rightarrow\infty}{\text{lim}}\frac{7}{n^{2}} + \underset{n\rightarrow\infty}{\text{lim}}\frac{5}{n^{4}}} \right)}{\left( {\underset{n\rightarrow\infty}{\text{lim}}\frac{6}{n^{4}} - \underset{n\rightarrow\infty}{\text{lim}}(4)} \right)}} \\ & {= \frac{\left( {\underset{n\rightarrow\infty}{\text{lim}}(3)\text{−}7 \cdot \underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{2}} + 5 \cdot \underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{4}}} \right)}{\left( {6 \cdot \underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{4}} - \underset{n\rightarrow\infty}{\text{lim}}(4)} \right)}} \\ & {= \frac{3 - 7 \cdot 0 + 5 \cdot 0}{6 \cdot 0 - 4} = - \frac{3}{4}.} \end{array}$$

The sequence converges and its limit is $-3\text{/}4.$

该数列收敛，其极限为 $-3\text{/}4$。

3. Consider the related function $f(x) = 2^{x}\text{/}x^{2}$ defined on all real numbers $x > 0.$ Since $2^{x}\rightarrow\infty$ and $x^{2}\rightarrow\infty$ as $x\rightarrow\infty,$ apply L’Hôpital’s rule and write

3. 考虑定义在全体实数 $x > 0$ 上的相关函数 $f(x) = 2^{x}\text{/}x^{2}$。由于当 $x\rightarrow\infty$ 时 $2^{x}\rightarrow\infty$ 且 $x^{2}\rightarrow\infty$，应用洛必达（L’Hôpital）法则，得到

$$\begin{array}{clccc} {\underset{x\rightarrow\infty}{\text{lim}}\frac{2^{x}}{x^{2}}} & {= \underset{x\rightarrow\infty}{\text{lim}}\frac{2^{x}\text{ln}\mspace{2mu} 2}{2x}} & & & \text{Take the derivatives of the numerator and denominator.} \\ & {= \underset{x\rightarrow\infty}{\text{lim}}\frac{2^{x}\left( {\text{ln}\mspace{2mu} 2} \right)^{2}}{2}} & & & \text{Take the derivatives again.} \\ & {= \infty.} & & & \end{array}$$

We conclude that the sequence diverges.

我们得出结论：该数列发散。

4. Consider the function $f(x) = \left( {1 + \frac{4}{x}} \right)^{x}$ defined on all real numbers $x > 0.$ This function has the indeterminate form $1^{\infty}$ as $x\rightarrow\infty.$ Let

4. 考虑定义在全体实数 $x > 0$ 上的函数 $f(x) = \left( {1 + \frac{4}{x}} \right)^{x}$。当 $x\rightarrow\infty$ 时，这个函数呈未定式 $1^{\infty}$。令

$$y = \underset{x\rightarrow\infty}{\text{lim}}\left( {1 + \frac{4}{x}} \right)^{x}.$$

Now taking the natural logarithm of both sides of the equation, we obtain

现在对等式两边同时取自然对数，我们得到

$$\text{ln}(y) = \text{ln}\left\lbrack {\underset{x\rightarrow\infty}{\text{lim}}\left( {1 + \frac{4}{x}} \right)^{x}} \right\rbrack.$$

Since the function $f(x) = \text{ln}\mspace{2mu} x$ is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,

由于函数 $f(x) = \text{ln}\mspace{2mu} x$ 在其定义域上连续，我们可以交换极限与自然对数的次序。因此，

$$\text{ln}(y) = \underset{x\rightarrow\infty}{\text{lim}}\left\lbrack {\text{ln}\left( {1 + \frac{4}{x}} \right)^{x}} \right\rbrack.$$

Using properties of logarithms, we write

利用对数的性质，我们写出

$$\underset{x\rightarrow\infty}{\text{lim}}\left\lbrack {\text{ln}\left( {1 + \frac{4}{x}} \right)^{x}} \right\rbrack = \underset{x\rightarrow\infty}{\text{lim}}x\mspace{2mu}\text{ln}\left( {1 + \frac{4}{x}} \right).$$

Since the right-hand side of this equation has the indeterminate form $\infty \cdot 0,$ rewrite it as a fraction to apply L’Hôpital’s rule. Write

由于该等式右边呈未定式 $\infty \cdot 0$，将其改写为分式以应用洛必达法则。写出

$$\underset{x\rightarrow\infty}{\text{lim}}x\mspace{2mu}\text{ln}\left( {1 + \frac{4}{x}} \right) = \underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\left( {1 + 4\text{/}x} \right)}{1\text{/}x}.$$

Since the right-hand side is now in the indeterminate form $0\text{/}0,$ we are able to apply L’Hôpital’s rule. We conclude that

由于右边现在呈未定式 $0\text{/}0$，我们可以应用洛必达法则。我们得出结论：

$$\underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\left( {1 + 4\text{/}x} \right)}{1\text{/}x} = \underset{x\rightarrow\infty}{\text{lim}}\frac{4}{1 + 4\text{/}x} = 4.$$

Therefore, $\text{ln}(y) = 4$ and $y = e^{4}.$ Therefore, since $\underset{x\rightarrow\infty}{\text{lim}}\left( {1 + \frac{4}{x}} \right)^{x} = e^{4},$ we can conclude that the sequence $\left\{ \left( {1 + \frac{4}{n}} \right)^{n} \right\}$ converges to $e^{4}.$

因此，$\text{ln}(y) = 4$ 且 $y = e^{4}$。于是由 $\underset{x\rightarrow\infty}{\text{lim}}\left( {1 + \frac{4}{x}} \right)^{x} = e^{4}$，我们可以得出结论：数列 $\left\{ \left( {1 + \frac{4}{n}} \right)^{n} \right\}$ 收敛于 $e^{4}$。

Consider the sequence $\left\{ \left( {5n^{2} + 1} \right)\text{/}e^{n} \right\}.$ Determine whether or not the sequence converges. If it converges, find its limit.

考虑数列 $\left\{ \left( {5n^{2} + 1} \right)\text{/}e^{n} \right\}$。判定该数列是否收敛。如果收敛，求出它的极限。

Recall that if $f$ is a continuous function at a value $L,$ then $f(x)\rightarrow f(L)$ as $x\rightarrow L.$ This idea applies to sequences as well. Suppose a sequence $a_{n}\rightarrow L,$ and a function $f$ is continuous at $L.$ Then $f\left( a_{n} \right)\rightarrow f(L).$ This property often enables us to find limits for complicated sequences. For example, consider the sequence $\sqrt{5 - \frac{3}{n^{2}}}.$ From Example 5.3a. we know the sequence $5 - \frac{3}{n^{2}}\rightarrow 5.$ Since $\sqrt{x}$ is a continuous function at $x = 5,$

回想一下，如果 $f$ 在值 $L$ 处连续，那么当 $x\rightarrow L$ 时 $f(x)\rightarrow f(L)$。这一思想对数列同样适用。假设数列 $a_{n}\rightarrow L$，且函数 $f$ 在 $L$ 处连续。那么 $f\left( a_{n} \right)\rightarrow f(L)$。这一性质常常使我们能够求复杂数列的极限。例如，考虑数列 $\sqrt{5 - \frac{3}{n^{2}}}$。由示例 5.3a. 可知，数列 $5 - \frac{3}{n^{2}}\rightarrow 5$。由于 $\sqrt{x}$ 在 $x = 5$ 处是连续函数，

$$\underset{n\rightarrow\infty}{\text{lim}}\sqrt{5 - \frac{3}{n^{2}}} = \sqrt{\underset{n\rightarrow\infty}{\text{lim}}\left( {5 - \frac{3}{n^{2}}} \right)} = \sqrt{5}.$$

Continuous Functions Defined on Convergent Sequences 定义在收敛数列上的连续函数

Consider a sequence $\left\{ a_{n} \right\}$ and suppose there exists a real number $L$ such that the sequence $\left\{ a_{n} \right\}$ converges to $L.$ Suppose $f$ is a continuous function at $L.$ Then there exists an integer $N$ such that $f$ is defined at all values $a_{n}$ for $n \geq N,$ and the sequence $\left\{ {f\left( a_{n} \right)} \right\}$ converges to $f(L)$ (Figure 5.5).

考虑数列 $\left\{ a_{n} \right\}$，并假设存在实数 $L$ 使得数列 $\left\{ a_{n} \right\}$ 收敛于 $L$。假设 $f$ 在 $L$ 处连续。那么存在整数 $N$，使得对一切 $n \geq N$，$f$ 在全部值 $a_{n}$ 处都有定义，且数列 $\left\{ {f\left( a_{n} \right)} \right\}$ 收敛于 $f(L)$（图 5.5）。

Proof 证明

Let $\epsilon > 0.$ Since $f$ is continuous at $L,$ there exists $\delta > 0$ such that $\left| {f(x) - f(L)} \right| < \varepsilon$ if $\left| {x - L} \right| < \delta.$ Since the sequence $\left\{ a_{n} \right\}$ converges to $L,$ there exists $N$ such that $\left| {a_{n} - L} \right| < \delta$ for all $n \geq N.$ Therefore, for all $n \geq N,$ $\left| {a_{n} - L} \right| < \delta,$ which implies $\left| {f(a_{n})\text{−}f(L)} \right| < \varepsilon.$ We conclude that the sequence $\left\{ {f\left( a_{n} \right)} \right\}$ converges to $f(L).$

令 $\epsilon > 0$。由于 $f$ 在 $L$ 处连续，存在 $\delta > 0$ 使得当 $\left| {x - L} \right| < \delta$ 时 $\left| {f(x) - f(L)} \right| < \varepsilon$。由于数列 $\left\{ a_{n} \right\}$ 收敛于 $L$，存在 $N$ 使得对所有 $n \geq N$ 有 $\left| {a_{n} - L} \right| < \delta$。因此，对一切 $n \geq N$，$\left| {a_{n} - L} \right| < \delta$，这意味着 $\left| {f(a_{n})\text{−}f(L)} \right| < \varepsilon$。我们得出结论：数列 $\left\{ {f\left( a_{n} \right)} \right\}$ 收敛于 $f(L)$。

□

Limits Involving Continuous Functions Defined on Convergent Sequences 涉及定义在收敛数列上的连续函数的极限

Determine whether the sequence $\left\{ \text{cos}\left( {3\text{/}n^{2}} \right) \right\}$ converges. If it converges, find its limit.

判定数列 $\left\{ \text{cos}\left( {3\text{/}n^{2}} \right) \right\}$ 是否收敛。如果收敛，求出它的极限。

Solution 解答

Since the sequence $\left\{ 3\text{/}n^{2} \right\}$ converges to $0$ and $\text{cos}\mspace{2mu} x$ is continuous at $x = 0,$ we can conclude that the sequence $\left\{ \text{cos}\left( {3\text{/}n^{2}} \right) \right\}$ converges and

由于数列 $\left\{ 3\text{/}n^{2} \right\}$ 收敛于 $0$ 且 $\text{cos}\mspace{2mu} x$ 在 $x = 0$ 处连续，我们可以得出结论：数列 $\left\{ \text{cos}\left( {3\text{/}n^{2}} \right) \right\}$ 收敛，且

$$\underset{n\rightarrow\infty}{\text{lim}}\text{cos}\left( \frac{3}{n^{2}} \right) = \text{cos}(0) = 1.$$

Determine if the sequence $\left\{ \sqrt{\frac{2n + 1}{3n + 5}} \right\}$ converges. If it converges, find its limit.

判定数列 $\left\{ \sqrt{\frac{2n + 1}{3n + 5}} \right\}$ 是否收敛。如果收敛，求出它的极限。

Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.

另一条涉及数列极限的定理是"极限导论"中所讨论的夹逼定理的推广。

Squeeze Theorem for Sequences 数列的夹逼定理

Consider sequences $\left\{ a_{n} \right\},$ $\left\{ b_{n} \right\},$ and $\left\{ c_{n} \right\}.$ Suppose there exists an integer $N$ such that

考虑数列 $\left\{ a_{n} \right\}$、$\left\{ b_{n} \right\}$ 与 $\left\{ c_{n} \right\}$。假设存在整数 $N$ 使得

$$a_{n} \leq b_{n} \leq c_{n}\ \text{for all}\ n \geq N.$$

If there exists a real number $L$ such that

如果存在实数 $L$ 使得

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n} = L = \underset{n\rightarrow\infty}{\text{lim}}c_{n},$$

then $\left\{ b_{n} \right\}$ converges and $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = L$ (Figure 5.6).

那么 $\left\{ b_{n} \right\}$ 收敛且 $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = L$（图 5.6）。

Proof 证明

Let $\varepsilon > 0.$ Since the sequence $\left\{ a_{n} \right\}$ converges to $L,$ there exists an integer $N_{1}$ such that $\left| {a_{n} - L} \right| < \varepsilon$ for all $n \geq N_{1}.$ Similarly, since $\left\{ c_{n} \right\}$ converges to $L,$ there exists an integer $N_{2}$ such that $\left| {c_{n} - L} \right| < \varepsilon$ for all $n \geq N_{2}.$ By assumption, there exists an integer $N$ such that $a_{n} \leq b_{n} \leq c_{n}$ for all $n \geq N.$ Let $M$ be the largest of $N_{1},N_{2},$ and $N.$ We must show that $\left| b_{n} - L \middle| < \varepsilon \right.$ for all $n \geq M.$ For all $n \geq M,$

令 $\varepsilon > 0$。由于数列 $\left\{ a_{n} \right\}$ 收敛于 $L$，存在整数 $N_{1}$ 使得对所有 $n \geq N_{1}$ 有 $\left| {a_{n} - L} \right| < \varepsilon$。类似地，由于 $\left\{ c_{n} \right\}$ 收敛于 $L$，存在整数 $N_{2}$ 使得对所有 $n \geq N_{2}$ 有 $\left| {c_{n} - L} \right| < \varepsilon$。由假设，存在整数 $N$ 使得对所有 $n \geq N$ 有 $a_{n} \leq b_{n} \leq c_{n}$。令 $M$ 为 $N_{1}$、$N_{2}$ 与 $N$ 中最大的一个。我们必须证明对所有 $n \geq M$ 有 $\left| b_{n} - L \middle| < \varepsilon \right.$。对一切 $n \geq M$，

$$\text{−}\varepsilon < \text{−}\left| {a_{n} - L} \right| \leq a_{n} - L \leq b_{n} - L \leq c_{n} - L \leq \left| {c_{n} - L} \right| < \varepsilon.$$

Therefore, $\text{−}\varepsilon < b_{n} - L < \varepsilon,$ and we conclude that $\left| {b_{n} - L} \right| < \varepsilon$ for all $n \geq M,$ and we conclude that the sequence $\left\{ b_{n} \right\}$ converges to $L.$

因此，$\text{−}\varepsilon < b_{n} - L < \varepsilon$，我们得出结论：对所有 $n \geq M$ 有 $\left| {b_{n} - L} \right| < \varepsilon$，并由此得出数列 $\left\{ b_{n} \right\}$ 收敛于 $L$。

□

Using the Squeeze Theorem 应用夹逼定理

Use the Squeeze Theorem to find the limit of each of the following sequences.

应用夹逼定理求下列每个数列的极限。

1. $\left\{ \frac{\text{cos}\mspace{2mu} n}{n^{2}} \right\}$

2. $\left\{ \left( {- \frac{1}{2}} \right)^{n} \right\}$

Solution 解答

1. Since $-1 \leq \text{cos}\mspace{2mu} n \leq 1$ for all integers $n,$ we have

1. 由于对所有整数 $n$ 有 $-1 \leq \text{cos}\mspace{2mu} n \leq 1$，我们得到

$$- \frac{1}{n^{2}} \leq \frac{\text{cos}\mspace{2mu} n}{n^{2}} \leq \frac{1}{n^{2}}.$$

Since $-1\text{/}n^{2}\rightarrow 0$ and $1\text{/}n^{2}\rightarrow 0,$ we conclude that $\text{cos}\mspace{2mu} n\text{/}n^{2}\rightarrow 0$ as well.

由于 $-1\text{/}n^{2}\rightarrow 0$ 且 $1\text{/}n^{2}\rightarrow 0$，我们得出结论：$\text{cos}\mspace{2mu} n\text{/}n^{2}\rightarrow 0$ 同样成立。

2. Since

2. 由于

$$- \frac{1}{2^{n}} \leq \left( {- \frac{1}{2}} \right)^{n} \leq \frac{1}{2^{n}}$$

for all positive integers $n,$ $-1\text{/}2^{n}\rightarrow 0$ and $1\text{/}2^{n}\rightarrow 0,$ we can conclude that ${(-1\text{/}2)}^{n}\rightarrow 0.$

对所有正整数 $n$，$-1\text{/}2^{n}\rightarrow 0$ 且 $1\text{/}2^{n}\rightarrow 0$，我们可以得出结论：${(-1\text{/}2)}^{n}\rightarrow 0$。

Find $\underset{n\rightarrow\infty}{\text{lim}}\frac{2n - \text{sin}\mspace{2mu} n}{n}.$

求 $\underset{n\rightarrow\infty}{\text{lim}}\frac{2n - \text{sin}\mspace{2mu} n}{n}$。

Using the idea from Example 5.5b. we conclude that $r^{n}\rightarrow 0$ for any real number $r$ such that $-1 < r < 0.$ If $r < \text{−}1,$ the sequence $\left\{ r^{n} \right\}$ diverges because the terms oscillate and become arbitrarily large in magnitude. If $r = -1,$ the sequence $\left\{ r^{n} \right\} = \left\{ (-1)^{n} \right\}$ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.

利用示例 5.5b. 的思想，我们得出结论：对任意满足 $-1 < r < 0$ 的实数 $r$，有 $r^{n}\rightarrow 0$。若 $r < \text{−}1$，数列 $\left\{ r^{n} \right\}$ 发散，因为其各项振荡且绝对值变得任意大。若 $r = -1$，如前面所讨论，数列 $\left\{ r^{n} \right\} = \left\{ (-1)^{n} \right\}$ 发散。下面是关于几何数列性质的总结。

$$r^{n}\rightarrow 0\ \text{if}\ |r| < 1$$ (5.1) $$r^{n}\rightarrow 1\ \text{if}\ r = 1$$ (5.2) $$r^{n}\rightarrow\infty\ \text{if}\ r > 1$$ (5.3) $$\left\{ r^{n} \right\}\ \text{diverges if}\ r \leq \text{−}1$$ (5.4)

Bounded Sequences 有界数列

We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.

我们现在把注意力转向涉及数列的最重要定理之一：单调收敛定理。在陈述该定理之前，我们需要引入一些术语和背景。我们首先定义数列有界的含义。

A sequence $\left\{ a_{n} \right\}$ is bounded above if there exists a real number $M$ such that

数列 $\left\{ a_{n} \right\}$ 是上有界的，如果存在实数 $M$ 使得

$$a_{n} \leq M$$

for all positive integers $n.$

对所有正整数 $n.$

A sequence $\left\{ a_{n} \right\}$ is bounded below if there exists a real number $M$ such that

数列 $\left\{ a_{n} \right\}$ 是下有界的，如果存在实数 $M$ 使得

$$M \leq a_{n}$$

for all positive integers $n.$

对所有正整数 $n.$

A sequence $\left\{ a_{n} \right\}$ is a bounded sequence if it is bounded above and bounded below.

若数列 $\left\{ a_{n} \right\}$ 既上有界又下有界，则它是（有界数列）。

If a sequence is not bounded, it is an unbounded sequence.

若一个数列无界，则它是无界数列。

For example, the sequence $\left\{ 1\text{/}n \right\}$ is bounded above because $1\text{/}n \leq 1$ for all positive integers $n.$ It is also bounded below because $1\text{/}n \geq 0$ for all positive integers *n.* Therefore, $\left\{ 1\text{/}n \right\}$ is a bounded sequence. On the other hand, consider the sequence $\left\{ 2^{n} \right\}.$ Because $2^{n} \geq 2$ for all $n \geq 1,$ the sequence is bounded below. However, the sequence is not bounded above. Therefore, $\left\{ 2^{n} \right\}$ is an unbounded sequence.

例如，数列 $\left\{ 1\text{/}n \right\}$ 是上有界的，因为对所有正整数 $n$ 都有 $1\text{/}n \leq 1$。它也是下有界的，因为对所有正整数 *n.* 都有 $1\text{/}n \geq 0$。因此 $\left\{ 1\text{/}n \right\}$ 是有界数列。另一方面，考虑数列 $\left\{ 2^{n} \right\}$。因为对所有 $n \geq 1$ 都有 $2^{n} \geq 2$，该数列是下有界的。然而，该数列不是上有界的。因此 $\left\{ 2^{n} \right\}$ 是无界数列。

We now discuss the relationship between boundedness and convergence. Suppose a sequence $\left\{ a_{n} \right\}$ is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $a_{n}$ that are arbitrarily large in magnitude as $n$ gets larger. As a result, the sequence $\left\{ a_{n} \right\}$ cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.

我们现在讨论有界性与收敛性之间的关系。设数列 $\left\{ a_{n} \right\}$ 无界。那么它不是上有界的，或不是下有界的，或两者都不是。无论哪种情况，随着 $n$ 增大，都会出现绝对值任意大的项 $a_{n}$。因此，数列 $\left\{ a_{n} \right\}$ 不可能收敛。所以，有界是数列收敛的必要条件。

Convergent Sequences Are Bounded 收敛数列必有界

If a sequence $\left\{ a_{n} \right\}$ converges, then it is bounded.

若数列 $\left\{ a_{n} \right\}$ 收敛，则它是有界的。

Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence $\left\{ (-1)^{n} \right\}$ is bounded, but the sequence diverges because the sequence oscillates between $1$ and $-1$ and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.

注意，数列有界并不是数列收敛的充分条件。例如，数列 $\left\{ (-1)^{n} \right\}$ 有界，但该数列发散，因为它在 $1$ 与 $-1$ 之间振荡，从不趋近于某个有限数。我们现在讨论有界数列收敛的一个充分（但非必要）条件。

Consider a bounded sequence $\left\{ a_{n} \right\}.$ Suppose the sequence $\left\{ a_{n} \right\}$ is increasing. That is, $a_{1} \leq a_{2} \leq a_{3}\text{…}.$ Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that $\left\{ a_{n} \right\}$ converges. For example, consider the sequence

考虑一个有界数列 $\left\{ a_{n} \right\}$。设数列 $\left\{ a_{n} \right\}$ 递增。即 $a_{1} \leq a_{2} \leq a_{3}\text{…}$。由于该数列递增，各项不振动。因此有两种可能：该数列可能发散到无穷大，也可能收敛。然而，由于该数列有界，它上有界，因而不能发散到无穷大。我们由此得出 $\left\{ a_{n} \right\}$ 收敛。例如，考虑数列

$$\left\{ {\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\text{,…}} \right\}.$$

Since this sequence is increasing and bounded above, it converges. Next, consider the sequence

由于该数列递增且有上界，它收敛。接下来，考虑数列

$$\left\{ {2,0,3,0,4,0,1, - \frac{1}{2}, - \frac{1}{3}, - \frac{1}{4}\text{,…}} \right\}.$$

Even though the sequence is not increasing for all values of $n,$ we see that $-1\text{/}2 < \text{−}1\text{/}3 < \text{−}1\text{/}4 < \text{⋯}.$ Therefore, starting with the eighth term, $a_{8} = -1\text{/}2,$ the sequence is increasing. In this case, we say the sequence is *eventually* increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.

尽管该数列并非对所有 $n$ 都递增，但我们注意到 $-1\text{/}2 < \text{−}1\text{/}3 < \text{−}1\text{/}4 < \text{⋯}$。因此，从第八项 $a_{8} = -1\text{/}2$ 开始，该数列递增。此时，我们说该数列是*最终*递增的。由于该数列有上界，它收敛。同样，若一个数列递减（或最终递减）且有下界，它也收敛。

A sequence $\left\{ a_{n} \right\}$ is increasing for all $n \geq n_{0}$ if

数列 $\left\{ a_{n} \right\}$ 对所有 $n \geq n_{0}$ 递增，如果

$$a_{n} \leq a_{n + 1}\ \text{for all}\ n \geq n_{0}.$$

A sequence $\left\{ a_{n} \right\}$ is decreasing for all $n \geq n_{0}$ if

数列 $\left\{ a_{n} \right\}$ 对所有 $n \geq n_{0}$ 递减，如果

$$a_{n} \geq a_{n + 1}\ \text{for all}\ n \geq n_{0}.$$

A sequence $\left\{ a_{n} \right\}$ is a monotone sequence for all $n \geq n_{0}$ if it is increasing for all $n \geq n_{0}$ or decreasing for all $n \geq n_{0}.$

数列 $\left\{ a_{n} \right\}$ 若对所有 $n \geq n_{0}$ 递增或对所有 $n \geq n_{0}$ 递减，则它是（单调数列）。

We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.

现在我们已经有了陈述单调收敛定理所需的全部定义，该定理给出了数列收敛的一个充分条件。

Monotone Convergence Theorem 单调收敛定理

If $\left\{ a_{n} \right\}$ is a bounded sequence and there exists a positive integer $n_{0}$ such that $\left\{ a_{n} \right\}$ is monotone for all $n \geq n_{0},$ then $\left\{ a_{n} \right\}$ converges.

若 $\left\{ a_{n} \right\}$ 是有界数列，且存在正整数 $n_{0}$ 使得 $\left\{ a_{n} \right\}$ 对所有 $n \geq n_{0}$ 单调，则 $\left\{ a_{n} \right\}$ 收敛。

The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).

本定理的证明超出了本书范围。作为替代，我们给出一幅图，从直观上说明该定理为何成立（图 5.7）。

In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.

在下面的示例中，我们展示如何利用单调收敛定理证明数列收敛。

Using the Monotone Convergence Theorem 应用单调收敛定理

For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.

对下列每个数列，使用单调收敛定理证明该数列收敛并求出其极限。

1. $\left\{ \frac{4^{n}}{n\text{!}} \right\}$

2. $\left\{ a_{n} \right\}$ defined recursively such that

2. $\left\{ a_{n} \right\}$ 由如下递推方式定义：

$$a_{1} = 2\ \text{and}\ a_{n + 1} = \frac{a_{n}}{2} + \frac{1}{2a_{n}}\ \text{for all}\ n \geq 2.$$

Solution 解答

1. Writing out the first few terms, we see that

1. 写出前几项，我们看到

$$\left\{ \frac{4^{n}}{n\text{!}} \right\} = \left\{ {4,8,\frac{32}{3},\frac{32}{3},\frac{128}{15}\text{,…}} \right\}.$$

At first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all $n \geq 3.$ We can show this as follows.

起初各项递增。然而，在第三项之后，各项递减。事实上，对所有 $n \geq 3$ 各项都递减。我们可以如下证明。

$$a_{n + 1} = \frac{4^{n + 1}}{\left( {n + 1} \right)\text{!}} = \frac{4}{n + 1} \cdot \frac{4^{n}}{n\text{!}} = \frac{4}{n + 1} \cdot a_{n} \leq a_{n}\ if\ n \geq 3.$$

Therefore, the sequence is decreasing for all $n \geq 3.$ Further, the sequence is bounded below by $0$ because $4^{n}\text{/}n\text{!} \geq 0$ for all positive integers $n.$ Therefore, by the Monotone Convergence Theorem, the sequence converges.

因此，该数列对所有 $n \geq 3$ 递减。此外，该数列以 $0$ 为下界，因为对所有正整数 $n$ 都有 $4^{n}\text{/}n\text{!} \geq 0$。因此，由单调收敛定理，该数列收敛。

To find the limit, we use the fact that the sequence converges and let $L = \underset{n\rightarrow\infty}{\text{lim}}a_{n}.$ Now note this important observation. Consider $\underset{n\rightarrow\infty}{\text{lim}}a_{n + 1}.$ Since

为求极限，我们利用该数列收敛这一事实，令 $L = \underset{n\rightarrow\infty}{\text{lim}}a_{n}$。现在注意这一重要观察。考虑 $\underset{n\rightarrow\infty}{\text{lim}}a_{n + 1}$。由于

$$\left\{ a_{n + 1} \right\} = \left\{ {a_{2,}a_{3},a_{4}\text{,…}} \right\},$$ the only difference between the sequences $\left\{ a_{n + 1} \right\}$ and $\left\{ a_{n} \right\}$ is that $\left\{ a_{n + 1} \right\}$ omits the first term. Since a finite number of terms does not affect the convergence of a sequence,

$$\left\{ a_{n + 1} \right\} = \left\{ {a_{2,}a_{3},a_{4}\text{,…}} \right\},$$ 数列 $\left\{ a_{n + 1} \right\}$ 与 $\left\{ a_{n} \right\}$ 的唯一区别在于 $\left\{ a_{n + 1} \right\}$ 略去了第一项。由于有限项不影响数列的收敛性，

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n + 1} = \underset{n\rightarrow\infty}{\text{lim}}a_{n} = L.$$

Combining this fact with the equation

将这一事实与方程

$$a_{n + 1} = \frac{4}{n + 1}a_{n}$$

and taking the limit of both sides of the equation

并对方程两边取极限

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n + 1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{4}{n + 1}a_{n},$$

we can conclude that

我们可以得出

$$L = 0 \cdot L = 0.$$

2. Writing out the first several terms,

2. 写出前几项，

$$\left\{ {2,\frac{5}{4},\frac{41}{40},\frac{3281}{3280}\text{,…}} \right\}.$$

we can conjecture that the sequence is decreasing and bounded below by $1.$ To show that the sequence is bounded below by $1,$ we can show that

我们可以猜想该数列递减且以 $1$ 为下界。为证明该数列以 $1$ 为下界，我们可以证明

$$\frac{a_{n}}{2} + \frac{1}{2a_{n}} \geq 1.$$

To show this, first rewrite

为证明这一点，先改写

$$\frac{a_{n}}{2} + \frac{1}{2a_{n}} = \frac{a_{n}^{2} + 1}{2a_{n}}.$$

Since $a_{1} > 0$ and $a_{2}$ is defined as a sum of positive terms, $a_{2} > 0.$ Similarly, all terms $a_{n} > 0.$ Therefore,

由于 $a_{1} > 0$ 且 $a_{2}$ 定义为正项之和，故 $a_{2} > 0$。同理，所有项都满足 $a_{n} > 0$。因此，

$$\frac{a_{n}^{2} + 1}{2a_{n}} \geq 1$$

if and only if

当且仅当

$$a_{n}^{2} + 1 \geq 2a_{n}.$$

Rewriting the inequality $a_{n}^{2} + 1 \geq 2a_{n}$ as $a_{n}^{2} - 2a_{n} + 1 \geq 0,$ and using the fact that

将不等式 $a_{n}^{2} + 1 \geq 2a_{n}$ 改写为 $a_{n}^{2} - 2a_{n} + 1 \geq 0$，并利用

$$a_{n}^{2} - 2a_{n} + 1 = {(a_{n} - 1)}^{2} \geq 0$$

because the square of any real number is nonnegative, we can conclude that

因为任意实数的平方非负，我们可以得出

$$\frac{a_{n}}{2} + \frac{1}{2a_{n}} \geq 1.$$

To show that the sequence is decreasing, we must show that $a_{n + 1} \leq a_{n}$ for all $n \geq 1.$ Since $1 \leq a_{n}^{2},$ it follows that

为证明该数列递减，我们必须证明对所有 $n \geq 1$ 都有 $a_{n + 1} \leq a_{n}$。由于 $1 \leq a_{n}^{2}$，于是

$$a_{n}^{2} + 1 \leq 2a_{n}^{2}.$$

Dividing both sides by $2a_{n},$ we obtain

两边同除以 $2a_{n}$，得到

$$\frac{a_{n}}{2} + \frac{1}{2a_{n}} \leq a_{n}.$$

Using the definition of $a_{n + 1},$ we conclude that

利用 $a_{n + 1}$ 的定义，我们得出

$$a_{n + 1} = \frac{a_{n}}{2} + \frac{1}{2a_{n}} \leq a_{n}.$$

Since $\left\{ a_{n} \right\}$ is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.

由于 $\left\{ a_{n} \right\}$ 有下界且递减，由单调收敛定理，它收敛。

To find the limit, let $L = \underset{n\rightarrow\infty}{\text{lim}}a_{n}.$ Then using the recurrence relation and the fact that $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = \underset{n\rightarrow\infty}{\text{lim}}a_{n + 1},$ we have

为求极限，令 $L = \underset{n\rightarrow\infty}{\text{lim}}a_{n}$。然后利用递推关系以及 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = \underset{n\rightarrow\infty}{\text{lim}}a_{n + 1}$ 这一事实，我们有

$$\underset{n\rightarrow\infty}{\text{lim}}a_{n + 1} = \underset{n\rightarrow\infty}{\text{lim}}\left( {\frac{a_{n}}{2} + \frac{1}{2a_{n}}} \right),$$

and therefore

因此

$$L = \frac{L}{2} + \frac{1}{2L}.$$

Multiplying both sides of this equation by $2L,$ we arrive at the equation

将方程两边同乘以 $2L$，得到方程

$$2L^{2} = L^{2} + 1.$$

Solving this equation for $L,$ we conclude that $L^{2} = 1,$ which implies $L = \text{±}1.$ Since all the terms are positive, the limit $L = 1.$

解这个方程求 $L$，我们得出 $L^{2} = 1$，即 $L = \text{±}1$。由于所有项均为正，极限 $L = 1$。

Consider the sequence $\left\{ a_{n} \right\}$ defined recursively such that $a_{1} = 1,$ $a_{n} = a_{n - 1}\text{/}2.$ Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.

考虑由递推定义的数列 $\left\{ a_{n} \right\}$，其中 $a_{1} = 1$，$a_{n} = a_{n - 1}\text{/}2$。使用单调收敛定理证明该数列收敛并求出其极限。

Fibonacci Numbers 斐波那契数

The Fibonacci numbers are defined recursively by the sequence $\left\{ F_{n} \right\}$ where $F_{0} = 0,$ $F_{1} = 1$ and for $n \geq 2,$

斐波那契数由数列 $\left\{ F_{n} \right\}$ 递推定义，其中 $F_{0} = 0$，$F_{1} = 1$，且对 $n \geq 2$，

$$F_{n} = F_{n - 1} + F_{n - 2}.$$

Here we look at properties of the Fibonacci numbers.

这里我们考察斐波那契数的性质。

1. Write out the first twenty Fibonacci numbers.

1. 写出前二十个斐波那契数。

2. Find a closed formula for the Fibonacci sequence by using the following steps.

2. 利用下列步骤求出斐波那契数列的闭式公式。

1. Consider the recursively defined sequence $\left\{ x_{n} \right\}$ where $x_{o} = c$ and $x_{n + 1} = ax_{n}.$ Show that this sequence can be described by the closed formula $x_{n} = ca^{n}$ for all $n \geq 0.$

1. 考虑由递推定义的数列 $\left\{ x_{n} \right\}$，其中 $x_{o} = c$ 且 $x_{n + 1} = ax_{n}$。证明该数列可用闭式公式 $x_{n} = ca^{n}$ 对所有 $n \geq 0$ 描述。

2. Using the result from part a. as motivation, look for a solution of the equation

2. 以 a 部分的结果作为启发，寻找方程

$$F_{n} = F_{n - 1} + F_{n - 2}$$

of the form $F_{n} = c\lambda^{n}.$ Determine what two values for $\lambda$ will allow $F_{n}$ to satisfy this equation.

形如 $F_{n} = c\lambda^{n}$ 的解。确定 $\lambda$ 的哪两个取值能使 $F_{n}$ 满足该方程。

3. Consider the two solutions from part b.: $\lambda_{1}$ and $\lambda_{2}.$ Let $F_{n} = c_{1}\lambda_{1}{}^{n} + c_{2}\lambda_{2}{}^{n}.$ Use the initial conditions $F_{0}$ and $F_{1}$ to determine the values for the constants $c_{1}$ and $c_{2}$ and write the closed formula $F_{n}.$

3. 考虑 b 部分的两个解：$\lambda_{1}$ 与 $\lambda_{2}$。令 $F_{n} = c_{1}\lambda_{1}{}^{n} + c_{2}\lambda_{2}{}^{n}$。利用初值条件 $F_{0}$ 与 $F_{1}$ 确定常数 $c_{1}$ 与 $c_{2}$ 的值，并写出闭式公式 $F_{n}$。

3. Use the answer in 2 c. to show that

3. 利用 2 c 中的答案证明

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{F_{n + 1}}{F_{n}} = \frac{1 + \sqrt{5}}{2}.$$

The number $\phi = \left( {1 + \sqrt{5}} \right)\text{/}2$ is known as the golden ratio (Figure 5.8 and Figure 5.9).

数 $\phi = \left( {1 + \sqrt{5}} \right)\text{/}2$ 被称为黄金比例（图 5.8 与图 5.9）。

Section 5.1 Exercises 5.1 节习题

Find the first six terms of each of the following sequences, starting with $n = 1.$

求下列各数列的前六项，从 $n = 1$ 开始。

$a_{n} = 1 + (-1)^{n}$ for $n \geq 1$

2\.

$a_{n} = n^{2} - 1$ for $n \geq 1$

$a_{1} = 1$ and $a_{n} = a_{n - 1} + n$ for $n \geq 2$

4\.

$a_{1} = 1,$ $a_{2} = 1$ and $a_{n + 2} = a_{n} + a_{n + 1}$ for $n \geq 1$

Find an explicit formula for $a_{n}$ where $a_{1} = 1$ and $a_{n} = a_{n - 1} + n$ for $n \geq 2.$

求 $a_{n}$ 的显式公式，其中 $a_{1} = 1$ 且对 $n \geq 2$ 有 $a_{n} = a_{n - 1} + n$。

6\.

Find a formula $a_{n}$ for the $n\text{th}$ term of the arithmetic sequence whose first term is $a_{1} = 1$ such that $a_{n + 1} - a_{n} = 17$ for $n \geq 1.$

求等差数列通项公式 $a_{n}$，其首项为 $a_{1} = 1$，且对 $n \geq 1$ 满足 $a_{n + 1} - a_{n} = 17$。

Find a formula $a_{n}$ for the $n\text{th}$ term of the arithmetic sequence whose first term is $a_{1} = -3$ such that $a_{n + 1} - a_{n} = 4$ for $n \geq 1.$

求等差数列通项公式 $a_{n}$，其首项为 $a_{1} = -3$，且对 $n \geq 1$ 满足 $a_{n + 1} - a_{n} = 4$。

8\.

Find a formula $a_{n}$ for the $n\text{th}$ term of the geometric sequence whose first term is $a_{1} = 1$ such that $\frac{a_{n + 1}}{a_{n}} = 10$ for $n \geq 1.$

求几何级数通项公式 $a_{n}$，其首项为 $a_{1} = 1$，且对 $n \geq 1$ 满足 $\frac{a_{n + 1}}{a_{n}} = 10$。

Find a formula $a_{n}$ for the $n\text{th}$ term of the geometric sequence whose first term is $a_{1} = 3$ such that $\frac{a_{n + 1}}{a_{n}} = 1\text{/}10$ for $n \geq 1.$

求几何级数通项公式 $a_{n}$，其首项为 $a_{1} = 3$，且对 $n \geq 1$ 满足 $\frac{a_{n + 1}}{a_{n}} = 1\text{/}10$。

10\.

Find an explicit formula for the $n\text{th}$ term of the sequence whose first several terms are $\left\{ {0,3,8,15,24,35,48,63,80,99\text{,…}} \right\}.$ (*Hint:* First add one to each term.)

求下列数列第 $n$ 项的显式公式，其前几项为 $\left\{ {0,3,8,15,24,35,48,63,80,99\text{,…}} \right\}$。（*提示：*先给每一项加一。）

11.

Find an explicit formula for the $n\text{th}$ term of the sequence satisfying $a_{1} = 0$ and $a_{n} = 2a_{n - 1} + 1$ for $n \geq 2.$

求满足 $a_{1} = 0$、且对 $n \geq 2$ 有 $a_{n} = 2a_{n - 1} + 1$ 的数列第 $n$ 项的显式公式。

Find a formula for the general term $a_{n}$ of each of the following sequences.

求下列各数列通项 $a_{n}$ 的公式。

12\.

$\left\{ {1,0,-1,0,1,0,-1,0\text{,…}} \right\}$ (*Hint:* Find where $\text{sin}\mspace{2mu} x$ takes these values)

13.

$\left\{ {1,\text{−}{1\text{/}{3,{1\text{/}5}}},\text{−}{1\text{/}{7\text{,…}}}} \right\}$

Find a function $f(n)$ that identifies the $n\text{th}$ term $a_{n}$ of the following recursively defined sequences, as $a_{n} = f(n).$

求一个函数 $f(n)$，使得 $a_{n} = f(n)$，用它给出下列递推定义数列的第 $n$ 项 $a_{n}$。

14\.

$a_{1} = 1$ and $a_{n + 1} = \text{−}a_{n}$ for $n \geq 1$

15.

$a_{1} = 2$ and $a_{n + 1} = 2a_{n}$ for $n \geq 1$

16\.

$a_{1} = 1$ and $a_{n + 1} = \left( {n + 1} \right)a_{n}$ for $n \geq 1$

17.

$a_{1} = 2$ and $a_{n + 1} = \left( {n + 1} \right)a_{n}\text{/}2$ for $n \geq 1$

18\.

$a_{1} = 1$ and $a_{n + 1} = a_{n}\text{/}2^{n}$ for $n \geq 1$

Plot the first $N$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.

画出每个数列的前 $N$ 项。说明图像证据是否表明该数列收敛或发散。

19.

\[T\] $a_{1} = 1,$ $a_{2} = 2,$ and for $n \geq 2,$ $a_{n} = \frac{1}{2}\left( {a_{n - 1} + a_{n - 2}} \right);$ $N = 30$

20\.

\[T\] $a_{1} = 1,$ $a_{2} = 2,$ $a_{3} = 3$ and for $n \geq 4,$ $a_{n} = \frac{1}{3}\left( {a_{n - 1} + a_{n - 2} + a_{n - 3}} \right),$ $N = 30$

21.

\[T\] $a_{1} = 1,$ $a_{2} = 2,$ and for $n \geq 3,$ $a_{n} = \sqrt{a_{n - 1}a_{n - 2}};$ $N = 30$

22\.

\[T\] $a_{1} = 1,$ $a_{2} = 2,$ $a_{3} = 3,$ and for $n \geq 4,$ $a_{n} = \sqrt{a_{n - 1}a_{n - 2}a_{n - 3}};$ $N = 30$

Suppose that $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 1,$ $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = -1,$ and $0 < \text{−}b_{n} < a_{n}$ for all $n.$ Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.

假设 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 1,$ $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = -1,$ 且对一切 $n$ 有 $0 < \text{−}b_{n} < a_{n}$。计算下列各个极限，或说明该极限不存在，或说明没有足够信息判断极限是否存在。

23.

$\underset{n\rightarrow\infty}{\text{lim}}\left( 3a_{n} - 4b_{n} \right)$

24\.

$\underset{n\rightarrow\infty}{\text{lim}}\left( \frac{1}{2}b_{n} - \frac{1}{2}a_{n} \right)$

25.

$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n} + b_{n}}{a_{n} - b_{n}}$

26\.

$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n} - b_{n}}{a_{n} + b_{n}}$

Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.

求下列每个数列的极限，必要时使用洛必达法则。

27.

$\frac{n^{2}}{2^{n}}$

28\.

$\frac{\left( {n - 1} \right)^{2}}{\left( {n + 1} \right)^{2}}$

29.

$\frac{\sqrt{n}}{\sqrt{n + 1}}$

30\.

$n^{1\text{/}n}$ (*Hint:* $n^{1\text{/}n} = e^{\frac{1}{n}\mspace{2mu}\text{ln}\ n})$

$n^{1\text{/}n}$（*提示：* $n^{1\text{/}n} = e^{\frac{1}{n}\mspace{2mu}\text{ln}\ n}$）

For each of the following sequences, whose $n\text{th}$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

对下列每个数列（已给出其第 $n$ 项），说明该数列是否有界，以及它最终是否为单调、递增或递减的。

31.

$n\text{/}2^{n},$ $n \geq 2$

32\.

$\text{ln}\left( {1 + \frac{1}{n}} \right)$

33.

$\text{sin}\mspace{2mu} n$

34\.

$\text{cos}\left( n^{2} \right)$

35.

$n^{1\text{/}n},$ $n \geq 3$

36\.

$n^{-1\text{/}n},$ $n \geq 3$

37.

$\text{tan}\mspace{2mu} n$

38\.

Determine whether the sequence defined as follows has a limit. If it does, find the limit.

判断如下定义的数列是否有极限。若有，求其极限。

$a_{1} = \sqrt{2},$ $a_{2} = \sqrt{2\sqrt{2}},$ $a_{3} = \sqrt{2\sqrt{2\sqrt{2}}}$ etc.

$a_{1} = \sqrt{2},$ $a_{2} = \sqrt{2\sqrt{2}},$ $a_{3} = \sqrt{2\sqrt{2\sqrt{2}}}$ 等等。

39.

Determine whether the sequence defined as follows has a limit. If it does, find the limit.

判断如下定义的数列是否有极限。若有，求其极限。

$a_{1} = 3,$ $a_{n} = \sqrt{2a_{n - 1}},$ $n = 2,3\text{,…}.$

Use the Squeeze Theorem to find the limit of each of the following sequences.

用夹逼定理求下列每个数列的极限。

40\.

$n\mspace{2mu}\text{sin}\left( {1\text{/}n} \right)$

41.

$\frac{\text{cos}\left( {1\text{/}n} \right) - 1}{1\text{/}n}$

42\.

$a_{n} = \frac{n\text{!}}{n^{n}}$

43.

$a_{n} = \text{sin}\mspace{2mu} n\mspace{2mu}\text{sin}\left( {1\text{/}n} \right)$

For the following sequences, plot the first $25$ terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.

对下列数列，画出其前 $25$ 项，并说明图像证据是否表明该数列收敛或发散。

44\.

\[T\] $a_{n} = \text{sin}\mspace{2mu} n$

45.

\[T\] $a_{n} = \text{cos}\mspace{2mu} n$

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

确定该数列的极限，或证明该数列发散。若收敛，求其极限。

46\.

$a_{n} = \text{tan}^{-1}(n^{2})$

47.

$a_{n} = {(2n)}^{1\text{/}n} - n^{1\text{/}n}$

48\.

$a_{n} = \frac{\text{ln}(n^{2})}{\text{ln}(2n)}$

49.

$a_{n} = \left( {1 - \frac{2}{n}} \right)^{n}$

50\.

$a_{n} = \text{ln}\left( \frac{n + 2}{n^{2} - 3} \right)$

51.

$a_{n} = \frac{2^{n} + 3^{n}}{4^{n}}$

52\.

$a_{n} = \frac{{(1000)}^{n}}{n\text{!}}$

53.

$a_{n} = \frac{{(n\text{!})}^{2}}{(2n)\text{!}}$

Newton’s method seeks to approximate a solution $f(x) = 0$ that starts with an initial approximation $x_{0}$ and successively defines a sequence $x_{n + 1} = x_{n} - \frac{f\left( x_{n} \right)}{f^{\prime}\left( x_{n} \right)}.$ For the given choice of $f$ and $x_{0},$ write out the formula for $x_{n + 1}.$ If the sequence appears to converge, give an exact formula for the solution $x,$ then identify the limit $x$ accurate to four decimal places and the smallest $n$ such that $x_{n}$ agrees with $x$ up to four decimal places.

牛顿法用于逼近方程 $f(x) = 0$ 的解：从一个初始近似值 $x_{0}$ 出发，依次定义数列 $x_{n + 1} = x_{n} - \frac{f\left( x_{n} \right)}{f^{\prime}\left( x_{n} \right)}$。对于给定的 $f$ 与 $x_{0}$，写出 $x_{n + 1}$ 的公式。若该数列似乎收敛，则给出解 $x$ 的精确公式，并找出精确到四位小数的极限 $x$，以及使 $x_{n}$ 与 $x$ 在四位小数上一致的最小 $n$。

54\.

\[T\] $f(x) = x^{2} - 2,$ $x_{0} = 1$

55.

\[T\] $f(x) = \left( {x - 1} \right)^{2} - 2,$ $x_{0} = 2$

56\.

\[T\] $f(x) = e^{x} - 2,$ $x_{0} = 1$

57.

\[T\] $f(x) = \text{ln}\mspace{2mu} x - 1,$ $x_{0} = 2$

58\.

\[T\] Suppose you start with one liter of vinegar and repeatedly remove $0.1\ \text{L,}$ replace with water, mix, and repeat.

\[T\] 假设你从 1 升醋开始，反复取出 $0.1\ \text{L}$，用水补满，搅匀，然后重复。

1. Find a formula for the concentration after $n$ steps.

1. 求经过 $n$ 步后浓度的公式。

2. After how many steps does the mixture contain less than $10\text{\%}$ vinegar?

2. 经过多少步后，混合液中醋的含量低于 $10\text{\%}$？

59.

\[T\] A lake initially contains $2000$ fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by $6\text{\%}$ each month. However, factoring in all causes, $150$ fish are lost each month.

\[T\] 一个湖中最初有 $2000$ 条鱼。假设在没有捕食者或其他减少因素的情况下，鱼群每月增长 $6\text{\%}$。但综合考虑各种因素后，每月会损失 $150$ 条鱼。

1. Explain why the fish population after $n$ months is modeled by $P_{n} = 1.06P_{n - 1} - 150$ with $P_{0} = 2000.$

1. 解释为什么 $n$ 个月后的鱼群数量由 $P_{n} = 1.06P_{n - 1} - 150$ 建模，其中 $P_{0} = 2000$。

2. How many fish will be in the pond after one year?

2. 一年后池塘里会有多少条鱼？

60\.

\[T\] A bank account earns $5\text{\%}$ interest compounded monthly. Suppose that $\text{\$}1000$ is initially deposited into the account, but that $\text{\$}10$ is withdrawn each month.

\[T\] 一个银行账户按月复利计息，年利率为 $5\text{\%}$。假设最初存入 $\text{\$}1000$，但每月取出 $\text{\$}10$。

1. Show that the amount in the account after $n$ months is $A_{n} = \left( {1 + .05\text{/}12} \right)A_{n - 1} - 10;$ $A_{0} = 1000.$

1. 证明 $n$ 个月后账户中的金额为 $A_{n} = \left( {1 + .05\text{/}12} \right)A_{n - 1} - 10;$ $A_{0} = 1000$。

2. How much money will be in the account after $1$ year?

2. 一年后账户里会有多少钱？

3. Is the amount increasing or decreasing?

3. 金额是在增加还是减少？

4. Suppose that instead of $\text{\$}10,$ a fixed amount $d$ dollars is withdrawn each month. Find a value of $d$ such that the amount in the account after each month remains $\text{\$}1000.$

4. 假设每月取出的不是 $\text{\$}10$，而是固定金额 $d$ 美元。求一个 $d$ 值，使得每月之后账户中的金额保持为 $\text{\$}1000$。

5. What happens if $d$ is greater than this amount?

5. 如果 $d$ 大于这个金额会怎样？

61.

\[T\] A student takes out a college loan of $\text{\$}10,000$ at an annual percentage rate of $6\text{\%},$ compounded monthly.

\[T\] 一名学生借了一笔大学贷款，金额为 $\text{\$}10,000$，年百分率为 $6\text{\%}$，按月复利计息。

1. If the student makes payments of $\text{\$}100$ per month, how much does the student owe after $12$ months?

1. 如果该学生每月还款 $\text{\$}100$，那么 $12$ 个月后还欠多少？

2. After how many months will the loan be paid off?

2. 多少个月后贷款能还清？

62\.

\[T\] Consider a series combining geometric growth and arithmetic decrease. Let $a_{1} = 1.$ Fix $a > 1$ and $0 < b < a.$ Set $a_{n + 1} = a.a_{n} - b.$ Find a formula for $a_{n + 1}$ in terms of $a^{n},$ $a,$ and $b$ and a relationship between $a$ and $b$ such that $a_{n}$ converges.

\[T\] 考虑一个结合几何增长与算术递减的数列。令 $a_{1} = 1$。取定 $a > 1$ 且 $0 < b < a$。设 $a_{n + 1} = a.a_{n} - b$。求用 $a^{n}$、$a$ 和 $b$ 表示的 $a_{n + 1}$ 的公式，并给出使 $a_{n}$ 收敛的 $a$ 与 $b$ 之间的关系。

63.

\[T\] The binary representation $x = 0.b_{1}b_{2}b_{3}...$ of a number $x$ between $0$ and $1$ can be defined as follows. Let $b_{1} = 0$ if $x < 1\text{/}2$ and $b_{1} = 1$ if $1\text{/}2 \leq x < 1.$ Let $x_{1} = 2x - b_{1}.$ Let $b_{2} = 0$ if $x_{1} < 1\text{/}2$ and $b_{2} = 1$ if $1\text{/}2 \leq x < 1.$ Let $x_{2} = 2x_{1} - b_{2}$ and in general, $x_{n} = 2x_{n - 1} - b_{n}$ and $b_{n - 1} = 0$ if $x_{n} < 1\text{/}2$ and $b_{n - 1} = 1$ if $1\text{/}2 \leq x_{n} < 1.$ Find the binary expansion of $1\text{/}3.$

\[T\] $0$ 与 $1$ 之间一个数 $x$ 的二进制表示 $x = 0.b_{1}b_{2}b_{3}...$ 可如下定义。若 $x < 1\text{/}2$ 则令 $b_{1} = 0$，若 $1\text{/}2 \leq x < 1$ 则令 $b_{1} = 1$。令 $x_{1} = 2x - b_{1}$。若 $x_{1} < 1\text{/}2$ 则令 $b_{2} = 0$，若 $1\text{/}2 \leq x < 1$ 则令 $b_{2} = 1$。令 $x_{2} = 2x_{1} - b_{2}$，一般地，$x_{n} = 2x_{n - 1} - b_{n}$，且若 $x_{n} < 1\text{/}2$ 则 $b_{n - 1} = 0$，若 $1\text{/}2 \leq x_{n} < 1$ 则 $b_{n - 1} = 1$。求 $1\text{/}3$ 的二进制展开。

64\.

\[T\] To find an approximation for $\pi,$ set $a_{0} = \sqrt{2 + 1},$ $a_{1} = \sqrt{2 + a_{0}},$ and, in general, $a_{n + 1} = \sqrt{2 + a_{n}}.$ Finally, set $p_{n} = 3.2^{n + 1}\sqrt{2 - a_{n}}.$ Find the first ten terms of $p_{n}$ and compare the values to $\pi.$

\[T\] 为求 $\pi$ 的近似值，令 $a_{0} = \sqrt{2 + 1},$ $a_{1} = \sqrt{2 + a_{0}},$ 一般地 $a_{n + 1} = \sqrt{2 + a_{n}}$。最后令 $p_{n} = 3.2^{n + 1}\sqrt{2 - a_{n}}$。求 $p_{n}$ 的前十项，并将其数值与 $\pi$ 比较。

For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of $N$ integers $a_{1},a_{2}\text{,…},a_{N}$ by fixing two special integers $K$ and $M$ and letting $a_{n + 1}$ be the remainder after dividing $K.a_{n}$ into $M,$ then creates a bit sequence of zeros and ones whose $n\text{th}$ term $b_{n}$ is equal to one if $a_{n}$ is odd and equal to zero if $a_{n}$ is even. If the bits $b_{n}$ are pseudorandom, then the behavior of their average $\left( {b_{1} + b_{2} + \text{⋯} + b_{N}} \right)\text{/}N$ should be similar to behavior of averages of truly randomly generated bits.

在接下来的两道习题中，假设你可以使用一个计算机程序或网络资源，能够生成任意长度的由 0 和 1 组成的列表。伪随机数生成器（PRNG）通过产生看似反复抛硬币结果的 0、1 序列，在模拟物理系统中的随机噪声方面起着重要作用。最简单的一类 PRNG 通过固定两个特殊整数 $K$ 和 $M$，并令 $a_{n + 1}$ 为 $K.a_{n}$ 除以 $M$ 的余数，递归地定义一个看似随机的 $N$ 个整数序列 $a_{1},a_{2}\text{,…},a_{N}$，然后生成一个由 0 和 1 组成的位序列，其第 $n$ 项 $b_{n}$ 在 $a_{n}$ 为奇数时等于 1，在 $a_{n}$ 为偶数时等于 0。如果各位 $b_{n}$ 是伪随机的，那么其平均值 $\left( {b_{1} + b_{2} + \text{⋯} + b_{N}} \right)\text{/}N$ 的行为应当类似于真正随机生成的位的平均值的行为。

65.

\[T\] Starting with $K = 16,807$ and $M = 2,147,483,647,$ using ten different starting values of $a_{1},$ compute sequences of bits $b_{n}$ up to $n = 1000,$ and compare their averages to ten such sequences generated by a random bit generator.

\[T\] 取 $K = 16,807$、$M = 2,147,483,647$，用十个不同的 $a_{1}$ 初值，计算位序列 $b_{n}$ 直到 $n = 1000$，并将其平均值与随机位生成器产生的十个这样的序列的平均值进行比较。

66\.

\[T\] Find the first $1000$ digits of $\pi$ using either a computer program or Internet resource. Create a bit sequence $b_{n}$ by letting $b_{n} = 1$ if the $n\text{th}$ digit of $\pi$ is odd and $b_{n} = 0$ if the $n\text{th}$ digit of $\pi$ is even. Compute the average value of $b_{n}$ and the average value of $d_{n} = \left| {b_{n + 1} - b_{n}} \right|,$ $n = 1\text{,...},999.$ Does the sequence $b_{n}$ appear random? Do the differences between successive elements of $b_{n}$ appear random?

\[T\] 用计算机程序或网络资源找出 $\pi$ 的前 $1000$ 位数字。构造位序列 $b_{n}$：若 $\pi$ 的第 $n$ 位数字是奇数则令 $b_{n} = 1$，若是偶数则令 $b_{n} = 0$。计算 $b_{n}$ 的平均值以及 $d_{n} = \left| {b_{n + 1} - b_{n}} \right|$ 的平均值，其中 $n = 1\text{,...},999$。序列 $b_{n}$ 看起来是随机的吗？$b_{n}$ 相邻元素之差看起来是随机的吗？

5.2 Infinite Series 5.2 无穷级数

5.2.1 Explain the meaning of the sum of an infinite series.
5.2.2 Calculate the sum of a geometric series.
5.2.3 Evaluate a telescoping series.

5.2.1 解释无穷级数的和的含义。
5.2.2 计算几何级数的和。
5.2.3 求裂项级数的值。

We have seen that a sequence is an ordered set of terms. If you add these terms together, you get a series. In this section we define an infinite series and show how series are related to sequences. We also define what it means for a series to converge or diverge. We introduce one of the most important types of series: the geometric series. We will use geometric series in the next chapter to write certain functions as polynomials with an infinite number of terms. This process is important because it allows us to evaluate, differentiate, and integrate complicated functions by using polynomials that are easier to handle. We also discuss the harmonic series, arguably the most interesting divergent series because it just fails to converge.

我们已经看到，数列是一组有序的项。如果把各项相加，就得到一个级数。本节我们定义无穷级数，并说明级数与数列之间的关系。我们还定义级数收敛或发散的含义。我们介绍最重要的一类级数之一：几何级数。下一章我们将利用几何级数把某些函数写成具有无穷多项的多项式。这一过程很重要，因为它使我们能够借助更易处理的多项式来求复杂函数的值、求导与积分。我们还讨论调和级数——可以说是最有趣的发散级数，因为它恰好未能收敛。

Sums and Series 和与级数

An infinite series is a sum of infinitely many terms and is written in the form

无穷级数是无穷多项的和，写作如下形式

$$\sum\limits_{n = 1}^{\infty}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯}.$$

But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the *limit* of partial sums. A partial sum of an infinite series is a finite sum of the form

但这意味着什么？我们无法像加有限项那样去加无穷多项。相反，无穷级数的值是通过部分和（partial sum）的*极限*来定义的。无穷级数的一个部分和具有如下有限和的形式

$${\sum\limits_{n = 1}^{k}a_{n}} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k}.$$

To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that $1000$ gallons enters the lake the first week. During the second week, an additional $500$ gallons of oil enters the lake. The third week, $250$ more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after $k$ weeks. Letting $S_{k}$ denote the amount of oil in the lake (measured in thousands of gallons) after $k$ weeks, we see that

为了看清如何用部分和来求无穷级数，考虑下面的例子。假设有石油渗入一个湖中，第一周有 $1000$ 加仑进入湖里。第二周又有 $500$ 加仑石油进入湖中。第三周再有 $250$ 加仑进入湖中。假设这一规律持续下去：每周进入湖中的石油是上一周的一半。如果永远持续下去，关于湖中石油的量我们能说什么？石油的量会无限制地任意增大，还是有可能趋近于某个有限值？为回答这个问题，我们考察 $k$ 周后湖中的石油量。令 $S_{k}$ 表示 $k$ 周后湖中石油的量（以千加仑计），我们看到

$$\begin{array}{l} {S_{1} = 1} \\ {S_{2} = 1 + 0.5 = 1 + \frac{1}{2}} \\ {S_{3} = 1 + 0.5 + 0.25 = 1 + \frac{1}{2} + \frac{1}{4}} \\ {S_{4} = 1 + 0.5 + 0.25 + 0.125 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}} \\ {S_{5} = 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}.} \end{array}$$

Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after $k$ weeks is

观察这个规律，我们看到湖中石油的量（以千加仑为单位）在第 $k$ 周后为

$$S_{k} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \text{⋯} + \frac{1}{2^{k - 1}} = {\sum\limits_{n = 1}^{k}\left( \frac{1}{2} \right)^{n - 1}}.$$

We are interested in what happens as $k\rightarrow\infty.$ Symbolically, the amount of oil in the lake as $k\rightarrow\infty$ is given by the infinite series

我们关心当 $k\rightarrow\infty$ 时会发生什么。用符号表示，湖中油量当 $k\rightarrow\infty$ 时由下面的无穷级数给出

$${\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 1}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \text{⋯}.$$

At the same time, as $k\rightarrow\infty,$ the amount of oil in the lake can be calculated by evaluating $\underset{k\rightarrow\infty}{\text{lim}}S_{k}.$ Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums $\left\{ S_{k} \right\}.$ If the sequence of partial sums $\left\{ S_{k} \right\}$ converges, we say that the infinite series converges, and its sum is given by $\underset{k\rightarrow\infty}{\text{lim}}S_{k}.$ If the sequence $\left\{ S_{k} \right\}$ diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence $\left\{ S_{k} \right\}.$

与此同时，当 $k\rightarrow\infty$ 时，湖中石油的量可以通过求 $\underset{k\rightarrow\infty}{\text{lim}}S_{k}$ 算得。因此，无穷级数的性态可以通过考察其部分和数列 $\left\{ S_{k} \right\}$ 的性态来确定。如果部分和数列 $\left\{ S_{k} \right\}$ 收敛，我们就说该无穷级数收敛，且其和由 $\underset{k\rightarrow\infty}{\text{lim}}S_{k}$ 给出。如果数列 $\left\{ S_{k} \right\}$ 发散，我们就说该无穷级数发散。现在我们着手确定这个数列 $\left\{ S_{k} \right\}$ 的极限。

First, simplifying some of these partial sums, we see that

首先，化简其中几个部分和，我们看到

$$\begin{array}{l} {S_{1} = 1} \\ {S_{2} = 1 + \frac{1}{2} = \frac{3}{2}} \\ {S_{3} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}} \\ {S_{4} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{15}{8}} \\ {S_{5} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{31}{16}.} \end{array}$$

Plotting some of these values in Figure 5.10, it appears that the sequence $\left\{ S_{k} \right\}$ could be approaching 2.

把其中一些数值画在图 5.10 中，数列 $\left\{ S_{k} \right\}$ 似乎趋近于 2。

Let’s look for more convincing evidence. In the following table, we list the values of $S_{k}$ for several values of $k.$

我们寻找更有说服力的证据。在下表中，我们列出了若干 $k$ 值对应的 $S_{k}$ 的值。

$k$	$5$	$10$	$15$	$20$
$S_{k}$	$1.9375$	$1.998$	$1.999939$	$1.999998$

$k$	$5$	$10$	$15$	$20$
$S_{k}$	$1.9375$	$1.998$	$1.999939$	$1.999998$

These data supply more evidence suggesting that the sequence $\left\{ S_{k} \right\}$ converges to $2.$ Later we will provide an analytic argument that can be used to prove that $\underset{k\rightarrow\infty}{\text{lim}}S_{k} = 2.$ For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to $2.$ Since this sequence of partial sums converges to $2,$ we say the infinite series converges to $2$ and write

这些数据提供了更多证据，表明数列 $\left\{ S_{k} \right\}$ 收敛到 $2$。稍后我们会给出可用于证明 $\underset{k\rightarrow\infty}{\text{lim}}S_{k} = 2$ 的分析论证。目前我们依靠数值与图形数据让自己相信，部分和数列确实收敛到 $2$。既然这个部分和数列收敛到 $2$，我们就说该无穷级数收敛到 $2$，并写作

$$\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 1} = 2.$$

Returning to the question about the oil in the lake, since this infinite series converges to $2,$ we conclude that the amount of oil in the lake will get arbitrarily close to $2000$ gallons as the amount of time gets sufficiently large.

回到湖中石油的问题，既然这个无穷级数收敛到 $2$，我们得出结论：当时间足够大时，湖中石油的量会任意接近某个值，即 $2000$ 加仑。

This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.

这个级数是几何级数（等比级数）的一个例子。我们将在本节后面更详细地讨论几何级数。首先，我们总结无穷级数收敛的含义。

An infinite series is an expression of the form

无穷级数是如下形式的表达式

$$\sum\limits_{n = 1}^{\infty}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯}.$$

For each positive integer $k,$ the sum

对每个正整数 $k$，和式

$$S_{k} = \sum\limits_{n = 1}^{k}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k}$$

is called the $k\text{th}$ partial sum of the infinite series. The partial sums form a sequence $\left\{ S_{k} \right\}.$ If the sequence of partial sums converges to a real number $S,$ the infinite series converges. If we can describe the convergence of a series to $S,$ we call $S$ the sum of the series, and we write

称为该无穷级数的第 $k$ 个部分和（partial sum）。这些部分和构成一个数列 $\left\{ S_{k} \right\}$。如果部分和数列收敛到一个实数 $S$，则该无穷级数收敛。如果我们能描述级数收敛到 $S$，就把 $S$ 称为该级数的和，并写作

$$\sum\limits_{n = 1}^{\infty}a_{n} = S.$$

If the sequence of partial sums diverges, we have the divergence of a series.

如果部分和数列发散，我们就说该级数发散。

Note that the index for a series need not begin with $n = 1$ but can begin with any value. For example, the series

注意，级数的下标不必从 $n = 1$ 开始，而可以从任意值开始。例如，级数

$${\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)}^{n - 1}$$

can also be written as

也可以写成

$${\sum\limits_{n = 0}^{\infty}\left( \frac{1}{2} \right)}^{n}\ \text{or}\ {\sum\limits_{n = 5}^{\infty}\left( \frac{1}{2} \right)}^{n - 5}.$$

Often it is convenient for the index to begin at $1,$ so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series

通常下标从 $1$ 开始会比较方便，因此如果由于某种原因它从别的值开始，我们可以通过变量代换重新设置下标。例如，考虑级数

$${\sum\limits_{n = 2}^{\infty}\frac{1}{n^{2}}}.$$

By introducing the variable $m = n - 1,$ so that $n = m + 1,$ we can rewrite the series as

引入变量 $m = n - 1$，从而 $n = m + 1$，我们可以把这个级数改写为

$${\sum\limits_{m = 1}^{\infty}\frac{1}{\left( {m + 1} \right)^{2}}}.$$

Evaluating Limits of Sequences of Partial Sums 求部分和数列的极限

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

对下列每个级数，利用部分和数列判断该级数收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}\frac{n}{n + 1}$

2. $\sum\limits_{n = 1}^{\infty}{(-1)}^{n}$

3. $\sum\limits_{n = 1}^{\infty}\frac{1}{n(n + 1)}$

Solution 解答

1. The sequence of partial sums $\left\{ S_{k} \right\}$ satisfies

1. 部分和数列 $\left\{ S_{k} \right\}$ 满足

$$\begin{array}{l} {S_{1} = \frac{1}{2}} \\ {S_{2} = \frac{1}{2} + \frac{2}{3}} \\ {S_{3} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4}} \\ {S_{4} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5}.} \end{array}$$

Notice that each term added is greater than $1\text{/}2.$ As a result, we see that

注意，每一项加上的量都大于 $1\text{/}2$。因此我们看到

$$\begin{array}{l} {S_{1} = \frac{1}{2}} \\ {S_{2} = \frac{1}{2} + \frac{2}{3} > \frac{1}{2} + \frac{1}{2} = 2\left( \frac{1}{2} \right)} \\ {S_{3} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} > \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 3\left( \frac{1}{2} \right)} \\ {S_{4} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} > \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 4\left( \frac{1}{2} \right).} \end{array}$$

From this pattern we can see that $S_{k} > k\left( \frac{1}{2} \right)$ for every integer $k.$ Therefore, $\left\{ S_{k} \right\}$ is unbounded and consequently, diverges. Therefore, the infinite series $\sum\limits_{n = 1}^{\infty}n\text{/}(n + 1)$ diverges.

由这个规律可见，对每个整数 $k$ 都有 $S_{k} > k\left( \frac{1}{2} \right)$。因此 $\left\{ S_{k} \right\}$ 无界，从而发散。于是无穷级数 $\sum\limits_{n = 1}^{\infty}n\text{/}(n + 1)$ 发散。

2. The sequence of partial sums $\left\{ S_{k} \right\}$ satisfies

2. 部分和数列 $\left\{ S_{k} \right\}$ 满足

$$\begin{array}{l} {S_{1} = -1} \\ {S_{2} = -1 + 1 = 0} \\ {S_{3} = -1 + 1 - 1 = -1} \\ {S_{4} = -1 + 1 - 1 + 1 = 0.} \end{array}$$

From this pattern we can see the sequence of partial sums is

由这个规律我们看到部分和数列是

$$\left\{ S_{k} \right\} = \left\{ {-1,0,-1,0\text{,…}} \right\}.$$

Since this sequence diverges, the infinite series $\sum\limits_{n = 1}^{\infty}(-1)^{n}$ diverges.

既然这个数列发散，无穷级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n}$ 也发散。

3. The sequence of partial sums $\left\{ S_{k} \right\}$ satisfies

3. 部分和数列 $\left\{ S_{k} \right\}$ 满足

$$\begin{array}{l} {S_{1} = \frac{1}{1 \cdot 2} = \frac{1}{2}} \\ {S_{2} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}} \\ {S_{3} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{3}{4}} \\ {S_{4} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} = \frac{4}{5}} \\ {S_{5} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} = \frac{5}{6}.} \end{array}$$

From this pattern, we can see that the $k\text{th}$ partial sum is given by the explicit formula

由这个规律可见，第 $k$ 个部分和由下面的显式公式给出

$$S_{k} = \frac{k}{k + 1}.$$

Since $k\text{/}\left( {k + 1} \right)\rightarrow 1,$ we conclude that the sequence of partial sums converges, and therefore the infinite series converges to $1.$ We have

由于 $k\text{/}\left( {k + 1} \right)\rightarrow 1$，我们得出结论：部分和数列收敛，因而该无穷级数收敛到 $1$。我们有

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)} = 1.$$

Determine whether the series $\sum\limits_{n = 1}^{\infty}(n + 1)\text{/}n$ converges or diverges.

判断级数 $\sum\limits_{n = 1}^{\infty}(n + 1)\text{/}n$ 收敛还是发散。

The Harmonic Series 调和级数

A useful series to know about is the harmonic series. The harmonic series is defined as

一个值得了解的很有用的级数是调和级数。调和级数定义为

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \text{⋯}.$$ (5.5)

This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums $\left\{ S_{k} \right\}$ approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence $\left\{ S_{k} \right\}$ in the following table.

这个级数很有意思，因为它发散，但发散得非常缓慢。我们的意思是，部分和数列 $\left\{ S_{k} \right\}$ 中的项趋向于无穷大，但过程非常缓慢。我们将证明该级数发散，不过先在下表中展示数列 $\left\{ S_{k} \right\}$ 中各项增长的缓慢程度。

$k$	$10$	$100$	$1000$	$10,000$	$100,000$	$1,000,000$
$S_{k}$	$2.92897$	$5.18738$	$7.48547$	$9.78761$	$12.09015$	$14.39273$

$k$	$10$	$100$	$1000$	$10,000$	$100,000$	$1,000,000$
$S_{k}$	$2.92897$	$5.18738$	$7.48547$	$9.78761$	$12.09015$	$14.39273$

Even after $1,000,000$ terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.

即便到了 $1,000,000$ 项，部分和仍然相当小。从这张表还看不出这个级数确实发散。然而，我们可以用分析的方法证明部分和数列发散，从而该级数发散。

To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:

为了证明部分和数列发散，我们证明部分和数列无界。我们先写出开头几个部分和：

$$\begin{array}{l} {S_{1} = 1} \\ {S_{2} = 1 + \frac{1}{2}} \\ {S_{3} = 1 + \frac{1}{2} + \frac{1}{3}} \\ {S_{4} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}.} \end{array}$$

Notice that for the last two terms in $S_{4},$

注意，对于 $S_{4}$ 的最后两项，

$$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4}.$$

Therefore, we conclude that

因此，我们得出结论

$$S_{4} > 1 + \frac{1}{2} + \left( {\frac{1}{4} + \frac{1}{4}} \right) = 1 + \frac{1}{2} + \frac{1}{2} = 1 + 2\left( \frac{1}{2} \right).$$

Using the same idea for $S_{8},$ we see that

对 $S_{8}$ 使用同样的思路，我们看到

$$\begin{array}{cl} S_{8} & {= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > 1 + \frac{1}{2} + \left( {\frac{1}{4} + \frac{1}{4}} \right) + \left( {\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}} \right)} \\ & {= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 + 3\left( \frac{1}{2} \right).} \end{array}$$

From this pattern, we see that $S_{1} = 1,$ $S_{2} = 1 + 1\text{/}2,$ $S_{4} > 1 + 2\left( {1\text{/}2} \right),$ and $S_{8} > 1 + 3\left( {1\text{/}2} \right).$ More generally, it can be shown that $S_{2^{j}} > 1 + j(1\text{/}2)$ for all $j > 1.$ Since $1 + j(1\text{/}2)\rightarrow\infty,$ we conclude that the sequence $\left\{ S_{k} \right\}$ is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since $\left\{ S_{k} \right\}$ is unbounded, it diverges. Thus, the harmonic series diverges.

由这个规律我们看到 $S_{1} = 1$、$S_{2} = 1 + 1\text{/}2$、$S_{4} > 1 + 2\left( {1\text{/}2} \right)$、$S_{8} > 1 + 3\left( {1\text{/}2} \right)$。更一般地，可以证明对一切 $j > 1$ 都有 $S_{2^{j}} > 1 + j(1\text{/}2)$。由于 $1 + j(1\text{/}2)\rightarrow\infty$，我们得出结论：数列 $\left\{ S_{k} \right\}$ 无界，从而发散。在上一节中我们说过，收敛数列必有界。因此，既然 $\left\{ S_{k} \right\}$ 无界，它就发散。于是，调和级数发散。

Algebraic Properties of Convergent Series 收敛级数的代数性质

Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.

由于收敛的无穷级数的和被定义为一个数列的极限，下面列出的级数代数性质直接由数列的代数性质推出。

Let $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ be convergent series. Then the following algebraic properties hold.

设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$ 都是收敛级数。则下面的代数性质成立。

1. The series $\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}$ converges and $\sum\limits_{n = 1}^{\infty}\left( {a_{n} + b_{n}} \right) = \sum\limits_{n = 1}^{\infty}a_{n} + \sum\limits_{n = 1}^{\infty}b_{n}.$ (Sum Rule)

1. 级数 $\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}$ 收敛，且 $\sum\limits_{n = 1}^{\infty}\left( {a_{n} + b_{n}} \right) = \sum\limits_{n = 1}^{\infty}a_{n} + \sum\limits_{n = 1}^{\infty}b_{n}$。

2. The series $\sum\limits_{n = 1}^{\infty}{(a_{n} - b_{n})}$ converges and $\sum\limits_{n = 1}^{\infty}\left( {a_{n} - b_{n}} \right) = \sum\limits_{n = 1}^{\infty}a_{n} - \sum\limits_{n = 1}^{\infty}b_{n}.$ (Difference Rule)

2. 级数 $\sum\limits_{n = 1}^{\infty}{(a_{n} - b_{n})}$ 收敛，且 $\sum\limits_{n = 1}^{\infty}\left( {a_{n} - b_{n}} \right) = \sum\limits_{n = 1}^{\infty}a_{n} - \sum\limits_{n = 1}^{\infty}b_{n}$。

3. For any real number $c,$ the series ${\sum\limits_{n = 1}^{\infty}c}a_{n}$ converges and $\sum\limits_{n = 1}^{\infty}ca_{n} = c\sum\limits_{n = 1}^{\infty}a_{n}.$ (Constant Multiple Rule)

3. 对任意实数 $c$，级数 ${\sum\limits_{n = 1}^{\infty}c}a_{n}$ 收敛，且 $\sum\limits_{n = 1}^{\infty}ca_{n} = c\sum\limits_{n = 1}^{\infty}a_{n}$。

Using Algebraic Properties of Convergent Series 利用收敛级数的代数性质

Evaluate

计算

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {\frac{3}{n\left( {n + 1} \right)} + \left( \frac{1}{2} \right)^{n - 2}} \right\rbrack.$$

Solution 解答

We showed earlier that

我们前面已经证明

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)}} = 1$$

and

以及

$$\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 1} = 2.$$

Since both of those series converge, we can apply the properties of Algebraic Properties of Convergent Series to evaluate

由于这两个级数都收敛，我们可以应用「收敛级数的代数性质」来求下面这个式子的值

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {\frac{3}{n\left( {n + 1} \right)} + \left( \frac{1}{2} \right)^{n - 2}} \right\rbrack.$$

Using the sum rule, write

利用求和法则，可写成

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {\frac{3}{n\left( {n + 1} \right)} + \left( \frac{1}{2} \right)^{n - 2}} \right\rbrack = \sum\limits_{n = 1}^{\infty}\frac{3}{n\left( {n + 1} \right)}\underset{n = 1}{\overset{\infty}{+ \sum}}\left( \frac{1}{2} \right)^{n - 2}.$$

Then, using the constant multiple rule and the sums above, we can conclude that

然后，利用常数倍法则以及上面的和式，我们可以得出结论

$$\begin{array}{ll} {\sum\limits_{n = 1}^{\infty}\frac{3}{n\left( {n + 1} \right)} + \sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 2}} & {= 3\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)} + \left( \frac{1}{2} \right)^{-1}\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 1}} \\ & {= 3(1) + \left( \frac{1}{2} \right)^{-1}(2) = 3 + 2(2) = 7.} \end{array}$$

Evaluate $\sum\limits_{n = 1}^{\infty}\frac{5}{2^{n - 1}}.$

计算 $\sum\limits_{n = 1}^{\infty}\frac{5}{2^{n - 1}}$。

Geometric Series 几何级数

A geometric series is any series that we can write in the form

几何级数（等比级数）是任何可以写成下列形式的级数

$$a + ar + ar^{2} + ar^{3} + \text{⋯} = \sum\limits_{n = 1}^{\infty}ar^{n - 1}.$$ (5.6)

Because the ratio of each term in this series to the previous term is *r*, the number *r* is called the ratio. We refer to *a* as the initial term because it is the first term in the series. For example, the series

由于该级数中每一项与前一项的比都是 *r*，这个数 *r* 称为公比（ratio）。我们把 *a* 称为首项（initial term），因为它是级数的第一项。例如，级数

$$\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n - 1} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \text{⋯}$$

is a geometric series with initial term $a = 1$ and ratio $r = 1\text{/}2.$

是一个首项为 $a = 1$、公比为 $r = 1\text{/}2$ 的几何级数。

In general, when does a geometric series converge? Consider the geometric series

一般而言，几何级数何时收敛？考虑几何级数

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1}$$

when $a > 0.$ Its sequence of partial sums $\left\{ S_{k} \right\}$ is given by

其中 $a > 0$。它的部分和数列 $\left\{ S_{k} \right\}$ 由下式给出

$$S_{k} = \sum\limits_{n = 1}^{k}ar^{n - 1} = a + ar + ar^{2} + \text{⋯} + ar^{k - 1}.$$

Consider the case when $r = 1.$ In that case,

考虑 $r = 1$ 的情形。此时

$$S_{k} = a + a(1) + a(1)^{2} + \text{⋯} + a(1)^{k - 1} = ak.$$

Since $a > 0,$ we know $ak\rightarrow\infty$ as $k\rightarrow\infty.$ Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for $r = 1.$ For $r \neq 1,$ to find the limit of $\left\{ S_{k} \right\},$ multiply Equation 5.6 by $1 - r.$ Doing so, we see that

由于 $a > 0$，我们知道当 $k\rightarrow\infty$ 时 $ak\rightarrow\infty$。因此部分和数列无界，从而发散。于是，当 $r = 1$ 时该无穷级数发散。对于 $r \neq 1$，为求 $\left\{ S_{k} \right\}$ 的极限，用 $1 - r$ 乘以公式 5.6。这样我们得到

$$\begin{array}{cl} {\left( {1 - r} \right)S_{k}} & {= a\left( {1 - r} \right)\left( {1 + r + r^{2} + r^{3} + \text{⋯} + r^{k - 1}} \right)} \\ & {= a\lbrack(1 + r + r^{2} + r^{3} + \text{⋯} + r^{k - 1}) - (r + r^{2} + r^{3} + \text{⋯} + r^{k})\rbrack} \\ & {= a\left( {1 - r^{k}} \right).} \end{array}$$

All the other terms cancel out.

其余各项全部消去。

Therefore,

因此

$$S_{k} = \frac{a\left( {1 - r^{k}} \right)}{1 - r}\ \text{for}\ r \neq 1.$$

From our discussion in the previous section, we know that the geometric sequence $r^{k}\rightarrow 0$ if $|r| < 1$ and that $r^{k}$ diverges if $|r| > 1$ or $r = \text{±}1.$ Therefore, for $|r| < 1,$ $S_{k}\rightarrow a\text{/}(1 - r)$ and we have

由上一节的讨论可知，当 $|r| < 1$ 时几何数列 $r^{k}\rightarrow 0$，而当 $|r| > 1$ 或 $r = \text{±}1$ 时 $r^{k}$ 发散。因此，对 $|r| < 1$，有 $S_{k}\rightarrow a\text{/}(1 - r)$，于是得到

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1} = \frac{a}{1 - r}\ \text{if}\ |r| < 1.$$

If $|r| \geq 1,$ $S_{k}$ diverges, and therefore

如果 $|r| \geq 1$，则 $S_{k}$ 发散，因此

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1}\text{diverges if}\ |r| \geq 1.$$

A geometric series is a series of the form

几何级数是如下形式的级数

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1} = a + ar + ar^{2} + ar^{3} + \text{⋯}.$$

If $|r| < 1,$ the series converges, and

如果 $|r| < 1$，该级数收敛，且

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1} = \frac{a}{1 - r}\ \text{for}\ |r| < 1.$$ (5.7)

If $|r| \geq 1,$ the series diverges.

如果 $|r| \geq 1$，该级数发散。

Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than $n = 1$ or the exponent involves a linear expression for $n$ other than $n - 1.$ As long as we can rewrite the series in the form given by Equation 5.5, it is a geometric series. For example, consider the series

几何级数有时以略有不同形式出现。例如，有时下标不是从 $n = 1$ 开始，或者指数中包含关于 $n$ 的、并非 $n - 1$ 的线性表达式。只要能把这个级数改写成公式 5.5 所给的形式，它就是几何级数。例如，考虑级数

$${\sum\limits_{n = 0}^{\infty}\left( \frac{2}{3} \right)^{n + 2}}.$$

To see that this is a geometric series, we write out the first several terms:

为了看清这是一个几何级数，我们把开头几项写出来：

$$\begin{array}{cl} {\sum\limits_{n = 0}^{\infty}\left( \frac{2}{3} \right)^{n + 2}} & {= \left( \frac{2}{3} \right)^{2} + \left( \frac{2}{3} \right)^{3} + \left( \frac{2}{3} \right)^{4} + \text{⋯}} \\ & {= \frac{4}{9} + \frac{4}{9} \cdot \left( \frac{2}{3} \right) + \frac{4}{9} \cdot \left( \frac{2}{3} \right)^{2} + \text{⋯}.} \end{array}$$

We see that the initial term is $a = 4\text{/}9$ and the ratio is $r = 2\text{/}3.$ Therefore, the series can be written as

我们看到首项是 $a = 4\text{/}9$，公比是 $r = 2\text{/}3$。因此，这个级数可以写成

$${\sum\limits_{n = 1}^{\infty}\frac{4}{9}} \cdot \left( \frac{2}{3} \right)^{n - 1}.$$

Since $r = 2\text{/}3 < 1,$ this series converges, and its sum is given by

由于 $r = 2\text{/}3 < 1$，这个级数收敛，其和为

$${\sum\limits_{n = 1}^{\infty}\frac{4}{9}} \cdot \left( \frac{2}{3} \right)^{n - 1} = \frac{4\text{/}9}{1 - 2\text{/}3} = \frac{4}{3}.$$

Determining Convergence or Divergence of a Geometric Series 判断几何级数的收敛或发散

Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.

判断下列每个几何级数收敛还是发散；若收敛，求其和。

1. $\sum\limits_{n = 1}^{\infty}\frac{{(-3)}^{n + 1}}{4^{n - 1}}$

2. $\sum\limits_{n = 1}^{\infty}e^{2n}$

Solution 解答

1. Writing out the first several terms in the series, we have

1. 把这个级数的开头几项写出来，我们有

$$\begin{array}{cl} {\sum\limits_{n = 1}^{\infty}\frac{{(-3)}^{n + 1}}{4^{n - 1}}} & {= \frac{{(-3)}^{2}}{4^{0}} + \frac{{(-3)}^{3}}{4} + \frac{{(-3)}^{4}}{4^{2}} + \text{⋯}} \\ & {= {(-3)}^{2} + {(-3)}^{2} \cdot \left( \frac{-3}{4} \right) + {(-3)}^{2} \cdot \left( \frac{-3}{4} \right)^{2} + \text{⋯}} \\ & {= 9 + 9 \cdot \left( \frac{-3}{4} \right) + 9 \cdot \left( \frac{-3}{4} \right)^{2} + \text{⋯}.} \end{array}$$

The initial term $a = 9$ and the ratio $r = -3\text{/}4.$ Since $\left| r \middle| = 3\text{/}4 < 1, \right.$ the series converges to

首项 $a = 9$，公比 $r = -3\text{/}4$。由于 $\left| r \middle| = 3\text{/}4 < 1, \right.$ 该级数收敛到

$$\frac{9}{1 - (-3\text{/}4)} = \frac{9}{7\text{/}4} = \frac{36}{7}.$$

2. Writing this series as

2. 把这个级数写成

$$e^{2}\sum\limits_{n = 1}^{\infty}\left( e^{2} \right)^{n - 1}$$

we can see that this is a geometric series where $r = e^{2} > 1.$ Therefore, the series diverges.

可以看出这是一个几何级数，其中 $r = e^{2} > 1$。因此该级数发散。

Determine whether the series $\sum\limits_{n = 1}^{\infty}\left( \frac{-2}{5} \right)^{n - 1}$ converges or diverges. If it converges, find its sum.

判断级数 $\sum\limits_{n = 1}^{\infty}\left( \frac{-2}{5} \right)^{n - 1}$ 收敛还是发散。若收敛，求其和。

We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.

现在我们来看几何级数的一个优美应用。我们说明如何用它们把循环小数写成分数。

Writing Repeating Decimals as Fractions of Integers 把循环小数写成分数

Use a geometric series to write $3.\overset{\text{—}}{26}$ as a fraction of integers.

利用几何级数把 $3.\overset{\text{—}}{26}$ 写成分数。

Solution 解答

Since $3.\overset{\text{—}}{26} = 3.262626\text{…},$ first we write

由于 $3.\overset{\text{—}}{26} = 3.262626\text{…}$，我们先写成

$$\begin{array}{cl} {3.262626\text{…}} & {= 3 + \frac{26}{100} + \frac{26}{10,000} + \frac{26}{1,000,000} + \text{⋯}} \\ & {= 3 + \frac{26}{10^{2}} + \frac{26}{10^{4}} + \frac{26}{10^{6}} + \text{⋯}.} \end{array}$$

Ignoring the term 3, the rest of this expression is a geometric series with initial term $a = 26\text{/}10^{2}$ and ratio $r = 1\text{/}10^{2}.$ Therefore, the sum of this series is

忽略掉 3 这一项，剩下部分是一个几何级数，其首项为 $a = 26\text{/}10^{2}$，公比为 $r = 1\text{/}10^{2}$。因此，这个级数的和为

$$\frac{26\text{/}10^{2}}{1 - (1\text{/}10^{2})} = \frac{26\text{/}10^{2}}{99\text{/}10^{2}} = \frac{26}{99}.$$

Thus,

于是

$$3.262626\text{…} = 3 + \frac{26}{99} = \frac{323}{99}.$$

Write $5.2\overset{\text{–}}{7}$ as a fraction of integers.

把 $5.2\overset{\text{–}}{7}$ 写成分数。

Chapter Opener: Finding the Area of the Koch Snowflake 章节开篇：求科赫雪花的面积

Define a sequence of figures $\left\{ F_{n} \right\}$ recursively as follows (Figure 5.11). Let $F_{0}$ be an equilateral triangle with sides of length $1.$ For $n \geq 1,$ let $F_{n}$ be the curve created by removing the middle third of each side of $F_{n - 1}$ and replacing it with an equilateral triangle pointing outward. The limiting figure as $n\rightarrow\infty$ is known as Koch’s snowflake.

如下递归地定义一列图形 $\left\{ F_{n} \right\}$（图 5.11）。令 $F_{0}$ 为边长 $1$ 的等边三角形。对 $n \geq 1$，令 $F_{n}$ 为这样的曲线：去掉 $F_{n - 1}$ 每条边的三等分中间段，并在该处向外补上一个等边三角形。当 $n\rightarrow\infty$ 时的极限图形就称为科赫雪花（Koch’s snowflake）。

1. Find the length $L_{n}$ of the perimeter of $F_{n}.$ Evaluate $\underset{n\rightarrow\infty}{\text{lim}}L_{n}$ to find the length of the perimeter of Koch’s snowflake.

1. 求 $F_{n}$ 周长的长度 $L_{n}$。计算 $\underset{n\rightarrow\infty}{\text{lim}}L_{n}$，得到科赫雪花的周长。

2. Find the area $A_{n}$ of figure $F_{n}.$ Evaluate $\underset{n\rightarrow\infty}{\text{lim}}A_{n}$ to find the area of Koch’s snowflake.

2. 求图形 $F_{n}$ 的面积 $A_{n}$。计算 $\underset{n\rightarrow\infty}{\text{lim}}A_{n}$，得到科赫雪花的面积。

Solution 解答

1. Let $N_{n}$ denote the number of sides of figure $F_{n}.$ Since $F_{0}$ is a triangle, $N_{0} = 3.$ Let $l_{n}$ denote the length of each side of $F_{n}.$ Since $F_{0}$ is an equilateral triangle with sides of length $l_{0} = 1,$ we now need to determine $N_{1}$ and $l_{1}.$ Since $F_{1}$ is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of $F_{0},$ we get four sides in $F_{1}.$ Therefore, the number of sides for $F_{1}$ is

1. 令 $N_{n}$ 表示图形 $F_{n}$ 的边数。由于 $F_{0}$ 是三角形，$N_{0} = 3$。令 $l_{n}$ 表示 $F_{n}$ 每条边的长度。由于 $F_{0}$ 是边长为 $l_{0} = 1$ 的等边三角形，我们需要确定 $N_{1}$ 和 $l_{1}$。因为 $F_{1}$ 是通过去掉每条边的三等分中间段、并用两条线段代替而得到的，所以 $F_{0}$ 的每条边在 $F_{1}$ 中变成四条边。因此，$F_{1}$ 的边数为

$$N_{1} = 4 \cdot 3.$$

Since the length of each of these new line segments is $1\text{/}3$ the length of the line segments in $F_{0},$ the length of the line segments for $F_{1}$ is given by

由于每条新线段的长度是 $F_{0}$ 中线段长度的 $1\text{/}3$，所以 $F_{1}$ 的线段长度由下式给出

$$l_{1} = \frac{1}{3} \cdot 1 = \frac{1}{3}.$$

Similarly, for $F_{2},$ since the middle third of each side of $F_{1}$ is removed and replaced with two line segments, the number of sides in $F_{2}$ is given by

类似地，对 $F_{2}$，由于 $F_{1}$ 每条边的三等分中间段被去掉并换成两条线段，$F_{2}$ 的边数由下式给出

$$N_{2} = 4N_{1} = 4(4 \cdot 3) = 4^{2} \cdot 3.$$

Since the length of each of these sides is $1\text{/}3$ the length of the sides of $F_{1},$ the length of each side of figure $F_{2}$ is given by

由于这些边每条的长度是 $F_{1}$ 边长的 $1\text{/}3$，图形 $F_{2}$ 每条边的长度由下式给出

$$l_{2} = \frac{1}{3} \cdot l_{1} = \frac{1}{3} \cdot \frac{1}{3} = \left( \frac{1}{3} \right)^{2}.$$

More generally, since $F_{n}$ is created by removing the middle third of each side of $F_{n - 1}$ and replacing that line segment with two line segments of length $\frac{1}{3}l_{n - 1}$ in the shape of an equilateral triangle, we know that $N_{n} = 4N_{n - 1}$ and $l_{n} = \frac{l_{n - 1}}{3}.$ Therefore, the number of sides of figure $F_{n}$ is

更一般地，由于 $F_{n}$ 是通过去掉 $F_{n - 1}$ 每条边的三等分中间段、并用长度为 $\frac{1}{3}l_{n - 1}$ 的两条线段以等边三角形的形状代替而得到的，我们知道 $N_{n} = 4N_{n - 1}$ 且 $l_{n} = \frac{l_{n - 1}}{3}$。因此，图形 $F_{n}$ 的边数为

$$N_{n} = 4^{n} \cdot 3$$

and the length of each side is

而每条边的长度为

$$l_{n} = \left( \frac{1}{3} \right)^{n}.$$

Therefore, to calculate the perimeter of $F_{n},$ we multiply the number of sides $N_{n}$ and the length of each side $l_{n}.$ We conclude that the perimeter of $F_{n}$ is given by

因此，为计算 $F_{n}$ 的周长，我们把边数 $N_{n}$ 与每条边的长度 $l_{n}$ 相乘。我们得出结论：$F_{n}$ 的周长由下式给出

$$L_{n} = N_{n} \cdot l_{n} = 3 \cdot \left( \frac{4}{3} \right)^{n}.$$

Therefore, the length of the perimeter of Koch’s snowflake is

因此，科赫雪花的周长为

$$L = \underset{n\rightarrow\infty}{\text{lim}}L_{n} = \infty.$$

2. Let $T_{n}$ denote the area of each new triangle created when forming $F_{n}.$ For $n = 0,$ $T_{0}$ is the area of the original equilateral triangle. Therefore, $T_{0} = A_{0} = \sqrt{3}\text{/}4.$ For $n \geq 1,$ since the lengths of the sides of the new triangle are $1\text{/}3$ the length of the sides of $F_{n - 1},$ we have

2. 令 $T_{n}$ 表示构造 $F_{n}$ 时生成的每个小三角形的面积。当 $n = 0$ 时，$T_{0}$ 是原来等边三角形的面积。因此 $T_{0} = A_{0} = \sqrt{3}\text{/}4$。对 $n \geq 1$，由于新三角形边长是 $F_{n - 1}$ 边长的 $1\text{/}3$，我们有

$$T_{n} = \left( \frac{1}{3} \right)^{2}T_{n - 1} = \frac{1}{9} \cdot T_{n - 1}.$$

Therefore, $T_{n} = \left( \frac{1}{9} \right)^{n} \cdot \frac{\sqrt{3}}{4}.$ Since a new triangle is formed on each side of $F_{n - 1},$

因此 $T_{n} = \left( \frac{1}{9} \right)^{n} \cdot \frac{\sqrt{3}}{4}$。由于在 $F_{n - 1}$ 的每条边上都会生成一个新三角形，

$$\begin{array}{cl} A_{n} & {= A_{n - 1} + N_{n - 1} \cdot T_{n}} \\ & {= A_{n - 1} + \left( {3 \cdot 4^{n - 1}} \right) \cdot \left( \frac{1}{9} \right)^{n} \cdot \frac{\sqrt{3}}{4}} \\ & {= A_{n - 1} + \frac{3}{4} \cdot \left( \frac{4}{9} \right)^{n} \cdot \frac{\sqrt{3}}{4}.} \end{array}$$

Writing out the first few terms $A_{0},A_{1},A_{2},$ we see that

把开头几项 $A_{0},A_{1},A_{2}$ 写出来，我们看到

$$\begin{array}{l} \\ {A_{0} = \frac{\sqrt{3}}{4}} \\ {A_{1} = A_{0} + \frac{3}{4} \cdot \left( \frac{4}{9} \right) \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} + \frac{3}{4} \cdot \left( \frac{4}{9} \right) \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{3}{4} \cdot \left( \frac{4}{9} \right)} \right\rbrack} \\ {A_{2} = A_{1} + \frac{3}{4} \cdot \left( \frac{4}{9} \right)^{2} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{3}{4} \cdot \left( \frac{4}{9} \right)} \right\rbrack + \frac{3}{4} \cdot \left( \frac{4}{9} \right)^{2} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{3}{4} \cdot \left( \frac{4}{9} \right) + \frac{3}{4} \cdot \left( \frac{4}{9} \right)^{2}} \right\rbrack.} \end{array}$$

More generally,

更一般地，

$$A_{n} = \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{3}{4}\left( {\frac{4}{9} + \left( \frac{4}{9} \right)^{2} + \text{⋯} + \left( \frac{4}{9} \right)^{n}} \right)} \right\rbrack.$$

Factoring $4\text{/}9$ out of each term inside the inner parentheses, we rewrite our expression as

把内层括号中每一项提取出 $4\text{/}9$，我们把表达式改写为

$$A_{n} = \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{1}{3}\left( {1 + \frac{4}{9} + \left( \frac{4}{9} \right)^{2} + \text{⋯} + \left( \frac{4}{9} \right)^{n - 1}} \right)} \right\rbrack.$$

The expression $1 + \left( \frac{4}{9} \right) + \left( \frac{4}{9} \right)^{2} + \text{⋯} + \left( \frac{4}{9} \right)^{n - 1}$ is a geometric sum. As shown earlier, this sum satisfies

表达式 $1 + \left( \frac{4}{9} \right) + \left( \frac{4}{9} \right)^{2} + \text{⋯} + \left( \frac{4}{9} \right)^{n - 1}$ 是一个几何和。如前所示，这个和满足

$$1 + \frac{4}{9} + \left( \frac{4}{9} \right)^{2} + \text{⋯} + \left( \frac{4}{9} \right)^{n - 1} = \frac{1 - {(4\text{/}9)}^{n}}{1 - (4\text{/}9)}.$$

Substituting this expression into the expression above and simplifying, we conclude that

把这个式子代入上面的表达式并化简，我们得出结论

$$\begin{array}{cl} A_{n} & {= \frac{\sqrt{3}}{4}\left\lbrack {1 + \frac{1}{3}\left( \frac{1 - {(4\text{/}9)}^{n}}{1 - (4\text{/}9)} \right)} \right\rbrack} \\ & {= \frac{\sqrt{3}}{4}\left\lbrack {\frac{8}{5} - \frac{3}{5}\left( \frac{4}{9} \right)^{n}} \right\rbrack.} \end{array}$$

Therefore, the area of Koch’s snowflake is

因此，科赫雪花的面积为

$$A = \underset{n\rightarrow\infty}{\text{lim}}A_{n} = \frac{2\sqrt{3}}{5}.$$

Analysis 分析

The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve $y = 1\text{/}x^{2}$ and the $x$-axis on the interval $\lbrack 1,\infty).$ Since the improper integral

科赫雪花很有意思，因为它面积有限而周长无限。虽然一开始这似乎不可能，但请回想在本书前面已经见过类似的例子。例如，考虑由曲线 $y = 1\text{/}x^{2}$ 与 $x$ 轴在区间 $\lbrack 1,\infty)$ 上所围成的区域。由于反常积分

$${\int_{1}^{\infty}\frac{1}{x^{2}}}dx$$

converges, the area of this region is finite, even though the perimeter is infinite.

收敛，该区域的面积是有限的，尽管其周长是无限的。

Telescoping Series 裂项级数

Consider the series ${\sum\limits_{n = 1}^{\infty}\frac{1}{n(n + 1)}}.$ We discussed this series in Example 5.7, showing that the series converges by writing out the first several partial sums $S_{1},S_{2}\text{,…},S_{6}$ and noticing that they are all of the form $S_{k} = \frac{k}{k + 1}.$ Here we use a different technique to show that this series converges. By using partial fractions, we can write

考虑级数 ${\sum\limits_{n = 1}^{\infty}\frac{1}{n(n + 1)}}.$ 我们在示例 5.7 中讨论过这个级数，通过写出前几个部分和 $S_{1},S_{2}\text{,…},S_{6}$ 并注意它们都具有形式 $S_{k} = \frac{k}{k + 1}$ 来说明该级数收敛。这里我们使用一种不同的技巧来说明这个级数收敛。利用部分分式，我们可以写出

$$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.$$

Therefore, the series can be written as

因此，该级数可以写成

$$\sum\limits_{n = 1}^{\infty}\left\lbrack \frac{1}{n} - \frac{1}{n + 1} \right\rbrack = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \text{⋯}.$$

Writing out the first several terms in the sequence of partial sums $\left\{ S_{k} \right\},$ we see that

写出部分和数列 $\left\{ S_{k} \right\}$ 中的前几项，我们看到

$$\begin{array}{l} {S_{1} = 1 - \frac{1}{2}} \\ {S_{2} = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) = 1 - \frac{1}{3}} \\ {S_{3} = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) = 1 - \frac{1}{4}.} \end{array}$$

In general,

一般地，

$$S_{k} = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \text{⋯} + \left( {\frac{1}{k} - \frac{1}{k + 1}} \right) = 1 - \frac{1}{k + 1}.$$

We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since $S_{k} = 1 - 1\text{/}(k + 1)$ and $1\text{/}(k + 1)\rightarrow 0$ as $k\rightarrow\infty,$ the sequence of partial sums converges to $1,$ and therefore the series converges to $1.$

我们注意到中间项相互抵消，只剩下首项和末项。在某种意义上，该级数像单筒望远镜一样收缩，镜筒彼此套叠以缩短镜身。因此，我们把具有这种性质的级数称为裂项级数。对于这个级数，由于 $S_{k} = 1 - 1\text{/}(k + 1)$ 且当 $k\rightarrow\infty$ 时 $1\text{/}(k + 1)\rightarrow 0$，部分和数列收敛于 $1$，因此该级数收敛于 $1.$

A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.

裂项级数是这样一种级数：在每一个部分和中，绝大多数项相互抵消，只留下部分首项和部分末项。

For example, any series of the form

例如，任何如下形式的级数

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {b_{n} - b_{n + 1}} \right\rbrack = \left( {b_{1} - b_{2}} \right) + \left( {b_{2} - b_{3}} \right) + \left( {b_{3} - b_{4}} \right) + \text{⋯}$$

is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that

都是裂项级数。我们可以通过写出几个部分和看出这一点。特别地，我们看到

$$\begin{array}{l} {S_{1} = b_{1} - b_{2}} \\ {S_{2} = \left( {b_{1} - b_{2}} \right) + \left( {b_{2} - b_{3}} \right) = b_{1} - b_{3}} \\ {S_{3} = \left( {b_{1} - b_{2}} \right) + \left( {b_{2} - b_{3}} \right) + \left( {b_{3} - b_{4}} \right) = b_{1} - b_{4}.} \end{array}$$

In general, the *k*th partial sum of this series is

一般地，该级数的第 *k* 个部分和为

$$S_{k} = b_{1} - b_{k + 1}.$$

Since the *k*th partial sum can be simplified to the difference of these two terms, the sequence of partial sums $\left\{ S_{k} \right\}$ will converge if and only if the sequence $\left\{ b_{k + 1} \right\}$ converges. Moreover, if the sequence $b_{k + 1}$ converges to some finite number $B,$ then the sequence of partial sums converges to $b_{1} - B,$ and therefore

由于第 *k* 个部分和可以简化为这两项之差，部分和数列 $\left\{ S_{k} \right\}$ 收敛当且仅当数列 $\left\{ b_{k + 1} \right\}$ 收敛。此外，如果数列 $b_{k + 1}$ 收敛于某个有限数 $B$，那么部分和数列收敛于 $b_{1} - B$，因此

$$\sum\limits_{n = 1}^{\infty}\lbrack b_{n} - b_{n + 1}\rbrack = b_{1} - B.$$

In the next example, we show how to use these ideas to analyze a telescoping series of this form.

在下面的示例中，我们展示如何用这些思想来分析这种形式的裂项级数。

Evaluating a Telescoping Series 求裂项级数

Determine whether the telescoping series

判断下列裂项级数

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {\text{cos}\left( \frac{1}{n} \right) - \text{cos}\left( \frac{1}{n + 1} \right)} \right\rbrack$$

converges or diverges. If it converges, find its sum.

收敛还是发散。如果收敛，求其和。

Solution 解答

By writing out terms in the sequence of partial sums, we can see that

通过写出部分和数列中的各项，我们可以看到

$$\begin{array}{cll} S_{1} & = & {\text{cos}(1) - \text{cos}\left( \frac{1}{2} \right)} \\ S_{2} & = & {\left( {\text{cos}(1) - \text{cos}\left( \frac{1}{2} \right)} \right) + \left( {\text{cos}\left( \frac{1}{2} \right) - \text{cos}\left( \frac{1}{3} \right)} \right) = \text{cos}(1) - \text{cos}\left( \frac{1}{3} \right)} \\ S_{3} & = & {\left( {\text{cos}(1) - \text{cos}\left( \frac{1}{2} \right)} \right) + \left( {\text{cos}\left( \frac{1}{2} \right) - \text{cos}\left( \frac{1}{3} \right)} \right) + \left( {\text{cos}\left( \frac{1}{3} \right) - \text{cos}\left( \frac{1}{4} \right)} \right)} \\ & = & {\text{cos}(1) - \text{cos}\left( \frac{1}{4} \right).} \end{array}$$

In general,

一般地，

$$S_{k} = \text{cos}(1) - \text{cos}\left( \frac{1}{k + 1} \right).$$

Since $1\text{/}(k + 1)\rightarrow 0$ as $k\rightarrow\infty$ and $\text{cos}\mspace{2mu} x$ is a continuous function, $\text{cos}(1\text{/}(k + 1))\rightarrow\text{cos}(0) = 1.$ Therefore, we conclude that $S_{k}\rightarrow\text{cos}(1) - 1.$ The telescoping series converges and the sum is given by

由于当 $k\rightarrow\infty$ 时 $1\text{/}(k + 1)\rightarrow 0$，且 $\text{cos}\mspace{2mu} x$ 是连续函数，故 $\text{cos}(1\text{/}(k + 1))\rightarrow\text{cos}(0) = 1.$ 因此，我们得出结论：$S_{k}\rightarrow\text{cos}(1) - 1.$ 该裂项级数收敛，其和为

$$\sum\limits_{n = 1}^{\infty}\left\lbrack {\text{cos}\left( \frac{1}{n} \right) - \text{cos}\left( \frac{1}{n + 1} \right)} \right\rbrack = \text{cos}(1) - 1.$$

Determine whether $\sum\limits_{n = 1}^{\infty}\left\lbrack {e^{1\text{/}n} - e^{1\text{/}(n + 1)}} \right\rbrack$ converges or diverges. If it converges, find its sum.

判断 $\sum\limits_{n = 1}^{\infty}\left\lbrack {e^{1\text{/}n} - e^{1\text{/}(n + 1)}} \right\rbrack$ 收敛还是发散。如果收敛，求其和。

Euler’s Constant 欧拉常数

We have shown that the harmonic series $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ diverges. Here we investigate the behavior of the partial sums $S_{k}$ as $k\rightarrow\infty.$ In particular, we show that they behave like the natural logarithm function by showing that there exists a constant $\gamma$ such that

我们已经证明调和级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 发散。这里我们来研究当 $k\rightarrow\infty$ 时部分和 $S_{k}$ 的行为。特别地，我们通过证明存在常数 $\gamma$ 使得

$$\sum\limits_{n = 1}^{k}\frac{1}{n} - \text{ln}\mspace{2mu} k\rightarrow\gamma\ \text{as}\ k\rightarrow\infty.$$

This constant $\gamma$ is known as Euler’s constant.

这个常数 $\gamma$ 称为欧拉常数。

1. Let $T_{k} = {\sum\limits_{n = 1}^{k}{\frac{1}{n} - \text{ln}\mspace{2mu} k}}.$ Evaluate $T_{k}$ for various values of $k.$

1. 设 $T_{k} = {\sum\limits_{n = 1}^{k}{\frac{1}{n} - \text{ln}\mspace{2mu} k}}.$ 对 $k$ 的若干取值计算 $T_{k}$。

2. For $T_{k}$ as defined in part 1. show that the sequence $\left\{ T_{k} \right\}$ converges by using the following steps.

2. 对第 1 部分所定义的 $T_{k}$，利用以下步骤证明数列 $\left\{ T_{k} \right\}$ 收敛。

1. Show that the sequence $\left\{ T_{k} \right\}$ is monotone decreasing. (*Hint:* Show that $\text{ln}{\left( {1 + 1\text{/}k > 1\text{/}\left( {k + 1} \right)} \right))}$

1. 证明数列 $\left\{ T_{k} \right\}$ 单调递减。（*提示：* 证明 $\text{ln}{\left( {1 + 1\text{/}k > 1\text{/}\left( {k + 1} \right)} \right))}$

2. Show that the sequence $\left\{ T_{k} \right\}$ is bounded below by zero. (*Hint:* Express $\text{ln}\mspace{2mu} k$ as a definite integral.)

2. 证明数列 $\left\{ T_{k} \right\}$ 下有界（下界为零）。（*提示：* 将 $\text{ln}\mspace{2mu} k$ 表示为一个定积分。）

3. Use the Monotone Convergence Theorem to conclude that the sequence $\left\{ T_{k} \right\}$ converges. The limit $\gamma$ is Euler’s constant.

3. 利用单调收敛定理得出数列 $\left\{ T_{k} \right\}$ 收敛。该极限 $\gamma$ 就是欧拉常数。

3. Now estimate how far $T_{k}$ is from $\gamma$ for a given integer $k.$ Prove that for $k \geq 1,$ $0 < T_{k} - \gamma \leq 1\text{/}k$ by using the following steps.

3. 现在估计对于给定的整数 $k$，$T_{k}$ 距离 $\gamma$ 有多远。利用以下步骤证明：当 $k \geq 1$ 时，$0 < T_{k} - \gamma \leq 1\text{/}k$。

1. Show that $\text{ln}\left( {k + 1} \right) - \text{ln}\mspace{2mu} k < 1\text{/}k.$

1. 证明 $\text{ln}\left( {k + 1} \right) - \text{ln}\mspace{2mu} k < 1\text{/}k.$

2. Use the result from part a. to show that for any integer $k,$

2. 利用 a 部分的结果证明：对任意整数 $k$，

$$T_{k} - T_{k + 1} < \frac{1}{k} - \frac{1}{k + 1}.$$

3. For any integers $k$ and $j$ such that $j > k,$ express $T_{k} - T_{j}$ as a telescoping sum by writing

3. 对任意满足 $j > k$ 的整数 $k$ 和 $j$，通过写出

$$T_{k} - T_{j} = \left( {T_{k} - T_{k + 1}} \right) + \left( {T_{k + 1} - T_{k + 2}} \right) + \left( {T_{k + 2} - T_{k + 3}} \right) + \text{⋯} + \left( {T_{j - 1} - T_{j}} \right).$$

Use the result from part b. combined with this telescoping sum to conclude that

结合 b 部分的结果与这个裂项和，得出结论：

$$T_{k} - T_{j} < \frac{1}{k} - \frac{1}{j}.$$

4. Apply the limit to both sides of the inequality in part c. to conclude that

4. 对 c 部分不等式两边同时取极限，得出结论：

$$T_{k} - \gamma \leq \frac{1}{k}.$$

5. Estimate $\gamma$ to an accuracy of within $0.001.$

5. 将 $\gamma$ 估计到精度在 $0.001$ 以内。

Section 5.2 Exercises 5.2 节习题

Using sigma notation, write the following expressions as infinite series.

用 sigma 记号将下列表达式写成无穷级数。

67.

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \text{⋯}$

68\.

$1 - 1 + 1 - 1 + \text{⋯}$

69.

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...$

70\.

$\text{sin}\mspace{2mu} 1 + \text{sin}\mspace{2mu}\frac{1}{2} + \text{sin}\mspace{2mu}\frac{1}{3} + \text{sin}\mspace{2mu}\frac{1}{4} + \text{⋯}$

Compute the first four partial sums $S_{1}\text{,…},S_{4}$ for the series having $n\text{th}$ term $a_{n}$ starting with $n = 1$ as follows.

计算下列级数的前四个部分和 $S_{1}\text{,…},S_{4}$，该级数的第 $n\text{th}$ 项 $a_{n}$ 从 $n = 1$ 开始。

71.

$a_{n} = n$

72\.

$a_{n} = 1\text{/}n$

73.

$a_{n} = \text{sin}\left( {n\pi\text{/}2} \right)$

74\.

$a_{n} = (-1)^{n}$

In the following exercises, compute the general term $a_{n}$ of the series with the given partial sum $S_{n}.$ If the sequence of partial sums converges, find its limit $S.$

在以下习题中，计算具有给定部分和 $S_{n}.$ 的级数的通项 $a_{n}$。若部分和的数列收敛，求其极限 $S.$

75.

$S_{n} = 1 - \frac{1}{n},$ $n \geq 2$

76\.

$S_{n} = \frac{n\left( {n + 1} \right)}{2},$ $n \geq 1$

77.

$S_{n} = \sqrt{n},n \geq 2$

78\.

$S_{n} = 2 - \left( {n + 2} \right)\text{/}2^{n},n \geq 1$

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

对于下列各级数，利用部分和的数列判断该级数是收敛还是发散。

79.

$\sum\limits_{n = 1}^{\infty}\frac{n}{n + 2}$

80\.

$\sum\limits_{n = 1}^{\infty}\left( {1 - (-1)^{n})} \right)$

81.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\left( {n + 1} \right)\left( {n + 2} \right)}$ (*Hint:* Use a partial fraction decomposition like that for ${\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)}}.$)

$\sum\limits_{n = 1}^{\infty}\frac{1}{\left( {n + 1} \right)\left( {n + 2} \right)}$ （*提示：* 使用类似于 ${\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)}}.$ 的部分分式分解）

82\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{2n + 1}$ (*Hint:* Follow the reasoning for ${\sum\limits_{n = 1}^{\infty}\frac{1}{n}}.$)

$\sum\limits_{n = 1}^{\infty}\frac{1}{2n + 1}$ （*提示：* 仿照 ${\sum\limits_{n = 1}^{\infty}\frac{1}{n}}.$ 的推理）

Suppose that ${\sum\limits_{n = 1}^{\infty}{a_{n} = 1}},$ that ${\sum\limits_{n = 1}^{\infty}{b_{n} = -1}},$ that $a_{1} = 2,$ and $b_{1} = -3.$ Find the sum of the indicated series.

设 ${\sum\limits_{n = 1}^{\infty}{a_{n} = 1}},$ 设 ${\sum\limits_{n = 1}^{\infty}{b_{n} = -1}},$ 且 $a_{1} = 2,$ $b_{1} = -3.$ 求所示级数的和。

83.

$\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}$

84\.

$\sum\limits_{n = 1}^{\infty}{(a_{n} - 2b_{n})}$

85.

$\sum\limits_{n = 2}^{\infty}{(a_{n} - b_{n})}$

86\.

$\sum\limits_{n = 1}^{\infty}{(3a_{n + 1} - 4b_{n + 1})}$

State whether the given series converges and explain why.

说明所给级数是否收敛，并解释原因。

87.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n + 1000}$ (*Hint:* Rewrite using a change of index.)

$\sum\limits_{n = 1}^{\infty}\frac{1}{n + 1000}$ （*提示：* 利用下标的改变重写。）

88\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n + 10^{80}}$ (*Hint:* Rewrite using a change of index.)

$\sum\limits_{n = 1}^{\infty}\frac{1}{n + 10^{80}}$ （*提示：* 利用下标的改变重写。）

89.

$1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \text{⋯}$

90\.

$1 + \frac{e}{\pi} + \frac{e^{2}}{\pi^{2}} + \frac{e^{3}}{\pi^{3}} + \text{⋯}$

91.

$1 + \frac{\pi}{e^{2}} + \frac{\pi^{2}}{e^{4}} + \frac{\pi^{3}}{e^{6}} + \frac{\pi^{4}}{e^{8}} + \text{⋯}$

92\.

$1 - \sqrt{\frac{\pi}{3}} + \sqrt{\frac{\pi^{2}}{9}} - \sqrt{\frac{\pi^{3}}{27}} + \text{⋯}$

For $a_{n}$ as follows, write the sum as a geometric series of the form ${\sum\limits_{n = 1}^{\infty}{ar^{n}}}.$ State whether the series converges and if it does, find the value of ${\sum a_{n}}.$

对于如下的 $a_{n}$，将和写成形如 ${\sum\limits_{n = 1}^{\infty}{ar^{n}}}$ 的几何级数。说明该级数是否收敛；若收敛，求 ${\sum a_{n}}.$ 的值。

93.

$a_{1} = -1$ and $a_{n}\text{/}a_{n + 1} = -5$ for $n \geq 1.$

$a_{1} = -1$ 且 $a_{n}\text{/}a_{n + 1} = -5$，对 $n \geq 1$ 成立。

94\.

$a_{1} = 2$ and $a_{n}\text{/}a_{n + 1} = 1\text{/}2$ for $n \geq 1.$

$a_{1} = 2$ 且 $a_{n}\text{/}a_{n + 1} = 1\text{/}2$，对 $n \geq 1$ 成立。

95.

$a_{1} = 10$ and $a_{n}\text{/}a_{n + 1} = 10$ for $n \geq 1.$

$a_{1} = 10$ 且 $a_{n}\text{/}a_{n + 1} = 10$，对 $n \geq 1$ 成立。

96\.

$a_{1} = 1\text{/}10$ and $a_{n}\text{/}a_{n + 1} = -10$ for $n \geq 1.$

$a_{1} = 1\text{/}10$ 且 $a_{n}\text{/}a_{n + 1} = -10$，对 $n \geq 1$ 成立。

Use the identity $\frac{1}{1 - y} = {\sum\limits_{n = 0}^{\infty}y^{n}}$ to express the function as a geometric series in the indicated term.

利用恒等式 $\frac{1}{1 - y} = {\sum\limits_{n = 0}^{\infty}y^{n}}$ 将该函数表示为所指项的几何级数。

97.

$\frac{x}{1 + x}$ in $x$

$\frac{x}{1 + x}$ 关于 $x$

98\.

$\frac{\sqrt{x}}{1 - x^{3\text{/}2}}$ in $\sqrt{x}$

$\frac{\sqrt{x}}{1 - x^{3\text{/}2}}$ 关于 $\sqrt{x}$

99.

$\frac{1}{1 + \text{sin}^{2}x}$ in $\text{sin}\mspace{2mu} x$

$\frac{1}{1 + \text{sin}^{2}x}$ 关于 $\text{sin}\mspace{2mu} x$

100\.

$\text{sec}^{2}x$ in $\text{sin}\mspace{2mu} x$

$\text{sec}^{2}x$ 关于 $\text{sin}\mspace{2mu} x$

Evaluate the following telescoping series or state whether the series diverges.

计算下列裂项级数，或说明该级数是否发散。

101.

${\sum\limits_{n = 1}^{\infty}{2^{1\text{/}n} -}}2^{1\text{/}{({n + 1})}}$

102\.

${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{13}}} - \frac{1}{\left( {n + 1} \right)^{13}}$

103.

$\sum\limits_{n = 1}^{\infty}\left( {\sqrt{n} - \sqrt{n + 1}} \right)$

104\.

$\sum\limits_{n = 1}^{\infty}\left( {\text{sin}\mspace{2mu} n - \text{sin}\left( {n + 1} \right)} \right)$

Express the following series as a telescoping sum and evaluate its *n*th partial sum.

将下列级数表示为裂项和，并求其 *n*th 部分和。

105.

$\sum\limits_{n = 1}^{\infty}{\text{ln}\left( \frac{n}{n + 1} \right)}$

106\.

$\sum\limits_{n = 1}^{\infty}\frac{2n + 1}{\left( {n^{2} + n} \right)^{2}}$ (*Hint:* Factor denominator and use partial fractions.)

$\sum\limits_{n = 1}^{\infty}\frac{2n + 1}{\left( {n^{2} + n} \right)^{2}}$ （*提示：* 分解分母并使用部分分式。）

107.

$\sum\limits_{n = 2}^{\infty}\frac{\text{ln}\left( {1 +_{n}^{1}} \right)}{\text{ln}\mspace{2mu} n\mspace{2mu}\text{ln}\left( {n + 1} \right)}$

108\.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {n + 2} \right)}{n\left( {n + 1} \right)2^{n + 1}}$ (*Hint:* Look at $\left. 1\text{/}\left( {n2^{n}} \right). \right)$

$\sum\limits_{n = 1}^{\infty}\frac{\left( {n + 2} \right)}{n\left( {n + 1} \right)2^{n + 1}}$ （*提示：* 考察 $\left. 1\text{/}\left( {n2^{n}} \right). \right)$

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

一般的裂项级数是指，在连续求和给定项数后，除最初少数几项外其余各项均被消去的级数。

109.

Let $a_{n} = f(n) - 2f\left( {n + 1} \right) + f\left( {n + 2} \right),$ in which $f(n)\rightarrow 0$ as $n\rightarrow\infty.$ Find ${\sum\limits_{n = 1}^{\infty}a_{n}}.$

设 $a_{n} = f(n) - 2f\left( {n + 1} \right) + f\left( {n + 2} \right),$ 其中 $f(n)\rightarrow 0$ 当 $n\rightarrow\infty.$ 求 ${\sum\limits_{n = 1}^{\infty}a_{n}}.$

110\.

$a_{n} = f(n) - f\left( {n + 1} \right) - f\left( {n + 2} \right) + f\left( {n + 3} \right),$ in which $f(n)\rightarrow 0$ as $n\rightarrow\infty.$ Find ${\sum\limits_{n = 1}^{\infty}a_{n}}.$

$a_{n} = f(n) - f\left( {n + 1} \right) - f\left( {n + 2} \right) + f\left( {n + 3} \right),$ 其中 $f(n)\rightarrow 0$ 当 $n\rightarrow\infty.$ 求 ${\sum\limits_{n = 1}^{\infty}a_{n}}.$

111.

Suppose that $a_{n} = c_{0}f(n) + c_{1}f\left( {n + 1} \right) + c_{2}f\left( {n + 2} \right) + c_{3}f\left( {n + 3} \right) + c_{4}f\left( {n + 4} \right),$ where $f(n)\rightarrow 0$ as $n\rightarrow\infty.$ Find a condition on the coefficients $c_{0}\text{,…},c_{4}$ that make this a general telescoping series.

设 $a_{n} = c_{0}f(n) + c_{1}f\left( {n + 1} \right) + c_{2}f\left( {n + 2} \right) + c_{3}f\left( {n + 3} \right) + c_{4}f\left( {n + 4} \right),$ 其中 $f(n)\rightarrow 0$ 当 $n\rightarrow\infty.$ 求使这成为一个一般裂项级数的系数 $c_{0}\text{,…},c_{4}$ 满足的条件。

112\.

Evaluate $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)}$ (*Hint:* $\left. \frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)} = \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2\left( {n + 2} \right)} \right)$

计算 $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)}$ （*提示：* $\left. \frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)} = \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2\left( {n + 2} \right)} \right)$

113.

Evaluate ${\sum\limits_{n = 2}^{\infty}\frac{2}{n^{3} - n}}.$

计算 ${\sum\limits_{n = 2}^{\infty}\frac{2}{n^{3} - n}}.$

114\.

Find a formula for $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + N} \right)}$ where $N$ is a positive integer.

求 $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + N} \right)}$ 的公式，其中 $N$ 为正整数。

115.

\[T\] Define a sequence $t_{k} = {\sum\limits_{n = 1}^{k - 1}{(1\text{/}n) - \text{ln}\mspace{2mu} k.}}$ Use the graph of $1\text{/}x$ to verify that $t_{k}$ is increasing. Plot $t_{k}$ for $k = 1\text{…}100$ and state whether it appears that the sequence converges.

\[T\] 定义数列 $t_{k} = {\sum\limits_{n = 1}^{k - 1}{(1\text{/}n) - \text{ln}\mspace{2mu} k.}}$ 利用 $1\text{/}x$ 的图像验证 $t_{k}$ 是递增的。对 $k = 1\text{…}100$ 描出 $t_{k}$，并说明该数列是否看起来收敛。

116\.

\[T\] Suppose that $N$ equal uniform rectangular blocks are stacked one on top of the other, allowing for some overhang. Archimedes’ law of the lever implies that the stack of $N$ blocks is stable as long as the center of mass of the top $\left( {N - 1} \right)$ blocks lies at the edge of the bottom block. Let $x$ denote the position of the edge of the bottom block, and think of its position as relative to the center of the next-to-bottom block. This implies that $(N - 1)x = \left( {\frac{1}{2} - x} \right)$ or $x = 1\text{/}\left( {2N} \right).$ Use this expression to compute the maximum overhang (the position of the edge of the top block over the edge of the bottom block.) See the following figure.

\[T\] 设 $N$ 个相同均匀长方体木块一个叠在另一个上面，允许有一定的悬挑。阿基米德杠杆定律表明，只要最上面 $\left( {N - 1} \right)$ 块木块的质心位于底块边缘，这 $N$ 块木块的叠堆就是稳定的。令 $x$ 表示底块边缘的位置，并将其位置视为相对于倒数第二块木块的中心。这意味着 $(N - 1)x = \left( {\frac{1}{2} - x} \right)$ 或 $x = 1\text{/}\left( {2N} \right).$ 利用此表达式计算最大悬挑（顶块边缘相对于底块边缘的位置）。见下图。

Each of the following infinite series converges to the given multiple of $\pi$ or $1\text{/}\pi.$

下列每个无穷级数都收敛到给定的 $\pi$ 的倍数或 $1\text{/}\pi.$

In each case, find the minimum value of $N$ such that the $N\text{th}$ partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to $15$ decimals place, $\pi = 3.141592653589793....$

对每种情形，求 $N$ 的最小值，使得该级数的 $N\text{th}$ 部分和在给定的小数位数内精确近似左边，并给出所求的近似值。取到 $15$ 位小数，$\pi = 3.141592653589793....$

117.

\[T\] $\pi = -3 + {\sum\limits_{n = 1}^{\infty}\frac{n2^{n}n\text{!}^{2}}{\left( {2n} \right)\text{!}}},$ error $< 0.0001$

\[T\] $\pi = -3 + {\sum\limits_{n = 1}^{\infty}\frac{n2^{n}n\text{!}^{2}}{\left( {2n} \right)\text{!}}},$ 误差 $< 0.0001$

118\.

\[T\] $\frac{\pi}{2} = {\sum\limits_{k = 0}^{\infty}\frac{k\text{!}}{\left( {2k + 1} \right)\text{!}\text{!}}} = {\sum\limits_{k = 0}^{\infty}\frac{2^{k}k\text{!}^{2}}{\left( {2k + 1} \right)\text{!}}},$ error $< 10^{-4}$

119.

\[T\] $\frac{9801}{2\pi} = \frac{\sqrt{2}}{9801}{\sum\limits_{k = 0}^{\infty}\frac{\left( {4k} \right)\text{!}\left( {1103 + 26390k} \right)}{\left( {k\text{!}} \right)^{4}396^{4k}}},$ error $< 10^{-12}$

120\.

\[T\] $\frac{1}{12\pi} = {\sum\limits_{k = 0}^{\infty}\frac{(-1)^{k}\left( {6k} \right)\text{!}\left( {13591409 + 545140134k} \right)}{\left( {3k} \right)\text{!}\left( {k\text{!}} \right)^{3}640320^{3k + 3\text{/}2}}},$ error $< 10^{-15}$

121.

\[T\] A fair coin is one that has probability $1\text{/}2$ of coming up heads when flipped.

\[T\] 一枚均匀的硬币是指抛掷时出现正面的概率为 $1\text{/}2$ 的硬币。

1. What is the probability that a fair coin will come up tails $n$ times in a row?

1. 一枚均匀的硬币连续 $n$ 次出现反面的概率是多少？

2. Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.

2. 求一枚硬币在偶数次抛掷的最后一次首次出现正面的概率。

122\.

\[T\] Find the probability that a fair coin is flipped a multiple of three times before coming up heads.

\[T\] 求一枚均匀的硬币在首次出现正面之前被抛掷了三的倍数次的概率。

123.

Find the probability that a fair coin will come up heads for the second time after an even number of flips.

求一枚均匀的硬币在偶数次抛掷之后第二次出现正面的概率。

124\.

\[T\] Find a series that expresses the probability that a fair coin will come up heads for the second time on a multiple of three flips.

\[T\] 求一个级数，表示一枚均匀的硬币在三的倍数次抛掷时第二次出现正面的概率。

125.

\[T\] The expected number of times that a fair coin will come up heads is defined as the sum over $n = 1,2\text{,…}$ of $n$ times the probability that the coin will come up heads exactly $n$ times in a row, or $n\text{/}2^{n + 1}.$ Compute the expected number of consecutive times that a fair coin will come up heads.

\[T\] 一枚均匀的硬币出现正面的期望次数定义为对 $n = 1,2\text{,…}$ 求和：$n$ 乘以该硬币恰好连续 $n$ 次出现正面的概率，即 $n\text{/}2^{n + 1}.$ 计算一枚均匀的硬币连续出现正面的期望次数。

126\.

\[T\] A person deposits $\text{\$}10$ at the beginning of each quarter into a bank account that earns $4\text{\%}$ annual interest compounded quarterly (four times a year).

\[T\] 某人在每个季度初向一个银行账户存入 $\text{\$}10$，该账户按 $4\text{\%}$ 的年利率每季度复利计息（每年四次）。

1. Show that the interest accumulated after $n$ quarters is $\text{\$}10\left( {\frac{1.01^{n + 1} - 1}{0.01} - n} \right).$

1. 证明经过 $n$ 个季度后累积的利息为 $\text{\$}10\left( {\frac{1.01^{n + 1} - 1}{0.01} - n} \right).$

2. Find the first eight terms of the sequence.

2. 求该数列的前八项。

3. How much interest has accumulated after $2$ years?

3. $2$ 年后累积了多少利息？

127.

\[T\] Suppose that the amount of a drug in a patient’s system diminishes by a multiplicative factor $r < 1$ each hour. Suppose that a new dose is administered every $N$ hours. Find an expression that gives the amount $A(n)$ in the patient’s system after $n$ hours for each $n$ in terms of the dosage $d$ and the ratio $r.$ (*Hint:* Write $n = mN + k,$ where $0 \leq k < N,$ and sum over values from the different doses administered.)

\[T\] 设某药物在病人体内的量每小时按乘性因子 $r < 1$ 减少。假设每 $N$ 小时给药一次。求一个表达式，给出经过 $n$ 小时后病人体内药物量 $A(n)$（对每个 $n$），用剂量 $d$ 与比值 $r$ 表示。（*提示：* 写出 $n = mN + k,$ 其中 $0 \leq k < N,$ 并对不同次给药的各值求和。）

128\.

\[T\] A certain drug is effective for an average patient only if there is at least $1$ mg per kg in the patient’s system, while it is safe only if there is at most $2$ mg per kg in an average patient’s system. Suppose that the amount in a patient’s system diminishes by a multiplicative factor of $0.9$ each hour after a dose is administered. Find the maximum interval $N$ of hours between doses, and corresponding dose range $d$ (in mg/kg) for this $N$ that will enable use of the drug to be both safe and effective in the long term.

\[T\] 某种药物只有病人体内至少含 $1$ mg/kg 时才对一般病人有效，而只有在一般病人体内至多含 $2$ mg/kg 时才安全。设给药后病人体内药量每小时按 $0.9$ 的乘性因子减少。求该药物在长期内既能安全又有效时，给药间隔的最大小时数 $N$，以及与此 $N$ 对应的剂量范围 $d$（单位 mg/kg）。

129.

Suppose that $a_{n} \geq 0$ is a sequence of numbers. Explain why the sequence of partial sums of $a_{n}$ is increasing.

设 $a_{n} \geq 0$ 是一个数列。解释为什么 $a_{n}$ 的部分和的数列是递增的。

130\.

\[T\] Suppose that $a_{n}$ is a sequence of positive numbers and the sequence $S_{n}$ of partial sums of $a_{n}$ is bounded above. Explain why $\sum\limits_{n = 1}^{\infty}a_{n}$ converges. Does the conclusion remain true if we remove the hypothesis $a_{n} \geq 0\text{?}$

\[T\] 设 $a_{n}$ 是一个正项数列，且 $a_{n}$ 的部分和数列 $S_{n}$ 上有界。解释为什么 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。若去掉假设 $a_{n} \geq 0\text{?}$ 结论是否仍然成立？

131.

\[T\] Suppose that $a_{1} = S_{1} = 1$ and that, for given numbers $S > 1$ and $0 < k < 1,$ one defines $a_{n + 1} = k(S - S_{n})$ and $S_{n + 1} = a_{n + 1} + S_{n}.$ Does $S_{n}$ converge? If so, to what? (*Hint:* First argue that $S_{n} < S$ for all $n$ and $S_{n}$ is increasing.)

\[T\] 设 $a_{1} = S_{1} = 1$，且对给定的数 $S > 1$ 与 $0 < k < 1$，定义 $a_{n + 1} = k(S - S_{n})$ 与 $S_{n + 1} = a_{n + 1} + S_{n}.$ 数列 $S_{n}$ 是否收敛？若收敛，收敛到什么？（*提示：* 先论证对所有 $n$ 有 $S_{n} < S$ 且 $S_{n}$ 递增。）

132\.

\[T\] A version of von Bertalanffy growth can be used to estimate the age of an individual in a homogeneous species from its length if the annual increase in year $n + 1$ satisfies $a_{n + 1} = k(S - S_{n}),$ with $S_{n}$ as the length at year $n,$ $S$ as a limiting length, and $k$ as a relative growth constant. If $S_{1} = 3,$ $S = 9,$ and $k = 1\text{/}2,$ numerically estimate the smallest value of $n$ such that $S_{n} \geq 8.$ Note that $S_{n + 1} = S_{n} + a_{n + 1}.$ Find the corresponding $n$ when $k = 1\text{/}4.$

\[T\] von Bertalanffy 生长模型的一个版本可用于根据同种群个体的体长估计其年龄，若第 $n + 1$ 年的年增长满足 $a_{n + 1} = k(S - S_{n}),$ 其中 $S_{n}$ 为第 $n$ 年的体长，$S$ 为极限体长，$k$ 为相对生长常数。若 $S_{1} = 3,$ $S = 9,$ 且 $k = 1\text{/}2,$ 数值估计使 $S_{n} \geq 8$ 成立的最小 $n$ 值。注意 $S_{n + 1} = S_{n} + a_{n + 1}.$ 当 $k = 1\text{/}4$ 时求对应的 $n$。

133.

\[T\] Suppose that $\sum\limits_{n = 1}^{\infty}a_{n}$ is a convergent series of positive terms. Explain why $\underset{N\rightarrow\infty}{\text{lim}}{\sum\limits_{n = N + 1}^{\infty}{a_{n} = 0}}.$

\[T\] 设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个正项收敛级数。解释为什么 $\underset{N\rightarrow\infty}{\text{lim}}{\sum\limits_{n = N + 1}^{\infty}{a_{n} = 0}}.$

134\.

\[T\] Find the length of the dashed zig-zag path in the following figure.

\[T\] 求下图中虚线锯齿路径的长度。

135.

\[T\] Find the total length of the dashed path in the following figure.

\[T\] 求下图中虚线路径的总长度。

136\.

\[T\] The Sierpinski triangle is obtained from a triangle by deleting the middle fourth as indicated in the first step, by deleting the middle fourths of the remaining three congruent triangles in the second step, and in general deleting the middle fourths of the remaining triangles in each successive step. Assuming that the original triangle is shown in the figure, find the areas of the remaining parts of the original triangle after $N$ steps and find the total length of all of the boundary triangles after $N$ steps.

\[T\] Sierpinski 三角形是这样得到的：如第一步所示删去一个三角形中间的四分之一，第二步删去剩余三个全等三角形各自中间的四分之一，一般地每一步删去剩余三角形各自中间的四分之一。假设原三角形如图所示，求经过 $N$ 步后原三角形剩余部分的面积，以及经过 $N$ 步后所有边界三角形的总长度。

137.

\[T\] The Sierpinski gasket is obtained by dividing the unit square into nine equal sub-squares, removing the middle square, then doing the same at each stage to the remaining sub-squares. The figure shows the remaining set after four iterations. Compute the total area removed after $N$ stages, and compute the length the total perimeter of the remaining set after $N$ stages.

\[T\] Sierpinski 垫圈是这样得到的：将单位正方形分成九个相等的小正方形，去掉中间一个，然后在每一阶段对剩余的小正方形作同样处理。图中显示四次迭代后剩余的点集。计算经过 $N$ 阶段后移去的总面积，以及经过 $N$ 阶段后剩余点集的总周长。

5.3 The Divergence and Integral Tests 5.3 发散判别法与积分判别法

5.3.1 Use the divergence test to demonstrate that a series diverges.
5.3.2 Use the integral test to determine the convergence of a series.
5.3.3 Estimate the value of a series by finding bounds on its remainder term.

5.3.1 使用发散判别法证明一个级数发散。
5.3.2 使用积分判别法判断一个级数的收敛性。
5.3.3 通过对余项给出界来估计级数的值。

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums $\left\{ S_{k} \right\}.$ In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. In this section, we discuss two of these tests: the divergence test and the integral test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

在上一节中，我们通过显式计算部分和的数列 $\left\{ S_{k} \right\}$ 的极限确定了几个级数的收敛或发散。实际上，显式计算这个极限可能很困难甚至不可能。幸运的是，存在若干判别法，使我们能判断许多类型级数的收敛或发散。在本节中，我们讨论其中两个判别法：发散判别法与积分判别法。我们将在本章余下部分考察其他几个判别法，然后总结何时以及如何使用它们。

Divergence Test 发散判别法

A series $\sum\limits_{n = 1}^{\infty}a_{n}$ being convergent is equivalent to the convergence of the sequence of partial sums ($S_{k}$) as $k\rightarrow\infty.$

级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，等价于其部分和数列（$S_{k}$）当 $k\rightarrow\infty$ 时收敛。

To verify this, notice that from the algebraic properties,

要验证这一点，注意由代数性质可得，

$$\underset{k\rightarrow\infty}{\text{lim}}a_{k} = \underset{k\rightarrow\infty}{\text{lim}}(S_{k} - S_{k - 1}) = \underset{k\rightarrow\infty}{\text{lim}}S_{k} - \underset{k\rightarrow\infty}{\text{lim}}S_{k - 1} = S - S = 0.$$

Therefore, if $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, the $n\text{th}$ term $a_{n}\rightarrow 0$ as $n\rightarrow\infty.$ An important consequence of this fact is the following statement:

因此，若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，则第 $n$ 项满足 $a_{n}\rightarrow 0$（$n\rightarrow\infty$ 时）。这一事实的一个重要推论如下：

$$\text{If}\ a_{n}\nrightarrow 0\ \text{as}\ n\rightarrow\infty,{\sum\limits_{n = 1}^{\infty}a_{n}}\ \text{diverges}.$$ (5.8)

This test is known as the divergence test because it provides a way of proving that a series diverges.

该判别法称为发散判别法，因为它给出了一种证明级数发散的方法。

Divergence Test 发散判别法

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = c \neq 0$ or $\underset{n\rightarrow\infty}{\text{lim}}a_{n}$ does not exist, then the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = c \neq 0$，或者 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}$ 不存在，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

It is important to note that the converse of this theorem is not true. That is, if $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ we cannot make any conclusion about the convergence of ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ For example, $\underset{n\rightarrow\infty}{\text{lim}}(1\text{/}n) = 0,$ but the harmonic series $\sum\limits_{n = 1}^{\infty}1\text{/}n$ diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if $a_{n}\rightarrow 0,$ the divergence test is inconclusive.

重要的是，该定理的逆命题并不成立。也就是说，若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0$，则无法就 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 的收敛性得出任何结论。例如，$\underset{n\rightarrow\infty}{\text{lim}}(1\text{/}n) = 0$，但调和级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n$ 发散。在本节及本章其余各节中，我们会给出更多这类级数的例子。因此，虽然可以用发散判别法证明某级数发散，但不能用它证明级数收敛。具体而言，若 $a_{n}\rightarrow 0$，则发散判别法无法判定。

Using the divergence test 使用发散判别法

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

对于下列各级数，应用发散判别法。若发散判别法证明该级数发散，则指明这一点；否则说明发散判别法无法判定。

1. $\sum\limits_{n = 1}^{\infty}\frac{n}{3n - 1}$

2. $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}$

3. $\sum\limits_{n = 1}^{\infty}e^{1\text{/}n^{2}}$

Solution 解答

1. Since $n\text{/}\left( {3n - 1} \right)\rightarrow 1\text{/}3 \neq 0,$ by the divergence test, we can conclude that

1. 由于 $n\text{/}\left( {3n - 1} \right)\rightarrow 1\text{/}3 \neq 0$，根据发散判别法，可以断定

$$\sum\limits_{n = 1}^{\infty}\frac{n}{3n - 1}$$

diverges.

发散。

2. Since $1\text{/}n^{3}\rightarrow 0,$ the divergence test is inconclusive.

2. 由于 $1\text{/}n^{3}\rightarrow 0$，发散判别法无法判定。

3. Since $e^{1\text{/}n^{2}}\rightarrow 1 \neq 0,$ by the divergence test, the series

3. 由于 $e^{1\text{/}n^{2}}\rightarrow 1 \neq 0$，根据发散判别法，该级数

$$\sum\limits_{n = 1}^{\infty}e^{1\text{/}n^{2}}$$

diverges.

发散。

What does the divergence test tell us about the series ${\sum\limits_{n = 1}^{\infty}\text{cos}}(1\text{/}n^{2})\text{?}$

对于级数 ${\sum\limits_{n = 1}^{\infty}\text{cos}}(1\text{/}n^{2})\text{?}$，发散判别法能够说明什么？

Integral Test 积分判别法

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums $\left\{ S_{k} \right\}$ and showing that $S_{2^{k}} > 1 + k\text{/}2$ for all positive integers $k.$ In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

在上一节中，我们通过考察部分和数列 $\left\{ S_{k} \right\}$ 并证明对所有正整数 $k$ 都有 $S_{2^{k}} > 1 + k\text{/}2$，从而证明了调和级数发散。本节我们采用另一种方法证明调和级数发散。这一方法很重要，因为它可用于证明许多其他级数的发散或收敛。这个称为积分判别法的检验，把无穷和与反常积分进行比较。应当注意，此判别法仅适用于各项均为正的级数。

To illustrate how the integral test works, use the harmonic series as an example. In Figure 5.12, we depict the harmonic series by sketching a sequence of rectangles with areas $1,1\text{/}2,1\text{/}3,1\text{/}4\text{,…}$ along with the function $f(x) = 1\text{/}x.$ From the graph, we see that

为说明积分判别法的工作原理，以调和级数为例。在图 5.12 中，我们画出一列面积分别为 $1,1\text{/}2,1\text{/}3,1\text{/}4\text{,…}$ 的矩形，以及函数 $f(x) = 1\text{/}x$，以此来描绘调和级数。由图可见，

$${\sum\limits_{n = 1}^{k}\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{k} > {\int_{1}^{k + 1}{\frac{1}{x}dx.}}$$

Therefore, for each $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

因此，对每个 $k$，第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = {\sum\limits_{n = 1}^{k}\frac{1}{n}} > {\int_{1}^{k + 1}{\frac{1}{x}dx}} = \text{ln}\mspace{2mu} x\ |_{1}^{k + 1} = \text{ln}(k + 1) - \text{ln}(1) = \text{ln}(k + 1).$$

Since $\underset{k\rightarrow\infty}{\text{lim}}\text{ln}(k + 1) = \infty,$ we see that the sequence of partial sums $\left\{ S_{k} \right\}$ is unbounded. Therefore, $\left\{ S_{k} \right\}$ diverges, and, consequently, the series $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ also diverges.

由于 $\underset{k\rightarrow\infty}{\text{lim}}\text{ln}(k + 1) = \infty$，可见部分和数列 $\left\{ S_{k} \right\}$ 无界。因此 $\left\{ S_{k} \right\}$ 发散，进而级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 也发散。

Now consider the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}.$ We show how an integral can be used to prove that this series converges. In Figure 5.13, we sketch a sequence of rectangles with areas $1,1\text{/}2^{2},1\text{/}3^{2}\text{,…}$ along with the function $f(x) = 1\text{/}x^{2}.$ From the graph we see that

现在考虑级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$。我们将展示如何用一个积分来证明该级数收敛。在图 5.13 中，我们画出一列面积分别为 $1,1\text{/}2^{2},1\text{/}3^{2}\text{,…}$ 的矩形，以及函数 $f(x) = 1\text{/}x^{2}$。由图可见，

$${\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \text{⋯} + \frac{1}{k^{2}} < 1 + {\int_{1}^{k}{\frac{1}{x^{2}}dx}}.$$

Therefore, for each $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

因此，对每个 $k$，第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = {\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}} < 1 + {\int_{1}^{k}{\frac{1}{x^{2}}dx}} = 1 - \left. \frac{1}{x} \right|_{1}^{k} = 1 - \frac{1}{k} + 1 = 2 - \frac{1}{k} < 2.$$

We conclude that the sequence of partial sums $\left\{ S_{k} \right\}$ is bounded. We also see that $\left\{ S_{k} \right\}$ is an increasing sequence:

我们断定部分和数列 $\left\{ S_{k} \right\}$ 有界。同时可见 $\left\{ S_{k} \right\}$ 是递增数列：

$$S_{k} = S_{k - 1} + \frac{1}{k^{2}}\ \text{for}\ k \geq 2.$$

Since $\left\{ S_{k} \right\}$ is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ converges.

由于 $\left\{ S_{k} \right\}$ 递增且有界，根据单调收敛定理，它收敛。因此级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ 收敛。

We can extend this idea to prove convergence or divergence for many different series. Suppose $\sum\limits_{n = 1}^{\infty}a_{n}$ is a series with positive terms $a_{n}$ such that there exists a continuous, positive, decreasing function $f$ where $f(n) = a_{n}$ for all positive integers. Then, as in Figure 5.14(a), for any integer $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

我们可以将这一思想推广，用以证明许多不同级数的收敛或发散。设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项为正的级数，且存在连续、正、递减的函数 $f$，使得对所有正整数都有 $f(n) = a_{n}$。那么，如图 5.14(a) 所示，对任意整数 $k$，第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k} < a_{1} + {\int_{1}^{k}f}(x)dx < a_{1} + {\int_{1}^{\infty}f}(x)dx.$$

Therefore, if $\int_{1}^{\infty}{f(x)dx}$ converges, then the sequence of partial sums $\left\{ S_{k} \right\}$ is bounded. Since $\left\{ S_{k} \right\}$ is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if $\int_{1}^{\infty}{f(x)dx}$ converges, then the series $\sum\limits_{n = 1}^{\infty}a_{n}$ also converges. On the other hand, from Figure 5.14(b), for any integer $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

因此，若 $\int_{1}^{\infty}{f(x)dx}$ 收敛，则部分和数列 $\left\{ S_{k} \right\}$ 有界。由于 $\left\{ S_{k} \right\}$ 是递增数列，若它同时是有界数列，则根据单调收敛定理，它收敛。我们断定：若 $\int_{1}^{\infty}{f(x)dx}$ 收敛，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 也收敛。另一方面，由图 5.14(b) 可知，对任意整数 $k$，第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k} > {\int_{1}^{k + 1}f}(x)dx.$$

If $\underset{k\rightarrow\infty}{\text{lim}}{\int_{1}^{k + 1}f}(x)dx = \infty,$ then $\left\{ S_{k} \right\}$ is an unbounded sequence and therefore diverges. As a result, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ also diverges. We conclude that if ${\int_{1}^{\infty}f}(x)dx$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

若 $\underset{k\rightarrow\infty}{\text{lim}}{\int_{1}^{k + 1}f}(x)dx = \infty$，则 $\left\{ S_{k} \right\}$ 是无界数列，从而发散。于是级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 也发散。我们断定：若 ${\int_{1}^{\infty}f}(x)dx$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

Integral Test 积分判别法

Suppose $\sum\limits_{n = 1}^{\infty}a_{n}$ is a series with positive terms $a_{n}.$ Suppose there exists a function $f$ and a positive integer $N$ such that the following three conditions are satisfied:

设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项为正的级数。假设存在函数 $f$ 和正整数 $N$，使得下列三个条件成立：

1. $f$ is continuous,

1. $f$ 连续，

2. $f$ is decreasing, and

2. $f$ 递减，且

3. $f(n) = a_{n}$ for all integers $n \geq N.$

3. 对所有整数 $n \geq N$，都有 $f(n) = a_{n}$。

Then

那么

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{N}^{\infty}{f(x)dx}}$$

both converge or both diverge (see Figure 5.14).

两者同时收敛或同时发散（见图 5.14）。

Although convergence of $\int_{N}^{\infty}{f(x)dx}$ implies convergence of the related series ${\sum\limits_{n = 1}^{\infty}a_{n}},$ it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

虽然 $\int_{N}^{\infty}{f(x)dx}$ 收敛蕴含相关级数 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 收敛，但它并不表示积分与级数的数值相同。两者可能不同，而且往往是不同的。例如，

$${\sum\limits_{n = 1}^{\infty}\left( \frac{1}{e} \right)^{n}} = \frac{1}{e} + \left( \frac{1}{e} \right)^{2} + \left( \frac{1}{e} \right)^{3} + \text{⋯}$$

is a geometric series with initial term $a = 1\text{/}e$ and ratio $r = 1\text{/}e,$ which converges to

是一个首项 $a = 1\text{/}e$、公比 $r = 1\text{/}e$ 的几何级数，它收敛于

$$\frac{1\text{/}e}{1 - (1\text{/}e)} = \frac{1\text{/}e}{(e - 1)\text{/}e} = \frac{1}{e - 1}.$$

However, the related integral ${\int_{1}^{\infty}{(1\text{/}e)}^{x}}dx$ satisfies

然而，相关积分 ${\int_{1}^{\infty}{(1\text{/}e)}^{x}}dx$ 满足

$${\int_{1}^{\infty}\left( \frac{1}{e} \right)^{x}}dx = {\int_{1}^{\infty}e^{\text{−}x}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}e^{\text{−}x}}dx = \underset{b\rightarrow\infty}{\text{lim}} - e^{\text{−}x}|_{1}^{b} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {\text{−}e^{\text{−}b} + e^{-1}} \right\rbrack = \frac{1}{e}.$$

Using the Integral Test 使用积分判别法

For each of the following series, use the integral test to determine whether the series converges or diverges. Assume that all conditions for the integral test are met.

对于下列各级数，使用积分判别法判断其收敛或发散。假设积分判别法的所有条件均已满足。

1. ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{3}$

2. $\sum\limits_{n = 1}^{\infty}1\text{/}\sqrt{2n - 1}$

Solution 解答

1. Compare

1. 比较

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}}\mspace{9mu}\text{and}\ {\int_{1}^{\infty}\frac{1}{x^{3}}}dx.$$

We have

我们有

$${\int_{1}^{\infty}\frac{1}{x^{3}}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}{\frac{1}{x^{3}}dx}} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2x^{2}}|_{1}^{b}} \right\rbrack = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2b^{2}} + \frac{1}{2}} \right\rbrack = \frac{1}{2}.$$

Thus the integral ${\int_{1}^{\infty}1}\text{/}x^{3}dx$ converges, and therefore so does the series

于是积分 ${\int_{1}^{\infty}1}\text{/}x^{3}dx$ 收敛，因此级数也收敛

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}}.$$

2. Compare

2. 比较

$$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{2n - 1}}\mspace{9mu}\text{and}\ {\int_{1}^{\infty}{\frac{1}{\sqrt{2x - 1}}dx}}.$$

Since

由于

$$\begin{array}{cl} {\int_{1}^{\infty}{\frac{1}{\sqrt{2x - 1}}dx}} & {= \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}{\frac{1}{\sqrt{2x - 1}}dx = \underset{b\rightarrow\infty}{\text{lim}}\sqrt{2x - 1}|_{1}^{b}}}} \\ & {= \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {\sqrt{2b - 1} - 1} \right\rbrack = \infty,} \end{array}$$

the integral $\int_{1}^{\infty}{1\text{/}\sqrt{2x - 1}\mspace{2mu} dx}$ diverges, and therefore

积分 $\int_{1}^{\infty}{1\text{/}\sqrt{2x - 1}\mspace{2mu} dx}$ 发散，因此

$$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{2n - 1}}$$

diverges.

发散。

Use the integral test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{n}{3n^{2} + 1}$ converges or diverges.

使用积分判别法判断级数 $\sum\limits_{n = 1}^{\infty}\frac{n}{3n^{2} + 1}$ 收敛还是发散。

The p-Series p 级数

The harmonic series $\sum\limits_{n = 1}^{\infty}1\text{/}n$ and the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}$ are both examples of a type of series called a *p*-series.

调和级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n$ 与级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}$ 都是一类称为 *p* 级数的级数的例子。

For any real number $p,$ the series

对任意实数 $p$，级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}$$

is called a *p*-series.

称为 *p* 级数。

We know the *p*-series converges if $p = 2$ and diverges if $p = 1.$ What about other values of $p\text{?}$ In general, it is difficult, if not impossible, to compute the exact value of most $p$-series. However, we can use the tests presented thus far to prove whether a $p$-series converges or diverges.

我们知道，当 $p = 2$ 时 *p* 级数收敛，当 $p = 1$ 时发散。其他 $p$ 的取值又如何？一般而言，要计算大多数 $p$ 级数的精确值是困难的，甚至不可能。不过，我们可以利用目前给出的判别法来证明 *p* 级数收敛还是发散。

If $p < 0,$ then $1\text{/}n^{p}\rightarrow\infty,$ and if $p = 0,$ then $1\text{/}n^{p}\rightarrow 1.$ Therefore, by the divergence test,

若 $p < 0$，则 $1\text{/}n^{p}\rightarrow\infty$；若 $p = 0$，则 $1\text{/}n^{p}\rightarrow 1$。因此，根据发散判别法，

$$\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}\ \text{diverges if}\ p \leq 0.$$

If $p > 0,$ then $f(x) = 1\text{/}x^{p}$ is a positive, continuous, decreasing function. Therefore, for $p > 0,$ we use the integral test, comparing

若 $p > 0$，则 $f(x) = 1\text{/}x^{p}$ 是一个正、连续、递减的函数。因此，对 $p > 0$，我们使用积分判别法，比较

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}\ \text{and}\ {\int_{1}^{\infty}{\frac{1}{x^{p}}dx}}.$$

We have already considered the case when $p = 1.$ Here we consider the case when $p > 0,p \neq 1.$ For this case,

我们已经考察过 $p = 1$ 的情形。这里考虑 $p > 0$ 且 $p \neq 1$ 的情形。此时，

$$\int_{1}^{\infty}{\frac{1}{x^{p}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}{\frac{1}{x^{p}}dx = \underset{b\rightarrow\infty}{\text{lim}}\frac{1}{1 - p}x^{1 - p}|_{1}^{b} = \underset{b\rightarrow\infty}{\text{lim}}\frac{1}{1 - p}\left\lbrack {b^{1 - p} - 1} \right\rbrack.}}}$$

Because

因为

$$b^{1 - p}\rightarrow 0\ \text{if}\ p > 1\ \text{and}\ b^{1 - p}\rightarrow\infty\ \text{if}\ p < 1,$$

we conclude that

我们断定

$${\int_{1}^{\infty}{\frac{1}{x^{p}}dx}} = \left\{ {\begin{matrix} {\frac{1}{p - 1}\ \text{if}\ p > 1} \\ {\infty\ \text{if}\ p \leq 1} \end{matrix}.} \right.$$

Therefore, $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ converges if $p > 1$ and diverges if $0 < p < 1.$

因此，当 $p > 1$ 时 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 收敛，当 $0 < p < 1$ 时发散。

In summary,

综上，

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}}\mspace{2mu}\left\{ \begin{array}{l} {\text{converges if}\ p > 1} \\ {\text{diverges if}\ p \leq 1} \end{array} \right..$$ (5.9)

Testing for Convergence of p-series p 级数收敛性的判别

For each of the following series, determine whether it converges or diverges.

对于下列各级数，判断其收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{4}}$

2. $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2\text{/}3}}$

Solution 解答

1. This is a *p*-series with $p = 4 > 1,$ so the series converges.

1. 这是一个 *p* 级数，其中 $p = 4 > 1$，故该级数收敛。

2. Since $p = 2\text{/}3 < 1,$ the series diverges.

2. 由于 $p = 2\text{/}3 < 1$，该级数发散。

Does the series $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{5\text{/}4}}$ converge or diverge?

级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{5\text{/}4}}$ 收敛还是发散？

Estimating the Value of a Series 估计级数的和

Suppose we know that a series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum $\sum\limits_{n = 1}^{N}a_{n}$ where $N$ is any positive integer. The question we address here is, for a convergent series $\sum\limits_{n = 1}^{\infty}a_{n},$ how good is the approximation $\sum\limits_{n = 1}^{N}a_{n}\text{?}$ More specifically, if we let

假设我们知道级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，并且想估计该级数的和。我们当然可以用任意有限和 $\sum\limits_{n = 1}^{N}a_{n}$（其中 $N$ 是任意正整数）来近似该和。这里我们讨论的问题是：对于一个收敛的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 近似 $\sum\limits_{n = 1}^{N}a_{n}\text{?}$ 的精度有多高？更具体地说，如果令

$$R_{N} = {\sum\limits_{n = 1}^{\infty}a_{n}} - {\sum\limits_{n = 1}^{N}a_{n}}$$

be the remainder when the sum of an infinite series is approximated by the $N\text{th}$ partial sum, how large is $R_{N}\text{?}$ For some types of series, we are able to use the ideas from the integral test to estimate $R_{N}.$

为无穷级数之和用第 $N\text{th}$ 个部分和近似时的余项，那么 $R_{N}\text{?}$ 有多大？对于某些类型的级数，我们能够运用积分判别法的思想来估计 $R_{N}.$

Remainder Estimate from the Integral Test 由积分判别法得到的余项估计

Suppose $\sum\limits_{n = 1}^{\infty}a_{n}$ is a convergent series with positive terms. Suppose there exists a function $f$ satisfying the following three conditions:

假设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项为正的收敛级数。假设存在函数 $f$ 满足以下三个条件：

1. $f$ is continuous,

1. $f$ 是连续的，

2. $f$ is decreasing, and

2. $f$ 是递减的，且

3. $f(n) = a_{n}$ for all integers $n \geq 1.$

3. 对所有整数 $n \geq 1,$ 有 $f(n) = a_{n}$。

Let $S_{N}$ be the *N*th partial sum of $\sum\limits_{n = 1}^{\infty}a_{n}.$ For all positive integers $N,$

令 $S_{N}$ 为 $\sum\limits_{n = 1}^{\infty}a_{n}$ 的第 *N* 个部分和。对所有正整数 $N,$

$$S_{N} + {\int_{N + 1}^{\infty}{f(x)dx < \sum\limits_{n = 1}^{\infty}a_{n} < S_{N} + {\int_{N}^{\infty}{f(x)dx}}}}.$$

In other words, the remainder $R_{N} = \sum\limits_{n = 1}^{\infty}a_{n} - S_{N} = \sum\limits_{n = N + 1}^{\infty}a_{n}$ satisfies the following estimate:

换句话说，余项 $R_{N} = \sum\limits_{n = 1}^{\infty}a_{n} - S_{N} = \sum\limits_{n = N + 1}^{\infty}a_{n}$ 满足以下估计：

$${\int_{N + 1}^{\infty}{f(x)dx}} < R_{N} < {\int_{N}^{\infty}{f(x)dx.}} $$ (5.10)

This is known as the remainder estimate.

这被称为余项估计。

We illustrate Remainder Estimate from the Integral Test in Figure 5.15. In particular, by representing the remainder $R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯}$ as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by $\int_{N}^{\infty}{f(x)dx}$ and bounded below by ${\int_{N + 1}^{\infty}{f(x)dx}}.$ In other words,

我们在图 5.15 中说明由积分判别法得到的余项估计。具体来说，把余项 $R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯}$ 表示为若干矩形面积之和，可见这些矩形的面积以上界 $\int_{N}^{\infty}{f(x)dx}$、下界 ${\int_{N + 1}^{\infty}{f(x)dx}}$ 为界。换言之，

$$R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯} > {\int_{N + 1}^{\infty}{f(x)dx}}$$

and

且

$$R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯} < {\int_{N}^{\infty}{f(x)dx}}.$$

We conclude that

我们由此得到

$${\int_{N + 1}^{\infty}{f(x)dx < R_{N} < {\int_{N}^{\infty}{f(x)dx}}}}.$$

Since

由于

$$\sum\limits_{n = 1}^{\infty}a_{n} = S_{N} + R_{N},$$

where $S_{N}$ is the $N\text{th}$ partial sum, we conclude that

其中 $S_{N}$ 是第 $N\text{th}$ 个部分和，我们由此得到

$$S_{N} + {\int_{N + 1}^{\infty}{f(x)dx < \sum\limits_{n = 1}^{\infty}a_{n} < S_{N} + {\int_{N}^{\infty}{f(x)dx}}}}.$$

Estimating the Value of a Series 估计级数的和

Consider the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{3}.$

考虑级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{3}.$

1. Calculate $S_{10} = \sum\limits_{n = 1}^{10}1\text{/}n^{3}$ and estimate the error.

1. 计算 $S_{10} = \sum\limits_{n = 1}^{10}1\text{/}n^{3}$ 并估计误差。

2. Determine the least value of $N$ necessary such that $S_{N}$ will estimate $\sum\limits_{n = 1}^{\infty}1\text{/}n^{3}$ to within $0.001.$

2. 确定所需的最小 $N$ 值，使得 $S_{N}$ 对 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{3}$ 的估计误差不超过 $0.001$。

Solution 解答

1. Using a calculating utility, we have

1. 使用计算工具，我们得到

$$S_{10} = 1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}} + \text{⋯} + \frac{1}{10^{3}} \approx 1.19753.$$

$$S_{10} = 1 + \frac{1}{2^{3}} + \text{⋯} + \frac{1}{10^{3}} \approx 1.19753.$$

By the remainder estimate, we know

根据余项估计，我们知道

$$R_{N} < {\int_{N}^{\infty}{\frac{1}{x^{3}}dx}}.$$

We have

我们有

$$\int_{10}^{\infty}{\frac{1}{x^{3}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{10}^{b}{\frac{1}{x^{3}}dx = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2x^{2}}}\rbrack_{N}^{b} \right.} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2b^{2}} + \frac{1}{2N^{2}}} \right\rbrack = \frac{1}{2N^{2}}.}}$$

Therefore, the error is $R_{10} < 1\text{/}2(10)^{2} = 0.005.$

因此，误差为 $R_{10} < 1\text{/}2(10)^{2} = 0.005$。

2. Find $N$ such that $R_{N} < 0.001.$ In part a. we showed that $R_{N} < 1\text{/}\left( {2N^{2}} \right).$ Therefore, the remainder $R_{N} < 0.001$ as long as $1\text{/}\left( {2N^{2}} \right) < 0.001.$ That is, we need $2N^{2} > 1000.$ Solving this inequality for $N,$ we see that we need $N > 22.36.$ To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is $N = 23.$

2. 求使得 $R_{N} < 0.001$ 的 $N$。在 (a) 部分中我们已证 $R_{N} < 1\text{/}\left( {2N^{2}} \right)$。因此，只要 $1\text{/}\left( {2N^{2}} \right) < 0.001$，就有余项 $R_{N} < 0.001$。也就是说，我们需要 $2N^{2} > 1000$。对该不等式解 $N$，得到 $N > 22.36$。为了保证余项落在所需范围内，需要向上取整到最近的整数。因此，所需的最小值为 $N = 23$。

For $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{4}},$ calculate $S_{5}$ and estimate the error $R_{5}.$

对于 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{4}},$ 计算 $S_{5}$ 并估计误差 $R_{5}$。

Section 5.3 Exercises 5.3 节习题

For each of the following series, if the divergence test applies, either state that $\underset{n\rightarrow\infty}{\text{lim}}a_{n}$ does not exist or find $\underset{n\rightarrow\infty}{\text{lim}}a_{n}.$ If the divergence test does not apply, state why.

对于下列各个级数，若发散判别法适用，则指出 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}$ 不存在或求出 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}$。若发散判别法不适用，说明原因。

138\.

$a_{n} = \frac{n}{n + 2}$

139.

$a_{n} = \frac{n}{5n^{2} - 3}$

140\.

$a_{n} = \frac{n}{\sqrt{3n^{2} + 2n + 1}}$

141.

$a_{n} = \frac{\left( {2n + 1} \right)\left( {n - 1} \right)}{\left( {n + 1} \right)^{2}}$

142\.

$a_{n} = \frac{\left( {2n + 1} \right)^{2n}}{\left( {3n^{2} + 1} \right)^{n}}$

143.

$a_{n} = \frac{2^{n}}{3^{n\text{/}2}}$

144\.

$a_{n} = \frac{2^{n} + 3^{n}}{10^{n\text{/}2}}$

145.

$a_{n} = e^{-2\text{/}n}$

146\.

$a_{n} = \text{cos}\mspace{2mu} n$

147.

$a_{n} = \text{tan}\mspace{2mu} n$

148\.

$a_{n} = \frac{1 - \text{cos}^{2}\left( {1\text{/}n} \right)}{\text{sin}^{2}\left( {2\text{/}n} \right)}$

149.

$a_{n} = \left( {1 - \frac{1}{n}} \right)^{2n}$

150\.

$a_{n} = \frac{\text{ln}\mspace{2mu} n}{n}$

151.

$a_{n} = \frac{\left( {\text{ln}\mspace{2mu} n} \right)^{2}}{\sqrt{n}}$

State whether the given $p$-series converges.

判断所给的 $p$ 级数是否收敛。

152\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$

153.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n\sqrt{n}}$

154\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt[3]{n^{2}}}$

155.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt[3]{n^{4}}}$

156\.

$\sum\limits_{n = 1}^{\infty}\frac{n^{e}}{n^{\pi}}$

157.

$\sum\limits_{n = 1}^{\infty}\frac{n^{\pi}}{n^{2e}}$

Use the integral test to determine whether the following sums converge.

用积分判别法判断下列各级数是否收敛。

158\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n + 5}}$

159.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt[3]{n + 5}}$

160\.

$\sum\limits_{n = 2}^{\infty}\frac{1}{n\mspace{2mu}\text{ln}\mspace{2mu} n}$

161.

$\sum\limits_{n = 1}^{\infty}\frac{n}{1 + n^{2}}$

162\.

$\sum\limits_{n = 1}^{\infty}\frac{e^{n}}{1 + e^{2n}}$

163.

$\sum\limits_{n = 1}^{\infty}\frac{2n}{1 + n^{4}}$

164\.

$\sum\limits_{n = 2}^{\infty}\frac{1}{n\mspace{2mu}\text{ln}^{2}n}$

Express the following sums as $p$-series and determine whether each converges.

将下列各级数表示为 $p$ 级数，并判断每个是否收敛。

165.

$\sum\limits_{n = 1}^{\infty}2^{\text{−}\text{ln}\mspace{2mu} n}$ (*Hint:* $2^{\text{−}\text{ln}\mspace{2mu} n} = 1\text{/}n^{\text{ln}\mspace{2mu} 2}$.)

166\.

$\sum\limits_{n = 1}^{\infty}3^{\text{−}\text{ln}\mspace{2mu} n}$ (*Hint:* $3^{\text{−}\text{ln}\mspace{2mu} n} = 1\text{/}n^{\text{ln}\mspace{2mu} 3}$.)

167.

$\sum\limits_{n = 1}^{\infty}{n2^{-2\mspace{2mu}\text{ln}\mspace{2mu} n}}$

168\.

$\sum\limits_{n = 1}^{\infty}{n3^{-2\mspace{2mu}\text{ln}\mspace{2mu} n}}$

Use the estimate $R_{N} \leq {\int_{N}^{\infty}f}(t)dt$ to find a bound for the remainder $R_{N} = {\sum\limits_{n = 1}^{\infty}a_{n}} - {\sum\limits_{n = 1}^{N}a_{n}}$ where $a_{n} = f(n).$

利用估计 $R_{N} \leq {\int_{N}^{\infty}f}(t)dt$ 为余项 $R_{N} = {\sum\limits_{n = 1}^{\infty}a_{n}} - {\sum\limits_{n = 1}^{N}a_{n}}$（其中 $a_{n} = f(n)$）找一个上界。

169.

$\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}}$

170\.

$\sum\limits_{n = 1}^{1000}\frac{1}{n^{3}}$

171.

$\sum\limits_{n = 1}^{1000}\frac{1}{1 + n^{2}}$

172\.

${\sum\limits_{n = 1}^{100}n}\text{/}2^{n}$

\[T\] Find the minimum value of $N$ such that the remainder estimate ${\int_{N + 1}^{\infty}{f < R_{N} <}}{\int_{N}^{\infty}f}$ guarantees that $\sum\limits_{n = 1}^{N}a_{n}$ estimates ${\sum\limits_{n = 1}^{\infty}a_{n}},$ accurate to within the given error.

\[T\] 求 $N$ 的最小值，使得余项估计 ${\int_{N + 1}^{\infty}{f < R_{N} <}}{\int_{N}^{\infty}f}$ 保证 $\sum\limits_{n = 1}^{N}a_{n}$ 对 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 的估计精确到给定误差以内。

173.

$a_{n} = \frac{1}{n^{2}},$ error $< 10^{-4}$

174\.

$a_{n} = \frac{1}{n^{1.1}},$ error $< 10^{-4}$

175.

$a_{n} = \frac{1}{n^{1.01}},$ error $< 10^{-4}$

176\.

$a_{n} = \frac{1}{n\mspace{2mu}\text{ln}^{2}n},$ error $< 10^{-3}$

177.

$a_{n} = \frac{1}{1 + n^{2}},$ error $< 10^{-3}$

In the following exercises, find a value of $N$ such that $R_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum\limits_{n = 1}^{N}a_{n}$ and compare it to the given estimate of the infinite series.

在以下习题中，求一个 $N$ 值，使得 $R_{N}$ 小于所需误差。计算相应的和 $\sum\limits_{n = 1}^{N}a_{n}$，并将其与所给无穷级数的估计值进行比较。

178\.

$a_{n} = \frac{1}{n^{11}},$ error $< 10^{-4},$ ${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{11}}} = 1.000494\text{…}$

179.

$a_{n} = \frac{1}{e^{n}},$ error $< 10^{-5},$ ${\sum\limits_{n = 1}^{\infty}\frac{1}{e^{n}}} = \frac{1}{e - 1} = 0.581976\text{…}$

180\.

$a_{n} = \frac{n}{e^{n^{2}}},$ error $< 10^{-5},$ ${\sum\limits_{n = 1}^{\infty}{n\text{/}e^{n^{2}}}} = 0.40488139857\text{…}$

181.

$a_{n} = 1\text{/}n^{4},$ error $< 10^{-4},$ ${\sum\limits_{n = 1}^{\infty}{1\text{/}n^{4}}} = \pi^{4}\text{/}90 = 1.08232...$

182\.

$a_{n} = 1\text{/}n^{6},$ error $< 10^{-6},$ ${\sum\limits_{n = 1}^{\infty}{1\text{/}n^{6}}} = \pi^{6}\text{/}945 = 1.01734306...,$

183.

Find the limit as $n\rightarrow\infty$ of $\frac{1}{n} + \frac{1}{n + 1} + \text{⋯} + \frac{1}{2n}.$ (*Hint:* Compare to ${\int_{n}^{2n}\frac{1}{t}}dt.\text{)}$

求当 $n\rightarrow\infty$ 时 $\frac{1}{n} + \frac{1}{n + 1} + \text{⋯} + \frac{1}{2n}$ 的极限。（*提示：* 与 ${\int_{n}^{2n}\frac{1}{t}}dt$ 比较。）

184\.

Find the limit as $n\rightarrow\infty$ of $\frac{1}{n} + \frac{1}{n + 1} + \text{⋯} + \frac{1}{3n}$

求当 $n\rightarrow\infty$ 时 $\frac{1}{n} + \frac{1}{n + 1} + \text{⋯} + \frac{1}{3n}$ 的极限

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

接下来的几道习题旨在让人感受调和级数部分和出现的若干应用。

185.

In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number $H_{k} = \left( {1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{k}} \right).$ Recall that $T_{k} = H_{k} - \text{ln}\mspace{2mu} k$ is decreasing. Compute $T = \underset{k\rightarrow\infty}{\text{lim}}T_{k}$ to four decimal places. (*Hint:* $\frac{1}{k + 1} < {\int_{k}^{k + 1}\frac{1}{x}}dx$.)

在概率论的某些应用（例如群体遗传学中用于预测突变率的所谓 Watterson 估计量）中，精确估计数 $H_{k} = \left( {1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{k}} \right)$ 十分重要。回想 $T_{k} = H_{k} - \text{ln}\mspace{2mu} k$ 是递减的。计算 $T = \underset{k\rightarrow\infty}{\text{lim}}T_{k}$，保留四位小数。（*提示：* $\frac{1}{k + 1} < {\int_{k}^{k + 1}\frac{1}{x}}dx$。）

186\.

\[T\] Complete sampling with replacement, sometimes called the *coupon collector’s problem*, is phrased as follows: Suppose you have $N$ unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps $E(N)$ that it takes to draw each unique item at least once. It turns out that $E(N) = N$. $H_{N} = N\left( {1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{N}} \right)$. Find $E(N)$ for $N = 10,20,\text{and}\ 50$.

\[T\] 有放回的完全抽样，有时称为* coupon collector’s problem（集券问题）*，表述如下：假设一个箱子里有 $N$ 件互不相同的物品。每一步随机取出一件，记下其种类，再放回箱中。问题问的是：要使每件物品都至少被抽到一次，所需的期望步数 $E(N)$ 是多少。结果是 $E(N) = N$。$H_{N} = N\left( {1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{N}} \right)$。求 $N = 10,20,\text{and}\ 50$ 时的 $E(N)$。

187.

\[T\] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has $n$ cards, then the probability that the insertion will be below the card initially at the bottom (call this card $B)$ is $1\text{/}n.$ Thus the expected number of top random insertions before $B$ is no longer at the bottom is *n*. Once one card is below $B,$ there are two places below $B$ and the probability that a randomly inserted card will fall below $B$ is $2\text{/}n.$ The expected number of top random insertions before this happens is $n\text{/}2.$ The two cards below $B$ are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

\[T\] 洗牌最简单的办法是取出顶牌，把它随机插到牌堆中的某个位置，称为顶部随机插入，然后重复。我们认为：当足够多次顶部随机插入后，原先在底部的牌已到达顶部并经历了一次随机插入，这副牌就算被随机洗过了。如果牌堆有 $n$ 张牌，那么插入位置落在最初位于底部的那张牌（称此牌为 $B$）之下的概率是 $1\text{/}n$。因此，在 $B$ 不再位于底部之前，顶部随机插入的期望次数是 *n*。一旦有一张牌位于 $B$ 之下，那么 $B$ 之下有两个位置，随机插入的牌落在 $B$ 之下的概率是 $2\text{/}n$。在此发生之前，顶部随机插入的期望次数是 $n\text{/}2$。这时位于 $B$ 之下的两张牌处于随机顺序。依此类推，求出为使整副牌被视为随机洗过所需的顶部随机插入期望次数的公式。

188\.

Suppose a scooter can travel $100$ km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel $100H_{N}$ km, where $H_{N} = 1 + 1\text{/}2 + \text{⋯} + 1\text{/}N.$

假设一辆踏板摩托车加满油能行驶 $100$ km。假定燃油可以在摩托车之间转移，但只能装在油箱里携带，请给出一种方案，使其中一辆摩托车能行驶 $100H_{N}$ km，其中 $H_{N} = 1 + 1\text{/}2 + \text{⋯} + 1\text{/}N$。

189.

Show that for the remainder estimate to apply on $\lbrack N,\infty)$ it is sufficient that $f(x)$ be decreasing on $\lbrack N,\infty),$ but $f$ need not be decreasing on $\lbrack 1,\infty).$

证明：要使余项估计在 $\lbrack N,\infty)$ 上成立，只需 $f(x)$ 在 $\lbrack N,\infty)$ 上递减即可，而 $f$ 不必在 $\lbrack 1,\infty)$ 上递减。

190\.

\[T\] Use the remainder estimate and integration by parts to approximate $\sum\limits_{n = 1}^{\infty}{n\text{/}e^{n}}$ within an error smaller than $0.0001.$

\[T\] 利用余项估计和分部积分，在误差小于 $0.0001$ 的范围内近似 $\sum\limits_{n = 1}^{\infty}{n\text{/}e^{n}}$。

191.

Does $\sum\limits_{n = 2}^{\infty}\frac{1}{n\left( {\text{ln}\mspace{2mu} n} \right)^{p}}$ converge if $p$ is large enough? If so, for which $p\text{?}$

若 $p$ 足够大，$\sum\limits_{n = 2}^{\infty}\frac{1}{n\left( {\text{ln}\mspace{2mu} n} \right)^{p}}$ 是否收敛？若收敛，是对哪些 $p$ 值？

192\.

\[T\] Suppose a computer can sum one million terms per second of the divergent series ${\sum\limits_{n = 1}^{N}\frac{1}{n}}.$ Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100.$

\[T\] 假设一台计算机每秒能给发散级数 ${\sum\limits_{n = 1}^{N}\frac{1}{n}}$ 求和一百万项。用积分判别法近似计算：要使部分和超过 $100$，需要累加多少秒。

193.

\[T\] A fast computer can sum one million terms per second of the divergent series ${\sum\limits_{n = 2}^{N}\frac{1}{n\mspace{2mu}\text{ln}\mspace{2mu} n}}.$ Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100.$

\[T\] 一台快速的计算机每秒能给发散级数 ${\sum\limits_{n = 2}^{N}\frac{1}{n\mspace{2mu}\text{ln}\mspace{2mu} n}}$ 求和一百万项。用积分判别法近似计算：要使部分和超过 $100$，需要累加多少秒。

5.4 Comparison Tests 5.4 比较判别法

5.4.1 Use the comparison test to test a series for convergence.
5.4.2 Use the limit comparison test to determine convergence of a series.

5.4.1 使用比较判别法检验级数的收敛性。
5.4.2 使用极限比较判别法判断级数的收敛性。

We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. In this section, we show how to use comparison tests to determine the convergence or divergence of a series by comparing it to a series whose convergence or divergence is known. Typically these tests are used to determine convergence of series that are similar to geometric series or *p*-series.

我们已经看到，积分判别法通过将级数与相关的广义积分比较，来判断级数的收敛或发散。本节我们说明如何使用比较判别法，把待判断的级数与一个已知收敛或发散的级数比较，从而确定其收敛或发散。通常这些判别法用于判断与几何级数或 *p* 级数相似的级数的收敛性。

Comparison Test 比较判别法

In the preceding two sections, we discussed two large classes of series: geometric series and *p*-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.

在前两节中，我们讨论了两类重要的级数：几何级数和 p 级数。我们确切地知道这些级数何时收敛、何时发散。这里我们将说明如何利用这些级数的收敛性或发散性，通过一种称为比较判别法的方法，来证明其他级数的收敛性或发散性。

For example, consider the series

例如，考虑级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} + 1}.$$

This series looks similar to the convergent series

这个级数看起来与收敛级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}.$$

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

由于每个级数的项都是正数，每个级数的部分和数列都是单调递增数列。此外，由于

$$0 < \frac{1}{n^{2} + 1} < \frac{1}{n^{2}}$$

for all positive integers $n,$ the $k\text{th}$ partial sum $S_{k}$ of $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} + 1}$ satisfies

对所有正整数 $n,$ 级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} + 1}$ 的第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = \sum\limits_{n = 1}^{k}\frac{1}{n^{2} + 1} < \sum\limits_{n = 1}^{k}\frac{1}{n^{2}} < \sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}.$$

(See Figure 5.16(a) and Table 5.1.) Since the series on the right converges, the sequence $\left\{ S_{k} \right\}$ is bounded above. We conclude that $\left\{ S_{k} \right\}$ is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, $\left\{ S_{k} \right\}$ converges, and thus

（见图 5.16(a) 和表 5.1。）由于右侧的级数收敛，数列 $\left\{ S_{k} \right\}$ 上有界。我们得出结论：$\left\{ S_{k} \right\}$ 是一个上有界的单调递增数列。因此，由单调收敛定理，$\left\{ S_{k} \right\}$ 收敛，从而

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} + 1}$$

converges.

收敛。

Similarly, consider the series

类似地，考虑级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n - 1\text{/}2}.$$

This series looks similar to the divergent series

这个级数看起来与发散级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n}.$$

The sequence of partial sums for each series is monotone increasing and

每个级数的部分和数列都是单调递增的，并且

$$\frac{1}{n - 1\text{/}2} > \frac{1}{n} > 0$$

for every positive integer $n.$ Therefore, the $k\text{th}$ partial sum $S_{k}$ of $\sum\limits_{n = 1}^{\infty}\frac{1}{n - 1\text{/}2}$ satisfies

对每一个正整数 $n.$ 因此，级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n - 1\text{/}2}$ 的第 $k$ 个部分和 $S_{k}$ 满足

$$S_{k} = \sum\limits_{n = 1}^{k}\frac{1}{n - 1\text{/}2} > \sum\limits_{n = 1}^{k}\frac{1}{n}.$$

(See Figure 5.16(b) and Table 5.2.) Since the series $\sum\limits_{n = 1}^{\infty}1\text{/}n\ $ diverges to infinity, the sequence of partial sums $\sum\limits_{n = 1}^{k}1\text{/}n$ is unbounded. Consequently, $\left\{ S_{k} \right\}$ is an unbounded sequence, and therefore diverges. We conclude that

（见图 5.16(b) 和表 5.2。）由于级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n\ $ 发散到无穷大，部分和数列 $\sum\limits_{n = 1}^{k}1\text{/}n$ 无界。因此，$\left\{ S_{k} \right\}$ 是无界数列，从而发散。我们得出结论：

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n - 1\text{/}2}\ $$

diverges.

发散。

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\sum\limits_{n = 1}^{k}\frac{1}{n^{2} + 1}$	$0.5$	$0.7$	$0.8$	$0.8588$	$0.8973$	$0.9243$	$0.9443$	$0.9597$
$\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}$	$1$	$\begin{array}{l} 1.25 \end{array}$	$1.3611$	$1.4236$	$1.4636$	$1.4914$	$1.5118$	$1.5274$

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\sum\limits_{n = 1}^{k}\frac{1}{n^{2} + 1}$	$0.5$	$0.7$	$0.8$	$0.8588$	$0.8973$	$0.9243$	$0.9443$	$0.9597$
$\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}$	$1$	$\begin{array}{l} 1.25 \end{array}$	$1.3611$	$1.4236$	$1.4636$	$1.4914$	$1.5118$	$1.5274$

Table 5.1 Comparing a series with a *p*-series (*p* = 2)

表 5.1 将某级数与 p 级数（p = 2）进行比较

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\sum\limits_{n = 1}^{k}\frac{1}{n - 1\text{/}2}$	$2$	$2.6667$	$3.0667$	$3.3524$	$3.5746$	$3.7564$	$3.9103$	$4.0436$
$\sum\limits_{n = 1}^{k}\frac{1}{n}$	$1$	$1.5$	$\begin{array}{l} 1.8333 \end{array}$	$2.0933$	$2.2833$	$2.45$	$2.5929$	$2.7179$

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\sum\limits_{n = 1}^{k}\frac{1}{n - 1\text{/}2}$	$2$	$2.6667$	$3.0667$	$3.3524$	$3.5746$	$3.7564$	$3.9103$	$4.0436$
$\sum\limits_{n = 1}^{k}\frac{1}{n}$	$1$	$1.5$	$\begin{array}{l} 1.8333 \end{array}$	$2.0933$	$2.2833$	$2.45$	$2.5929$	$2.7179$

Table 5.2 Comparing a series with the harmonic series

表 5.2 将某级数与调和级数进行比较

Comparison Test 比较判别法

1. Suppose there exists an integer $N$ such that $0 \leq a_{n} \leq b_{n}$ for all $n \geq N.$ If $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

1. 假设存在整数 $N$，使得对所有 $n \geq N$ 都有 $0 \leq a_{n} \leq b_{n}$。如果 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

2. Suppose there exists an integer $N$ such that $a_{n} \geq b_{n} \geq 0$ for all $n \geq N.$ If $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

2. 假设存在整数 $N$，使得对所有 $n \geq N$ 都有 $a_{n} \geq b_{n} \geq 0$。如果 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

Proof 证明

We prove part i. The proof of part ii. is the contrapositive of part i. Let $\left\{ S_{k} \right\}$ be the sequence of partial sums associated with $\sum\limits_{n = 1}^{\infty}a_{n},$ and let $L = \sum\limits_{n = 1}^{\infty}b_{n}.$ Since the terms $a_{n} \geq 0,$

我们证明第 i 部分。第 ii 部分的证明是第 i 部分的逆否命题。令 $\left\{ S_{k} \right\}$ 为与 $\sum\limits_{n = 1}^{\infty}a_{n}$ 关联的部分和数列，并令 $L = \sum\limits_{n = 1}^{\infty}b_{n}$。由于各项 $a_{n} \geq 0,$

$$S_{k} = a_{1} + a_{2} + \text{⋯} + a_{k} \leq a_{1} + a_{2} + \text{⋯} + a_{k} + a_{k + 1} = S_{k + 1}.$$

Therefore, the sequence of partial sums is increasing. Further, since $a_{n} \leq b_{n}$ for all $n \geq N,$ then

因此，部分和数列是递增的。此外，由于对所有 $n \geq N$ 都有 $a_{n} \leq b_{n}$，则

$${\sum\limits_{n = N}^{k}{a_{n} \leq}}{\sum\limits_{n = N}^{k}{b_{n} \leq}}{\sum\limits_{n = 1}^{\infty}{b_{n} = L.}}$$

Therefore, for all $k \geq 1,$

因此，对所有 $k \geq 1,$

$$S_{k} = \left( {a_{1} + a_{2} + \text{⋯} + a_{N - 1}} \right) + {\sum\limits_{n = N}^{k}{a_{n} \leq}}\left( {a_{1} + a_{2} + \text{⋯} + a_{N - 1}} \right) + L.$$

Since $a_{1} + a_{2} + \text{⋯} + a_{N - 1}$ is a finite number, we conclude that the sequence $\left\{ S_{k} \right\}$ is bounded above. Therefore, $\left\{ S_{k} \right\}$ is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that $\left\{ S_{k} \right\}$ converges, and therefore the series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

由于 $a_{1} + a_{2} + \text{⋯} + a_{N - 1}$ 是一个有限数，我们得出结论：数列 $\left\{ S_{k} \right\}$ 上有界。因此，$\left\{ S_{k} \right\}$ 是一个上有界的递增数列。由单调收敛定理，我们得出结论：$\left\{ S_{k} \right\}$ 收敛，从而级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

□

To use the comparison test to determine the convergence or divergence of a series ${\sum\limits_{n = 1}^{\infty}a_{n}},$ it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and *p*-series, these series are often used. If there exists an integer $N$ such that for all $n \geq N,$ each term $a_{n}$ is less than each corresponding term of a known convergent series, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges. Similarly, if there exists an integer $N$ such that for all $n \geq N,$ each term $a_{n}$ is greater than each corresponding term of a known divergent series, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

要使用比较判别法确定级数 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 的收敛性或发散性，需要找到一个合适的级数与之比较。由于我们已经知道几何级数和 p 级数的收敛性质，这些级数常被用作比较。如果存在整数 $N$，使得对所有 $n \geq N$，每一项 $a_{n}$ 都小于某个已知收敛级数的对应项，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。类似地，如果存在整数 $N$，使得对所有 $n \geq N$，每一项 $a_{n}$ 都大于某个已知发散级数的对应项，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

Using the Comparison Test 使用比较判别法

For each of the following series, use the comparison test to determine whether the series converges or diverges.

对于下列各个级数，使用比较判别法判断该级数是收敛还是发散。

$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3} + 3n + 1}$
$\sum\limits_{n = 1}^{\infty}\frac{1}{2^{n} + 1}$
$\sum\limits_{n = 2}^{\infty}\frac{1}{\text{ln}(n)}$

$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3} + 3n + 1}$
$\sum\limits_{n = 1}^{\infty}\frac{1}{2^{n} + 1}$
$\sum\limits_{n = 2}^{\infty}\frac{1}{\text{ln}(n)}$

Solution 解答

1. Compare to $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}$ Since $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}$ is a *p*-series with $p = 3,$ it converges. Further,

1. 与 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}$ 比较。由于 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}$ 是 $p = 3$ 的 p 级数，它收敛。此外，

$$\frac{1}{n^{3} + 3n + 1} < \frac{1}{n^{3}}$$

for every positive integer $n.$ Therefore, we can conclude that $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3} + 3n + 1}$ converges.

对每一个正整数 $n$ 都成立。因此，我们可以得出结论：$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3} + 3n + 1}$ 收敛。

2. Compare to $\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n}.$ Since $\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n}$ is a geometric series with $r = 1\text{/}2$ and $\left| 1\text{/}2 \middle| < 1, \right.$ it converges. Also,

2. 与 $\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n}$ 比较。由于 $\sum\limits_{n = 1}^{\infty}\left( \frac{1}{2} \right)^{n}$ 是公比 $r = 1\text{/}2$ 且 $\left| 1\text{/}2 \middle| < 1, \right.$ 的几何级数，它收敛。另外，

$$\frac{1}{2^{n} + 1} < \frac{1}{2^{n}}$$

for every positive integer $n.$ Therefore, we see that $\sum\limits_{n = 1}^{\infty}\frac{1}{2^{n} + 1}$ converges.

对每一个正整数 $n$ 都成立。因此，我们看到 $\sum\limits_{n = 1}^{\infty}\frac{1}{2^{n} + 1}$ 收敛。

3. Compare to $\sum\limits_{n = 2}^{\infty}\frac{1}{n}.$ Since

3. 与 $\sum\limits_{n = 2}^{\infty}\frac{1}{n}$ 比较。由于

$$\frac{1}{\text{ln}(n)} > \frac{1}{n}$$

for every integer $n \geq 2$ and $\sum\limits_{n = 2}^{\infty}1\text{/}n$ diverges, we have that $\sum\limits_{n = 2}^{\infty}\frac{1}{\text{ln}(n)}$ diverges.

对每一个整数 $n \geq 2$ 成立，且 $\sum\limits_{n = 2}^{\infty}1\text{/}n$ 发散，所以 $\sum\limits_{n = 2}^{\infty}\frac{1}{\text{ln}(n)}$ 发散。

Use the comparison test to determine if the series $\sum\limits_{n = 1}^{\infty}\frac{n}{n^{3} + n + 1}$ converges or diverges.

使用比较判别法判断级数 $\sum\limits_{n = 1}^{\infty}\frac{n}{n^{3} + n + 1}$ 是收敛还是发散。

Limit Comparison Test 极限比较判别法

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

如果我们能找到满足判别法假设的、可作比较的级数，比较判别法便能很好地发挥作用。然而，有时找到合适的级数比较困难。考虑级数

$$\sum\limits_{n = 2}^{\infty}\frac{1}{n^{2} - 1}.$$

It is natural to compare this series with the convergent series

很自然地将此级数与收敛级数

$$\sum\limits_{n = 2}^{\infty}\frac{1}{n^{2}}.$$

However, this series does not satisfy the hypothesis necessary to use the comparison test because

然而，该级数并不满足使用比较判别法所必需的假设，因为

$$\frac{1}{n^{2} - 1} > \frac{1}{n^{2}}$$

for all integers $n \geq 2.$ Although we could look for a different series with which to compare $\sum\limits_{n = 2}^{\infty}1\text{/}(n^{2} - 1),$ instead we show how we can use the limit comparison test to compare

对所有整数 $n \geq 2$ 成立。尽管我们可以另找一个级数来与 $\sum\limits_{n = 2}^{\infty}1\text{/}(n^{2} - 1)$ 比较，但我们改为展示如何用极限比较判别法来比较

$$\sum\limits_{n = 2}^{\infty}\frac{1}{n^{2} - 1}\ \text{and}\ \sum\limits_{n = 2}^{\infty}\frac{1}{n^{2}}.$$

Let us examine the idea behind the limit comparison test. Consider two series $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}.$ with positive terms $a_{n}\ \text{and}\ b_{n}$ and evaluate

我们来考察极限比较判别法背后的思想。考虑两个级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$，其正项分别为 $a_{n}\ \text{and}\ b_{n}$，并求

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}}.$$

若

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}} = L \neq 0,$$

then, for $n$ sufficiently large, $a_{n} \approx Lb_{n}.$ Therefore, either both series converge or both series diverge. For the series $\sum\limits_{n = 2}^{\infty}1\text{/}(n^{2} - 1)$ and $\sum\limits_{n = 2}^{\infty}1\text{/}n^{2},$ we see that

那么当 $n$ 充分大时，$a_{n} \approx Lb_{n}$。因此，两个级数要么同时收敛，要么同时发散。对于级数 $\sum\limits_{n = 2}^{\infty}1\text{/}(n^{2} - 1)$ 与 $\sum\limits_{n = 2}^{\infty}1\text{/}n^{2}$，我们有

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}(n^{2} - 1)}{1\text{/}n^{2}} = \underset{n\rightarrow\infty}{\text{lim}}\ \frac{n^{2}}{n^{2} - 1} = 1.$$

Since $\sum\limits_{n = 2}^{\infty}1\text{/}n^{2}$ converges, we conclude that

由于 $\sum\limits_{n = 2}^{\infty}1\text{/}n^{2}$ 收敛，我们断定

$$\sum\limits_{n = 2}^{\infty}\frac{1}{n^{2} - 1}\ $$

converges.

收敛。

The limit comparison test can be used in two other cases. Suppose

极限比较判别法还可用于另外两种情形。假设

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}} = 0.$$

In this case, $\left\{ a_{n}\text{/}b_{n} \right\}$ is a bounded sequence. As a result, there exists a constant $M$ such that $a_{n} \leq Mb_{n}.$ Therefore, if $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges. On the other hand, suppose

此时 $\left\{ a_{n}\text{/}b_{n} \right\}$ 是有界数列。于是存在常数 $M$ 使得 $a_{n} \leq Mb_{n}$。因此，若 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。另一方面，假设

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}} = \infty.$$

In this case, $\left\{ a_{n}\text{/}b_{n} \right\}$ is an unbounded sequence. Therefore, for every constant $M$ there exists an integer $N$ such that $a_{n} \geq Mb_{n}$ for all $n \geq N.$ Therefore, if $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges as well.

此时 $\left\{ a_{n}\text{/}b_{n} \right\}$ 是无界数列。因此，对每个常数 $M$，都存在整数 $N$，使得对所有 $n \geq N$ 有 $a_{n} \geq Mb_{n}$。于是，若 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 也发散。

Limit Comparison Test 极限比较判别法

Let $a_{n},\ b_{n} \geq 0$ for all $n \geq 1.$

设对所有 $n \geq 1$ 有 $a_{n},\ b_{n} \geq 0$。

1. If $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = L \neq 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ both converge or both diverge.

1. 若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = L \neq 0$，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$ 同收敛或同发散。

2. If $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = 0$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

2. 若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = 0$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

3. If $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = \infty$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

3. 若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n} = \infty$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

Note that if $a_{n}\text{/}b_{n}\rightarrow 0$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, the limit comparison test gives no information. Similarly, if $a_{n}\text{/}b_{n}\rightarrow\infty$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, the test also provides no information. For example, consider the two series ${\sum\limits_{n = 1}^{\infty}1}\text{/}\sqrt{n}$ and ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}.$ These series are both *p*-series with $p = 1\text{/}2$ and $p = 2,$ respectively. Since $p = 1\text{/}2 < 1,$ the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}\sqrt{n}$ diverges. On the other hand, since $p = 2 > 1,$ the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ converges. However, suppose we attempted to apply the limit comparison test, using the convergent $p - \text{series}$ ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{3}$ as our comparison series. First, we see that

注意，若 $a_{n}\text{/}b_{n}\rightarrow 0$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则极限比较判别法不提供任何信息。类似地，若 $a_{n}\text{/}b_{n}\rightarrow\infty$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，该判别法同样不提供信息。例如，考虑两个级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}\sqrt{n}$ 与 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$。这两个级数均为 *p*-级数，其 $p$ 分别为 $p = 1\text{/}2$ 与 $p = 2$。由于 $p = 1\text{/}2 < 1$，级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}\sqrt{n}$ 发散。另一方面，由于 $p = 2 > 1$，级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ 收敛。然而，假设我们尝试使用收敛的 $p - \text{series}$ ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{3}$ 作为比较级数来应用极限比较判别法。首先，我们有

$$\frac{1\text{/}\sqrt{n}}{1\text{/}n^{3}} = \frac{n^{3}}{\sqrt{n}} = n^{5\text{/}2}\rightarrow\infty\ \text{as}\ n\rightarrow\infty.$$

Similarly, we see that

类似地，我们有

$$\frac{1\text{/}n^{2}}{1\text{/}n^{3}} = n\rightarrow\infty\ \text{as}\ n\rightarrow\infty.$$

Therefore, if $a_{n}\text{/}b_{n}\rightarrow\infty$ when $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, we do not gain any information on the convergence or divergence of ${\sum\limits_{n = 1}^{\infty}a_{n}}.$

因此，当 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛而 $a_{n}\text{/}b_{n}\rightarrow\infty$ 时，我们无法获得关于 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 收敛或发散的任何信息。

Using the Limit Comparison Test 使用极限比较判别法

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

对下列每个级数，使用极限比较判别法判断其收敛还是发散。若判别法不适用，请说明。

1. $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n} + 1}$

2. $\sum\limits_{n = 1}^{\infty}\frac{2^{n} + 1}{3^{n}}$

3. $\sum\limits_{n = 1}^{\infty}\frac{\text{ln}(n)}{n^{2}}$

Solution 解答

1. Compare this series to $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}.$ Calculate

1. 将此级数与 $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ 比较。计算

$$\underset{n\rightarrow\infty}{\text{lim}}\ \frac{1\text{/}(\sqrt{n} + 1)}{1\text{/}\sqrt{n}} = \underset{n\rightarrow\infty}{\text{lim}}\begin{matrix} \frac{\sqrt{n}}{\sqrt{n} + 1} \\ \ \end{matrix} = \underset{n\rightarrow\infty}{\text{lim}}\ \frac{1}{1 + 1\text{/}\sqrt{n}} = 1.$$ By the limit comparison test, since $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ diverges, then $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n} + 1}$ diverges.

$$\underset{n\rightarrow\infty}{\text{lim}}\ \frac{1\text{/}(\sqrt{n} + 1)}{1\text{/}\sqrt{n}} = \underset{n\rightarrow\infty}{\text{lim}}\begin{matrix} \frac{\sqrt{n}}{\sqrt{n} + 1} \\ \ \end{matrix} = \underset{n\rightarrow\infty}{\text{lim}}\ \frac{1}{1 + 1\text{/}\sqrt{n}} = 1.$$ 由极限比较判别法，由于 $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ 发散，故 $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n} + 1}$ 发散。

2. Compare this series to $\sum\limits_{n = 1}^{\infty}\left( \frac{2}{3} \right)^{n}.$ We see that

2. 将此级数与 $\sum\limits_{n = 1}^{\infty}\left( \frac{2}{3} \right)^{n}$ 比较。我们有

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(2^{n} + 1)\text{/}3^{n}}{2^{n}\text{/}3^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n} + 1}{3^{n}} \cdot \frac{3^{n}}{2^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n} + 1}{2^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\left\lbrack {1 + \left( \frac{1}{2} \right)^{n}} \right\rbrack = 1.$$

Therefore,

因此，

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(2^{n} + 1)\text{/}3^{n}}{2^{n}\text{/}3^{n}} = 1.$$

Since $\sum\limits_{n = 1}^{\infty}\left( \frac{2}{3} \right)^{n}$ converges, we conclude that $\sum\limits_{n = 1}^{\infty}\frac{2^{n} + 1}{3^{n}}$ converges.

由于 $\sum\limits_{n = 1}^{\infty}\left( \frac{2}{3} \right)^{n}$ 收敛，我们断定 $\sum\limits_{n = 1}^{\infty}\frac{2^{n} + 1}{3^{n}}$ 收敛。

3. Since $\text{ln}\mspace{2mu} n < n,$ compare with $\sum\limits_{n = 1}^{\infty}\frac{1}{n}.$ We see that

3. 由于 $\text{ln}\mspace{2mu} n < n$，与 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 比较。我们有

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n^{2}} \cdot \frac{n}{1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n}.$$

In order to evaluate $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}n,$ evaluate the limit as $x\rightarrow\infty$ of the real-valued function $\text{ln}(x)\text{/}x.$ These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain

为了计算 $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}n$，我们求该实值函数 $\text{ln}(x)\text{/}x$ 当 $x\rightarrow\infty$ 时的极限。这两个极限相等，作此改变后我们便可使用洛必达法则。得到

$$\underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} x}{x} = \underset{x\rightarrow\infty}{\text{lim}}\frac{1}{x} = 0.$$

Therefore, $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}n = 0,$ and, consequently,

因此，$\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}n = 0$，进而

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n} = 0.$$

Since the limit is $0$ but $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ diverges, the limit comparison test does not provide any information.

由于极限为 $0$，而 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 发散，极限比较判别法不提供任何信息。

Compare with $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ instead. In this case,

改为与 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ 比较。此时，

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\ n\text{/}n^{2}}{1\text{/}n^{2}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\ n}{n^{2}} \cdot \frac{n^{2}}{1} = \underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n = \infty.$$

Since the limit is $\infty$ but $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ converges, the test still does not provide any information.

由于极限为 $\infty$，而 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ 收敛，该判别法仍不提供任何信息。

So now we try a series between the two we already tried. Choosing the series $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3\text{/}2}},$ we see that

于是我们尝试一个介于前两者之间（已试过）的级数。选取级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3\text{/}2}}$，我们有

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n^{3\text{/}2}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n^{2}} \cdot \frac{n^{3\text{/}2}}{1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{\sqrt{n}}.$$

As above, in order to evaluate $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}\sqrt{n},$ evaluate the limit as $x\rightarrow\infty$ of the real-valued function $\text{ln}\mspace{2mu} x\text{/}\sqrt{x}.$ Using L’Hôpital’s rule,

同理，为了计算 $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}\sqrt{n}$，我们求该实值函数 $\text{ln}\mspace{2mu} x\text{/}\sqrt{x}$ 当 $x\rightarrow\infty$ 时的极限。使用洛必达法则，

$$\underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} x}{\sqrt{x}} = \underset{x\rightarrow\infty}{\text{lim}}\frac{2\sqrt{x}}{x} = \underset{x\rightarrow\infty}{\text{lim}}\frac{2}{\sqrt{x}} = 0.$$

Since the limit is $0$ and $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3\text{/}2}}$ converges, we can conclude that $\sum\limits_{n = 1}^{\infty}\frac{\text{ln}\mspace{2mu} n}{n^{2}}$ converges.

由于极限为 $0$ 且 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3\text{/}2}}$ 收敛，我们可以断定 $\sum\limits_{n = 1}^{\infty}\frac{\text{ln}\mspace{2mu} n}{n^{2}}$ 收敛。

Use the limit comparison test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{5^{n}}{3^{n} + 2}$ converges or diverges.

使用极限比较判别法判断级数 $\sum\limits_{n = 1}^{\infty}\frac{5^{n}}{3^{n} + 2}$ 收敛还是发散。

Section 5.4 Exercises 5.4 节习题

Use the comparison test to determine whether the following series converge.

使用比较判别法判断下列级数是否收敛。

194\.

$\sum\limits_{n = 1}^{\infty}a_{n}$ where $a_{n} = \frac{2}{n\left( {n + 1} \right)}$

$\sum\limits_{n = 1}^{\infty}a_{n}$ 其中 $a_{n} = \frac{2}{n\left( {n + 1} \right)}$

195.

$\sum\limits_{n = 1}^{\infty}a_{n}$ where $a_{n} = \frac{1}{n\left( {n + {1\text{/}2}} \right)}$

$\sum\limits_{n = 1}^{\infty}a_{n}$ 其中 $a_{n} = \frac{1}{n\left( {n + {1\text{/}2}} \right)}$

196\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{2\left( {n + 1} \right)}$

197.

$\sum\limits_{n = 2}^{\infty}\frac{1}{2\left( {n - 1} \right)}$

198\.

$\sum\limits_{n = 2}^{\infty}\frac{1}{\left( {n\mspace{2mu}\text{ln}\mspace{2mu} n} \right)^{2}}$

199.

$\sum\limits_{n = 1}^{\infty}\frac{n\text{!}}{\left( {n + 2} \right)\text{!}}$

200\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n\text{!}}$

201.

$\sum\limits_{n = 1}^{\infty}\frac{\text{sin}\left( {1\text{/}n} \right)}{n^{2}}$

202\.

$\sum\limits_{n = 1}^{\infty}\frac{\text{sin}^{2}n}{n^{2}}$

203.

$\sum\limits_{n = 1}^{\infty}\frac{\text{sin}\left( 1\text{/}n \right)}{\left( \sqrt{n} \right)^{3}}$

204\.

$\sum\limits_{n = 1}^{\infty}\frac{n^{1.2} - 1}{n^{2.3} + 1}$

205.

$\sum\limits_{n = 1}^{\infty}\frac{\sqrt{n + 1} - \sqrt{n}}{n}$

206\.

$\sum\limits_{n = 1}^{\infty}\frac{\sqrt[4]{n}}{\sqrt[3]{n^{4} + n^{2}}}$

Use the limit comparison test to determine whether each of the following series converges or diverges.

使用极限比较判别法判断下列各级数收敛还是发散。

207.

$\sum\limits_{n = 1}^{\infty}\left( \frac{\text{ln}\mspace{2mu} n}{n} \right)^{2}$

208\.

$\sum\limits_{n = 1}^{\infty}\left( \frac{\text{ln}\mspace{2mu} n}{n^{0.6}} \right)^{2}$

209.

$\sum\limits_{n = 1}^{\infty}\frac{\text{ln}\left( {1 + \frac{1}{n}} \right)}{n}$

210\.

$\sum\limits_{n = 1}^{\infty}\text{ln}\left( {1 + \frac{1}{n^{2}}} \right)$

211.

$\sum\limits_{n = 1}^{\infty}\frac{1}{4^{n} - 3^{n}}$

212\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} - n\mspace{2mu}\text{sin}\mspace{2mu} n}$

213.

$\sum\limits_{n = 1}^{\infty}\frac{1}{e^{{(1.1)}n} - 3^{n}}$

214\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{e^{{(1.01)}n} - 3^{n}}$

215.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{1 + 1\text{/}n}}$

216\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{2^{1 + 1\text{/}n}n^{1 + 1\text{/}n}}$

217.

$\sum\limits_{n = 1}^{\infty}\left( {\frac{1}{n} - \text{sin}\left( \frac{1}{n} \right)} \right)$

218\.

$\sum\limits_{n = 1}^{\infty}\left( {1 - \text{cos}\left( \frac{1}{n} \right)} \right)$

219.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n}\left( \frac{\pi}{2} - {\text{tan}^{-1}n} \right)$

220\.

$\sum\limits_{n = 1}^{\infty}\left( {1 - \frac{1}{n}} \right)^{n}$ (*Hint:*$\left( {1 - \frac{1}{n}} \right)^{n}\rightarrow 1\text{/}e.)$

$\sum\limits_{n = 1}^{\infty}\left( {1 - \frac{1}{n}} \right)^{n}$ （*提示：*$\left( {1 - \frac{1}{n}} \right)^{n}\rightarrow 1\text{/}e.$）

221.

$\sum\limits_{n = 1}^{\infty}\left( {1 - e^{-1\text{/}n}} \right)$ (*Hint:*$1\text{/}e \approx \left( {1 - 1\text{/}n} \right)^{n},$ so $1 - e^{-1\text{/}n} \approx 1\text{/}n.)$

$\sum\limits_{n = 1}^{\infty}\left( {1 - e^{-1\text{/}n}} \right)$ （*提示：*$1\text{/}e \approx \left( {1 - 1\text{/}n} \right)^{n},$ 于是 $1 - e^{-1\text{/}n} \approx 1\text{/}n.$）

222\.

Does $\sum\limits_{n = 2}^{\infty}\frac{1}{\left( {\text{ln}\mspace{2mu} n} \right)^{p}}$ converge if $p$ is large enough? If so, for which $p\text{?}$

若 $p$ 足够大，则 $\sum\limits_{n = 2}^{\infty}\frac{1}{\left( {\text{ln}\mspace{2mu} n} \right)^{p}}$ 是否收敛？若收敛，哪些 $p\text{?}$

223.

Does $\sum\limits_{n = 1}^{\infty}\left( \frac{\left( {\text{ln}\mspace{2mu} n} \right)}{n} \right)^{p}$ converge if $p$ is large enough? If so, for which $p\text{?}$

若 $p$ 足够大，则 $\sum\limits_{n = 1}^{\infty}\left( \frac{\left( {\text{ln}\mspace{2mu} n} \right)}{n} \right)^{p}$ 是否收敛？若收敛，哪些 $p\text{?}$

224\.

For which $p$ does the series $\sum\limits_{n = 1}^{\infty}2^{pn}\text{/}3^{n}$ converge?

对于哪些 $p$，级数 $\sum\limits_{n = 1}^{\infty}2^{pn}\text{/}3^{n}$ 收敛？

225.

For which $p > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{2^{n}}$ converge?

对于哪些 $p > 0$，级数 $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{2^{n}}$ 收敛？

226\.

For which $r > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{r^{n^{2}}}{2^{n}}$ converge?

对于哪些 $r > 0$，级数 $\sum\limits_{n = 1}^{\infty}\frac{r^{n^{2}}}{2^{n}}$ 收敛？

227.

For which $r > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{r^{n^{2}}}$ converge?

对于哪些 $r > 0$，级数 $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{r^{n^{2}}}$ 收敛？

228\.

Find all values of $p$ and $q$ such that $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{\left( {n\text{!}} \right)^{q}}$ converges.

求所有使得 $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{\left( {n\text{!}} \right)^{q}}$ 收敛的 $p$ 与 $q$ 的值。

229.

Does $\underset{n = 1}{\overset{\infty}{}}\left( \frac{n\pi}{2} \right)$ converge or diverge? Explain.

$\underset{n = 1}{\overset{\infty}{}}\left( \frac{n\pi}{2} \right)$ 是收敛还是发散？说明理由。

230\.

Explain why, for each $n,$ at least one of $\left\{ \left| {\text{sin}\mspace{2mu} n} \right|,\left| {\text{sin}\left( {n + 1} \right)} \right|\text{,...},\left| {\text{sin}\mspace{2mu} n + 6} \right| \right\}$ is larger than $1\text{/}2.$ Use this relation to test convergence of $\sum\limits_{n = 1}^{\infty}\frac{\left| {\text{sin}\mspace{2mu} n} \right|}{\sqrt{n}}.$

说明为什么：对每个 $n,$ 集合 $\left\{ \left| {\text{sin}\mspace{2mu} n} \right|,\left| {\text{sin}\left( {n + 1} \right)} \right|\text{,...},\left| {\text{sin}\mspace{2mu} n + 6} \right| \right\}$ 中至少有一个大于 $1\text{/}2.$ 利用这一关系检验 $\sum\limits_{n = 1}^{\infty}\frac{\left| {\text{sin}\mspace{2mu} n} \right|}{\sqrt{n}}$ 的收敛性。

231.

Suppose that $a_{n} \geq 0$ and $b_{n} \geq 0$ and that $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ and $\sum\limits_{n = 1}^{\infty}{b^{2}{}_{n}}$ converge. Prove that $\sum\limits_{n = 1}^{\infty}{a_{n}b_{n}}$ converges and ${\sum\limits_{n = 1}^{\infty}a_{n}}\ b_{n} \leq \frac{1}{2}\left( {{\sum\limits_{n = 1}^{\infty}a_{n}^{2}} + {\sum\limits_{n = 1}^{\infty}b_{n}^{2}}} \right).$

设 $a_{n} \geq 0$ 且 $b_{n} \geq 0$，且 $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ 与 $\sum\limits_{n = 1}^{\infty}{b^{2}{}_{n}}$ 都收敛。证明 $\sum\limits_{n = 1}^{\infty}{a_{n}b_{n}}$ 收敛，且 ${\sum\limits_{n = 1}^{\infty}a_{n}}\ b_{n} \leq \frac{1}{2}\left( {{\sum\limits_{n = 1}^{\infty}a_{n}^{2}} + {\sum\limits_{n = 1}^{\infty}b_{n}^{2}}} \right).$

232\.

Does $\sum\limits_{n = 1}^{\infty}2^{\text{−}\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ converge? (*Hint:* Write $2^{\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ as a power of $\text{ln}\mspace{2mu} n.)$

$\sum\limits_{n = 1}^{\infty}2^{\text{−}\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ 是否收敛？（*提示：*将 $2^{\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ 写成 $\text{ln}\mspace{2mu} n$ 的幂。）

233.

Does $\sum\limits_{n = 1}^{\infty}\left( {\text{ln}\mspace{2mu} n} \right)^{\text{−}\text{ln}\mspace{2mu} n}$ converge? (*Hint:* Use $n = e^{\text{ln}{(n)}}$ to compare to a $p - \text{series}\text{.})$

$\sum\limits_{n = 1}^{\infty}\left( {\text{ln}\mspace{2mu} n} \right)^{\text{−}\text{ln}\mspace{2mu} n}$ 是否收敛？（*提示：*利用 $n = e^{\text{ln}{(n)}}$ 与 $p - \text{series}\text{.}$ 比较。）

234\.

Does $\sum\limits_{n = 2}^{\infty}\left( {\text{ln}\mspace{2mu} n} \right)^{\text{−}\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ converge? (*Hint:* Compare $a_{n}$ to $1\text{/}n.)$

$\sum\limits_{n = 2}^{\infty}\left( {\text{ln}\mspace{2mu} n} \right)^{\text{−}\text{ln}\mspace{2mu}\text{ln}\mspace{2mu} n}$ 是否收敛？（*提示：*将 $a_{n}$ 与 $1\text{/}n$ 比较。）

235.

Show that if $a_{n} \geq 0$ and $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ converges. If $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ converges, does $\sum\limits_{n = 1}^{\infty}a_{n}$ necessarily converge?

证明：若 $a_{n} \geq 0$ 且 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ 收敛。若 $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ 收敛，$\sum\limits_{n = 1}^{\infty}a_{n}$ 是否必然收敛？

236\.

Suppose that $a_{n} > 0$ for all $n$ and that $\sum\limits_{n = 1}^{\infty}a_{n}$ converges. Suppose that $b_{n}$ is an arbitrary sequence of zeros and ones. Does ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ necessarily converge?

设对所有 $n$ 都有 $a_{n} > 0$，且 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。设 $b_{n}$ 是任意由 0 和 1 组成的数列。问 ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ 是否必然收敛？

237.

Suppose that $a_{n} > 0$ for all $n$ and that $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges. Suppose that $b_{n}$ is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ necessarily diverge?

设对所有 $n$ 都有 $a_{n} > 0$，且 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。设 $b_{n}$ 是任意由 0 和 1 组成的数列，且有无穷多项等于 1。问 ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ 是否必然发散？

238\.

Complete the details of the following argument: If $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ converges to a finite sum $s,$ then $\frac{1}{2}s = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \text{⋯}$ and $s - \frac{1}{2}s = 1 + \frac{1}{3} + \frac{1}{5} + \text{⋯}.$ Why does this lead to a contradiction?

补全下列论证的细节：若 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 收敛到有限和 $s,$ 则 $\frac{1}{2}s = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \text{⋯}$ 且 $s - \frac{1}{2}s = 1 + \frac{1}{3} + \frac{1}{5} + \text{⋯}.$ 为什么这会导致矛盾？

239.

Show that if $a_{n} \geq 0$ and $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ converges, then $\sum\limits_{n = 1}^{\infty}{\text{sin}^{2}\left( a_{n} \right)}$ converges.

证明：若 $a_{n} \geq 0$ 且 $\sum\limits_{n = 1}^{\infty}{a^{2}{}_{n}}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}{\text{sin}^{2}\left( a_{n} \right)}$ 收敛。

240\.

Suppose that $a_{n}\text{/}b_{n}\rightarrow 0$ in the comparison test, where $a_{n} \geq 0$ and $b_{n} \geq 0.$ Prove that if $\sum b_{n}$ converges, then $\sum a_{n}$ converges.

设比较判别法中 $a_{n}\text{/}b_{n}\rightarrow 0$，其中 $a_{n} \geq 0$ 且 $b_{n} \geq 0.$ 证明：若 $\sum b_{n}$ 收敛，则 $\sum a_{n}$ 收敛。

241.

Let $b_{n}$ be an infinite sequence of zeros and ones. What is the largest possible value of $x = {\sum\limits_{n = 1}^{\infty}{b_{n}\text{/}2^{n}}}\text{?}$

设 $b_{n}$ 是一个由 0 和 1 组成的无穷数列。$x = {\sum\limits_{n = 1}^{\infty}{b_{n}\text{/}2^{n}}}$ 的最大可能值是多少？

242\.

Let $d_{n}$ be an infinite sequence of digits, meaning $d_{n}$ takes values in $\left\{ 0,\ 1\text{,…},9 \right\}.$ What is the largest possible value of $x = {\sum\limits_{n = 1}^{\infty}{d_{n}\text{/}10^{n}}}$ that converges?

设 $d_{n}$ 是一个由数字组成的无穷数列，即 $d_{n}$ 取值于 $\left\{ 0,\ 1\text{,…},9 \right\}.$ 使该级数收敛的 $x = {\sum\limits_{n = 1}^{\infty}{d_{n}\text{/}10^{n}}}$ 的最大可能值是多少？

243.

Explain why, if $x > 1\text{/}2,$ then $x$ cannot be written $x = {\sum\limits_{n = 2}^{\infty}\frac{b_{n}}{2^{n}}}\left( {b_{n} = 0\ \text{or}\ 1,\ b_{1} = 0} \right).$

说明为什么：若 $x > 1\text{/}2,$ 则 $x$ 不能写成 $x = {\sum\limits_{n = 2}^{\infty}\frac{b_{n}}{2^{n}}}\left( {b_{n} = 0\ \text{or}\ 1,\ b_{1} = 0} \right).$

244\.

\[T\] Evelyn has a perfect balancing scale, an unlimited number of $1\ \text{-kg}$ weights, and one each of $1\text{/}2\ \text{-kg},\ 1\text{/}4\ \text{-kg},\ 1\text{/}8\ \text{-kg},$ and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

\[T\] Evelyn 有一架完美的天平、数量不限的 $1\ \text{-kg}$ 砝码，以及各一个的 $1\text{/}2\ \text{-kg}、$ $1\text{/}4\ \text{-kg}、$ $1\text{/}8\ \text{-kg}$ 等砝码。她希望以任意精度称量一块来源不明的陨石。假设天平足够大，她能做到吗？这与无穷级数有什么关系？

245.

\[T\] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1\ \text{-kg}$ weights, and nine each of $0.1\ \text{-kg,}$ $0.01\ \text{-kg},\ 0.001\ \text{-kg,}$ and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

\[T\] Robert 想以任意精度知道自己的体重。他有一架完美运作的大天平、数量不限的 $1\ \text{-kg}$ 砝码，以及各九个的 $0.1\ \text{-kg}、$ $0.01\ \text{-kg}、$ $0.001\ \text{-kg}$ 等砝码。假设天平足够大，他能做到吗？这与无穷级数有什么关系？

246\.

The series $\sum\limits_{n = 1}^{\infty}\frac{1}{2n}$ is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which $n$ is odd. Let $m > 1$ be fixed. Show, more generally, that deleting all terms $1\text{/}n$ where $n = mk$ for some integer $k$ also results in a divergent series.

级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{2n}$ 是调和级数的一半，因而发散。它是由调和级数删去所有 $n$ 为奇数的项得到的。固定 $m > 1$。更一般地证明：删去所有满足 $n = mk$（$k$ 为某整数）的项 $1\text{/}n$，所得级数仍然发散。

247.

In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A *depleted harmonic series* is a series obtained from $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ by removing any term $1\text{/}n$ if a given digit, say $9,$ appears in the decimal expansion of $n.$ Argue that this depleted harmonic series converges by answering the following questions.

鉴于前一题，一个调和级数的子级数（其中约每五项删去一项）竟可能收敛，这或许令人惊讶。*稀疏调和级数* 是指从 $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ 中删去所有在十进制展开中含有某给定数字（例如 $9,$）的项 $1\text{/}n$ 后得到的级数。通过回答下列诸问题来说明该级数收敛。

1. How many whole numbers $n$ have $d$ digits?

1. 有多少个整数 $n$ 具有 $d$ 位数字？

2. How many $d\text{-digit}$ whole numbers $h(d).$ do not contain $9$ as one or more of their digits?

2. 有多少个 $d\text{-digit}$ 整数不含数字 $9$？将其个数记为 $h(d)$．

3. What is the smallest $d\text{-digit}$ number $m(d)\text{?}$

3. 最小的 $d\text{-digit}$ 数 $m(d)\text{?}$ 是多少？

4. Explain why the deleted harmonic series is bounded by ${\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}}.$

4. 说明为什么删去项后的调和级数被 ${\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}}$ 所界。

5. Show that $\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}$ converges.

5. 证明 $\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}$ 收敛。

248\.

Suppose that a sequence of numbers $a_{n} > 0$ has the property that $a_{1} = 1$ and $a_{n + 1} = \frac{1}{n + 1}S_{n},$ where $S_{n} = a_{1} + \text{⋯} + a_{n}.$ Can you determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges? (*Hint:* $S_{n}$ is monotone.)

设正数数列 $a_{n} > 0$ 满足 $a_{1} = 1$ 且 $a_{n + 1} = \frac{1}{n + 1}S_{n},$ 其中 $S_{n} = a_{1} + \text{⋯} + a_{n}.$ 你能否判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛？（*提示：*$S_{n}$ 是单调的。）

249.

Suppose that a sequence of numbers $a_{n} > 0$ has the property that $a_{1} = 1$ and $a_{n + 1} = \frac{1}{\left( {n + 1} \right)^{2}}S_{n},$ where $S_{n} = a_{1} + \text{⋯} + a_{n}.$ Can you determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges? (*Hint:* $S_{2} = a_{2} + a_{1} = a_{2} + S_{1} = a_{2} + 1 = 1 + 1\text{/}4 = \left( {1 + 1\text{/}4} \right)S_{1},$ $S_{3} = \frac{1}{3^{2}}S_{2} + S_{2} = \left( {1 + 1\text{/}9} \right)S_{2} = \left( {1 + 1\text{/}9} \right)\left( {1 + 1\text{/}4} \right)S_{1},$ etc. Look at $\text{ln}\left( S_{n} \right),$ and use $\text{ln}\left( {1 + t} \right) \leq t,$ $t > 0.)$

设正数数列 $a_{n} > 0$ 满足 $a_{1} = 1$ 且 $a_{n + 1} = \frac{1}{\left( {n + 1} \right)^{2}}S_{n},$ 其中 $S_{n} = a_{1} + \text{⋯} + a_{n}.$ 你能否判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛？（*提示：*$S_{2} = a_{2} + a_{1} = a_{2} + S_{1} = a_{2} + 1 = 1 + 1\text{/}4 = \left( {1 + 1\text{/}4} \right)S_{1},$ $S_{3} = \frac{1}{3^{2}}S_{2} + S_{2} = \left( {1 + 1\text{/}9} \right)S_{2} = \left( {1 + 1\text{/}9} \right)\left( {1 + 1\text{/}4} \right)S_{1},$ 依此类推。考察 $\text{ln}\left( S_{n} \right),$ 并利用 $\text{ln}\left( {1 + t} \right) \leq t,$ $t > 0.$）

5.5 Alternating Series 5.5 交错级数

5.5.1 Use the alternating series test to test an alternating series for convergence.
5.5.2 Estimate the sum of an alternating series.
5.5.3 Explain the meaning of absolute convergence and conditional convergence.

5.5.1 使用交错级数判别法检验交错级数的收敛性。
5.5.2 估计交错级数的和。
5.5.3 解释绝对收敛与条件收敛的含义。

So far in this chapter, we have primarily discussed series with positive terms. In this section we introduce alternating series—those series whose terms alternate in sign. We will show in a later chapter that these series often arise when studying power series. After defining alternating series, we introduce the alternating series test to determine whether such a series converges.

在本章到目前为止，我们主要讨论了各项为正的级数。本节我们引入交错级数——即各项正负交替的级数。我们将在后面的章节中说明，研究幂级数时常常会遇到这类级数。在定义交错级数之后，我们引入交错级数判别法来判断这类级数是否收敛。

The Alternating Series Test 交错级数判别法

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

各项在正值与负值之间交替出现的级数称为交错级数。例如，下列级数

$$\sum\limits_{n = 1}^{\infty}\left( {- \frac{1}{2}} \right)^{n} = - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \text{⋯}$$ (5.11)

and

以及

$$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \text{⋯}$$ (5.12)

are both alternating series.

都是交错级数。

Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form

任何各项在正值与负值之间交替出现的级数都称为交错级数。交错级数可以写成如下形式

$${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n + 1}b_{n}}} = b_{1} - b_{2} + b_{3} - b_{4} + \text{⋯}$$ (5.13)

或者

$$\sum\limits_{n = 1}^{\infty}{(-1)^{n}b_{n} = \text{−}b_{1} + b_{2} - b_{3} + b_{4} - \text{⋯}}$$ (5.14)

Where $b_{n} > 0$ for all positive integers *n*.

其中对所有正整数 *n* 都有 $b_{n} > 0$。

Series (1), shown in Equation 5.11, is a geometric series. Since $\left. |r| = \middle| \text{−}1\text{/}2 \middle| < 1, \right.$ the series converges. Series (2), shown in Equation 5.12, is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.

式 (1)（即式 5.11）是一个几何级数。由于 $\left. |r| = \middle| \text{−}1\text{/}2 \middle| < 1, \right.$，该级数收敛。式 (2)（即式 5.12）称为交错调和级数。我们将证明：调和级数发散，而交错调和级数收敛。

To prove this, we look at the sequence of partial sums $\left\{ S_{k} \right\}$ (Figure 5.17).

为证明这一点，我们考察部分和数列 $\left\{ S_{k} \right\}$（图 5.17）。

Proof 证明

Consider the odd terms $S_{2k + 1}$ for $k \geq 0.$ Since $1\text{/}(2k + 1) < 1\text{/}2k,$

考虑奇数项 $S_{2k + 1}$（$k \geq 0$）。由于 $1\text{/}(2k + 1) < 1\text{/}2k$，

$$S_{2k + 1} = S_{2k - 1} - \frac{1}{2k} + \frac{1}{2k + 1} < S_{2k - 1}.$$

Therefore, $\left\{ S_{2k + 1} \right\}$ is a decreasing sequence. Also,

因此，$\left\{ S_{2k + 1} \right\}$ 是一个递减数列。此外，

$$S_{2k + 1} = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \text{⋯} + \left( {\frac{1}{2k - 1} - \frac{1}{2k}} \right) + \frac{1}{2k + 1} > 0.$$

Therefore, $\left\{ S_{2k + 1} \right\}$ is bounded below. Since $\left\{ S_{2k + 1} \right\}$ is a decreasing sequence that is bounded below, by the Monotone Convergence Theorem, $\left\{ S_{2k + 1} \right\}$ converges. Similarly, the even terms $\left\{ S_{2k} \right\}$ form an increasing sequence that is bounded above because

因此，$\left\{ S_{2k + 1} \right\}$ 下有界。由于 $\left\{ S_{2k + 1} \right\}$ 是递减且有下界的数列，根据单调收敛定理，$\left\{ S_{2k + 1} \right\}$ 收敛。类似地，偶数项 $\left\{ S_{2k} \right\}$ 构成一个递增且上有界的数列，因为

$$S_{2k} = S_{2k - 2} + \frac{1}{2k - 1} - \frac{1}{2k} > S_{2k - 2}$$

and

并且

$$S_{2k} = 1 + \left( {- \frac{1}{2} + \frac{1}{3}} \right) + \text{⋯} + \left( {- \frac{1}{2k - 2} + \frac{1}{2k - 1}} \right) - \frac{1}{2k} < 1.$$

Therefore, by the Monotone Convergence Theorem, the sequence $\left\{ S_{2k} \right\}$ also converges. Since

因此，根据单调收敛定理，数列 $\left\{ S_{2k} \right\}$ 也收敛。由于

$$S_{2k + 1} = S_{2k} + \frac{1}{2k + 1},$$

we know that

我们知道

$$\underset{k\rightarrow\infty}{\text{lim}}S_{2k + 1} = \underset{k\rightarrow\infty}{\text{lim}}S_{2k} + \underset{k\rightarrow\infty}{\text{lim}}\frac{1}{2k + 1}.$$

Letting $S = \underset{k\rightarrow\infty}{\text{lim}}S_{2k + 1}$ and using the fact that $1\text{/}\left( {2k + 1} \right)\rightarrow 0,$ we conclude that $\underset{k\rightarrow\infty}{\text{lim}}S_{2k} = S.$ Since the odd terms and the even terms in the sequence of partial sums converge to the same limit $S,$ it can be shown that the sequence of partial sums converges to $S,$ and therefore the alternating harmonic series converges to $S.$

令 $S = \underset{k\rightarrow\infty}{\text{lim}}S_{2k + 1}$，并利用 $1\text{/}\left( {2k + 1} \right)\rightarrow 0$ 这一事实，我们断定 $\underset{k\rightarrow\infty}{\text{lim}}S_{2k} = S$。由于部分和数列中的奇数项与偶数项都收敛到同一极限 $S$，可以证明部分和数列收敛于 $S$，因此交错调和级数收敛于 $S$。

It can also be shown that $S = \text{ln}\mspace{2mu} 2,$ and we can write

还可以证明 $S = \text{ln}\mspace{2mu} 2$，从而可以写成

$$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \text{⋯} = \text{ln}(2).$$

□

More generally, any alternating series of form (3) (Equation 5.13) or (4) (Equation 5.14) converges as long as $b_{1} \geq b_{2} \geq b_{3} \geq \text{⋯}$ and $b_{n}\rightarrow 0$ (Figure 5.18). The proof is similar to the proof for the alternating harmonic series.

更一般地，只要 $b_{1} \geq b_{2} \geq b_{3} \geq \text{⋯}$ 且 $b_{n}\rightarrow 0$（图 5.18），形式为 (3)（式 5.13）或 (4)（式 5.14）的任何交错级数都收敛。其证明与交错调和级数的证明类似。

Alternating Series Test 交错级数判别法

An alternating series of the form

形如

$$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}\ \text{or}\ \sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n}$$

converges if

在满足下列条件时收敛：

1. $0 < b_{n + 1} \leq b_{n}$ for all $n \geq 1$ and

1. 对所有 $n \geq 1$ 都有 $0 < b_{n + 1} \leq b_{n}$，且

2. $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0.$

2. $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0$。

This is known as the alternating series test.

这就是所谓的交错级数判别法。

We remark that this theorem is true more generally as long as there exists some integer $N$ such that $0 < b_{n + 1} \leq b_{n}$ for all $n \geq N.$

我们指出，只要存在某个整数 $N$，使得对所有 $n \geq N$ 都有 $0 < b_{n + 1} \leq b_{n}$，该定理在更一般的意义上也成立。

Convergence of Alternating Series 交错级数的收敛性

For each of the following alternating series, determine whether the series converges or diverges.

对于下列各个交错级数，判断其收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n^{2}$

2. $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}(n + 1)$

Solution 解答

1. Since

1. 由于

$\frac{1}{\left( {n + 1} \right)^{2}} < \frac{1}{n^{2}}\quad\text{and}\quad\frac{1}{n^{2}}\rightarrow 0,$

the series converges.

该级数收敛。

2. Since $n\text{/}(n + 1)\nrightarrow 0$ as $n\rightarrow\infty,$ we cannot apply the alternating series test. Instead, we use the *n*th term test for divergence. Since

2. 由于当 $n\rightarrow\infty$ 时 $n\text{/}(n + 1)\nrightarrow 0$，我们不能应用交错级数判别法。转而使用第 *n* 项发散判别法。由于

$\underset{n\rightarrow\infty}{\text{lim}}\frac{(-1)^{n + 1}n}{n + 1} \neq 0,$

the series diverges.

该级数发散。

Determine whether the series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}2^{n}$ converges or diverges.

判断级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}2^{n}$ 是收敛还是发散。

Remainder of an Alternating Series 交错级数的余项

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

大多数交错级数的和很难显式计算，因此通常用部分和来近似其和。这样做时，我们关心近似的误差大小。考虑一个交错级数

$$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$$

satisfying the hypotheses of the alternating series test. Let $S$ denote the sum of this series and $\left\{ S_{k} \right\}$ be the corresponding sequence of partial sums. From Figure 5.18, we see that for any integer $N \geq 1,$ the remainder $R_{N}$ satisfies

它满足交错级数判别法的假设。设 $S$ 表示该级数的和，$\left\{ S_{k} \right\}$ 为相应的部分和数列。由图 5.18 可见，对任意整数 $N \geq 1$，余项 $R_{N}$ 满足

$$\left| R_{N} \right| = \left| {S - S_{N}} \right| \leq \left| {S_{N + 1} - S_{N}} \right| = b_{n + 1}.$$

Remainders in Alternating Series 交错级数中的余项

Consider an alternating series of the form

考虑形如

$$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}\text{or}\sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n}$$

that satisfies the hypotheses of the alternating series test. Let $S$ denote the sum of the series and $S_{N}$ denote the $N\text{th}$ partial sum. For any integer $N \geq 1,$ the remainder $R_{N} = S - S_{N}$ satisfies

它满足交错级数判别法的假设。设 $S$ 表示该级数的和，$S_{N}$ 表示第 $N\text{th}$ 个部分和。对任意整数 $N \geq 1$，余项 $R_{N} = S - S_{N}$ 满足

$$\left| R_{N} \middle| \leq b_{N + 1}. \right.$$

In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the $N\text{th}$ partial sum $S_{N}$ is in magnitude at most the size of the next term $b_{N + 1}.$

换言之，若交错级数判别法的条件成立，则用第 $N\text{th}$ 个部分和 $S_{N}$ 来近似该无穷级数时，其误差的绝对值至多为下一项 $b_{N + 1}$ 的大小。

Estimating the Remainder of an Alternating Series 估计交错级数的余项

Consider the alternating series

考虑交错级数

$${\sum\limits_{n = 1}^{\infty}\frac{{(-1)}^{n + 1}}{n^{2}}}.$$

Use the remainder estimate to determine a bound on the error $R_{10}$ if we approximate the sum of the series by the partial sum $S_{10}.$

若用部分和 $S_{10}$ 来近似该级数的和，利用余项估计确定误差 $R_{10}$ 的一个界。

Solution 解答

From the theorem stated above,

由上述定理可知，

$|R_{10}| \leq b_{11} = \frac{1}{11^{2}} \approx 0.008265.$

Find a bound for $R_{20}$ when approximating $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n$ by $S_{20}.$

当用 $S_{20}$ 近似 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n$ 时，求 $R_{20}$ 的一个界。

Absolute and Conditional Convergence 绝对收敛与条件收敛

Consider a series $\sum\limits_{n = 1}^{\infty}a_{n}$ and the related series ${\sum\limits_{n = 1}^{\infty}|}a_{n}|.$ Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n.$ The series whose terms are the absolute value of these terms is the harmonic series, since ${\sum\limits_{n = 1}^{\infty}|}{(-1)}^{n + 1}\text{/}n| = {\sum\limits_{n = 1}^{\infty}1}\text{/}n.$ Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

考虑级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 以及与之相关的级数 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|.$ 这里讨论这两个级数的收敛性之间关系的可能性。例如，考虑交错调和级数 ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n.$ 其各项取绝对值后所得的级数就是调和级数，因为 ${\sum\limits_{n = 1}^{\infty}|}{(-1)}^{n + 1}\text{/}n| = {\sum\limits_{n = 1}^{\infty}1}\text{/}n.$ 由于交错调和级数收敛，而调和级数发散，我们说交错调和级数表现出条件收敛。

By comparison, consider the series ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n^{2}.$ The series whose terms are the absolute values of the terms of this series is the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}.$ Since both of these series converge, we say the series ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n^{2}$ exhibits absolute convergence.

作为对比，考虑级数 ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n^{2}.$ 该级数各项取绝对值后所得的级数是级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}.$ 由于这两个级数都收敛，我们说级数 ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n + 1}}\text{/}n^{2}$ 表现出绝对收敛。

A series $\sum\limits_{n = 1}^{\infty}a_{n}$ exhibits absolute convergence if $\sum\limits_{n = 1}^{\infty}|a_{n}|$ converges. A series $\sum\limits_{n = 1}^{\infty}a_{n}$ exhibits conditional convergence if $\sum\limits_{n = 1}^{\infty}a_{n}$ converges but $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ diverges.

若 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 收敛，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 表现出绝对收敛。若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛但 $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ 发散，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 表现出条件收敛。

As shown by the alternating harmonic series, a series $\sum\limits_{n = 1}^{\infty}a_{n}$ may converge, but $\sum\limits_{n = 1}^{\infty}|a_{n}|$ may diverge. In the following theorem, however, we show that if $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

正如交错调和级数所示，级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 可能收敛，而 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 可能发散。然而，在下面的定理中，我们将证明：若 $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

Absolute Convergence Implies Convergence 绝对收敛蕴含收敛

If $\sum\limits_{n = 1}^{\infty}|a_{n}|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

若 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

Proof 证明

Suppose that ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ converges. We show this by using the fact that $\left. a_{n} = \middle| a_{n} \right|$ or $\left. a_{n} = \text{−} \middle| a_{n} \right|$ and therefore $\left| a_{n} \right| + a_{n} = 2\left| a_{n} \right|$ or $\left| a_{n} \right| + a_{n} = 0.$ Therefore, $0 \leq \left| a_{n} \right| + a_{n} \leq 2\left| a_{n} \right|.$ Consequently, by the comparison test, since $2\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ converges, the series

假设 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ 收敛。我们利用 $\left. a_{n} = \middle| a_{n} \right|$ 或 $\left. a_{n} = \text{−} \middle| a_{n} \right|$ 这一事实来证明，因此 $\left| a_{n} \right| + a_{n} = 2\left| a_{n} \right|$ 或 $\left| a_{n} \right| + a_{n} = 0.$ 于是 $0 \leq \left| a_{n} \right| + a_{n} \leq 2\left| a_{n} \right|.$ 因此，由比较判别法，由于 $2\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ 收敛，级数

$$\sum\limits_{n = 1}^{\infty}(\left| a_{n} \right| + a_{n})$$

converges. By using the algebraic properties for convergent series, we conclude that

收敛。利用收敛级数的代数性质，我们得出结论

$$\sum\limits_{n = 1}^{\infty}a_{n} = \sum\limits_{n = 1}^{\infty}(\left| a_{n} \right| + a_{n})\text{−}\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$$

converges.

收敛。

□

Absolute versus Conditional Convergence 绝对收敛与条件收敛的比较

For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.

对于下列各个级数，判断该级数是绝对收敛、条件收敛，还是发散。

1. $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}(3n + 1)$

2. $\sum\limits_{n = 1}^{\infty}\text{cos}(n)\text{/}n^{2}$

Solution 解答

1. We can see that

1. 我们可以看出

$$\sum\limits_{n = 1}^{\infty}\left| \frac{(-1)^{n + 1}}{3n + 1} \right| = \sum\limits_{n = 1}^{\infty}\frac{1}{3n + 1}$$

diverges by using the limit comparison test with the harmonic series.

通过与调和级数作极限比较判别法可知其发散。

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}(3n + 1)}{1\text{/}n} = \frac{1}{3}$$

Thus, applying Theorem 5.13, the series cannot converge absolutely. Moreover, because of the alternating series test, we can see that the series converges.

因此，应用定理 5.13，该级数不可能绝对收敛。此外，由于交错级数判别法，我们可以看出该级数收敛。

$$\frac{1}{3\left( {n + 1} \right) + 1} < \frac{1}{3n + 1}\ \text{and}\ \frac{1}{3n + 1}\rightarrow 0,$$

We can conclude that $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}(3n + 1)$ converges conditionally.

我们可以得出结论：$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}(3n + 1)$ 条件收敛。

2. Noting that $\left| \text{cos}\mspace{2mu} n \middle| \leq 1, \right.$ to determine whether the series converges absolutely, compare

2. 注意到 $\left| \text{cos}\mspace{2mu} n \middle| \leq 1, \right.$ 为了判断该级数是否绝对收敛，比较

$$\sum\limits_{n = 1}^{\infty}\left| \frac{\text{cos}\mspace{2mu} n}{n^{2}} \right|$$

with the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}.$ Since $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}$ converges, by the comparison test, $\sum\limits_{n = 1}^{\infty}|\text{cos}\mspace{2mu} n\text{/}n^{2}|$ converges, and therefore $\sum\limits_{n = 1}^{\infty}\text{cos}\mspace{2mu} n\text{/}n^{2}$ converges absolutely.

与级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}$ 比较。由于 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{2}$ 收敛，由比较判别法，$\sum\limits_{n = 1}^{\infty}|\text{cos}\mspace{2mu} n\text{/}n^{2}|$ 收敛，因此 $\sum\limits_{n = 1}^{\infty}\text{cos}\mspace{2mu} n\text{/}n^{2}$ 绝对收敛。

Determine whether the series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}(2n^{3} + 1)$ converges absolutely, converges conditionally, or diverges.

判断级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}(2n^{3} + 1)$ 是绝对收敛、条件收敛，还是发散。

To see the difference between absolute and conditional convergence, look at what happens when we *rearrange* the terms of the alternating harmonic series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n.$ We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.

为了看清绝对收敛与条件收敛之间的区别，来看当我们*重排*交错调和级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n$ 的各项时会发生什么。我们证明可以重排各项，使得新级数发散。当然，如果重排一个有限和的各项，其和不会改变。然而，当我们处理无穷和时，有趣的事情就可能发生。

Begin by adding enough of the positive terms to produce a sum that is larger than some real number $M > 0.$ For example, let $M = 10,$ and find an integer $k$ such that

先加入足够多的正项，使所得的和大于某个实数 $M > 0.$ 例如，令 $M = 10,$ 并求出整数 $k$ 使得

$$1 + \frac{1}{3} + \frac{1}{5} + \text{⋯} + \frac{1}{2k - 1} > 10.$$

(We can do this because the series $\sum\limits_{n = 1}^{\infty}1\text{/}(2n - 1)$ diverges to infinity.) Then subtract $1\text{/}2.$ Then add more positive terms until the sum reaches 100. That is, find another integer $j > k$ such that

（我们可以做到这一点，因为级数 $\sum\limits_{n = 1}^{\infty}1\text{/}(2n - 1)$ 发散到无穷大。）然后减去 $1\text{/}2.$ 再加入更多正项，直到和达到 100。也就是说，求出另一个整数 $j > k$ 使得

$$1 + \frac{1}{3} + \text{⋯} + \frac{1}{2k - 1} - \frac{1}{2} + \frac{1}{2k + 1} + \text{⋯} + \frac{1}{2j + 1} > 100.$$

Then subtract $1\text{/}4.$ Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.

然后减去 $1\text{/}4.$ 照此继续下去，我们找到了一种重排交错调和级数各项的方法，使得重排后级数的部分和数列无界，因而发散。

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In Example 5.22, we show how to rearrange the terms to create a new series that converges to $3\mspace{2mu}\text{ln}(2)\text{/}2.$ We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number $r;$ however, the proof of that fact is beyond the scope of this text.

交错调和级数的各项也可以重排，使得新级数收敛到一个不同的值。在示例 5.22 中，我们展示如何重排各项，以构造一个收敛到 $3\mspace{2mu}\text{ln}(2)\text{/}2$ 的新级数。我们指出，交错调和级数可以重排成收敛到任意实数 $r$ 的级数；然而，这一事实的证明超出了本书的范围。

In general, any series $\sum\limits_{n = 1}^{\infty}a_{n}$ that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series $\sum\limits_{n = 1}^{\infty}a_{n}$ that converges absolutely, the value of $\sum\limits_{n = 1}^{\infty}a_{n}$ is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.

一般地，任何条件收敛的级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 都可以重排，使得新级数发散或收敛到一个不同的实数。绝对收敛的级数不具有这一性质。对于任何绝对收敛的级数 $\sum\limits_{n = 1}^{\infty}a_{n}$，无论怎样重排各项，$\sum\limits_{n = 1}^{\infty}a_{n}$ 的值都相同。这一结果称为黎曼重排定理，超出了本书的范围。

Rearranging Series 重排级数

Use the fact that

利用如下事实

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \text{⋯} = \text{ln}\mspace{2mu} 2$$

to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is $3\mspace{2mu}\text{ln}(2)\text{/}2.$

重排交错调和级数的各项，使得重排后级数的和为 $3\mspace{2mu}\text{ln}(2)\text{/}2.$

Solution 解答

Let

令

$${\sum\limits_{n = 1}^{\infty}a_{n}} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \text{⋯}.$$

Since ${\sum\limits_{n = 1}^{\infty}a_{n}} = \text{ln}(2),$ by the algebraic properties of convergent series,

由于 ${\sum\limits_{n = 1}^{\infty}a_{n}} = \text{ln}(2),$ 由收敛级数的代数性质，

$${\sum\limits_{n = 1}^{\infty}\frac{1}{2}}a_{n} = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \text{⋯} = \frac{1}{2}{\sum\limits_{n = 1}^{\infty}a_{n}} = \frac{\text{ln}\mspace{2mu} 2}{2}.$$

Now introduce the series $\sum\limits_{n = 1}^{\infty}b_{n}$ such that for all $n \geq 1,$ $b_{2n - 1} = 0$ and $b_{2n} = a_{n}\text{/}2.$ Then

现在引入级数 $\sum\limits_{n = 1}^{\infty}b_{n}$，使得对所有 $n \geq 1,$ $b_{2n - 1} = 0$ 且 $b_{2n} = a_{n}\text{/}2.$ 于是

$${\sum\limits_{n = 1}^{\infty}b_{n}} = 0 + \frac{1}{2} + 0 - \frac{1}{4} + 0 + \frac{1}{6} + 0 - \frac{1}{8} + \text{⋯} = \frac{\text{ln}\mspace{2mu} 2}{2}.$$

Then using the algebraic limit properties of convergent series, since $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ converge, the series $\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}$ converges and

然后利用收敛级数的代数极限性质，由于 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$ 都收敛，级数 $\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}$ 收敛且

$${\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}} = {\sum\limits_{n = 1}^{\infty}a_{n}} + {\sum\limits_{n = 1}^{\infty}b_{n}} = \text{ln}\mspace{2mu} 2 + \frac{\text{ln}\mspace{2mu} 2}{2} = \frac{3\mspace{2mu}\text{ln}\mspace{2mu} 2}{2}.$$

Now adding the corresponding terms, $a_{n}$ and $b_{n},$ we see that

现在把对应项 $a_{n}$ 与 $b_{n}$ 相加，我们看到

$$\begin{matrix} {\sum\limits_{n = 1}^{\infty}\left( a_{n} + b_{n} \right)} & {= (1 + 0) + \left( - \frac{1}{2} + \frac{1}{2} \right) + \left( \frac{1}{3} + 0 \right) + \left( - \frac{1}{4} - \frac{1}{4} \right) + \left( \frac{1}{5} + 0 \right) + \left( - \frac{1}{6} + \frac{1}{6} \right)} \\ & \\ & {\mspace{27mu} + \left( \frac{1}{7} + 0 \right) + \left( - \frac{1}{8} - \frac{1}{8} \right) + \text{⋯}} \\ & {= 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \text{⋯}.} \end{matrix}$$

We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since ${\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}} = 3\mspace{2mu}\text{ln}(2)\text{/}2,$ we conclude that

我们注意到等号右侧的级数是交错调和级数的一个重排。由于 ${\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}} = 3\mspace{2mu}\text{ln}(2)\text{/}2,$ 我们得出结论

$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \text{⋯} = \frac{3\mspace{2mu}\text{ln}(2)}{2}.$$

Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

因此，我们找到了一个具有所需性质的交错调和级数的重排。

Section 5.5 Exercises 5.5 节习题

State whether each of the following series converges absolutely, conditionally, or not at all.

判断下列各级数是绝对收敛、条件收敛，还是根本不收敛。

250\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{n}{n + 3}$

251.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{\sqrt{n} + 1}{\sqrt{n} + 3}$

252\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{1}{\sqrt{n + 3}}$

253.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{\sqrt{n + 3}}{n}$

254\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{1}{n\text{!}}$

255.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{3^{n}}{n\text{!}}$

256\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( \frac{n - 1}{n} \right)^{n}$

257.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( \frac{n + 1}{n} \right)^{n}$

258\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{sin}^{2}n$

259.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{cos}^{2}n$

260\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{sin}^{2}\left( {1\text{/}n} \right)$

261.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{cos}^{2}\left( {1\text{/}n} \right)$

262\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{ln}\left( {1\text{/}n} \right)$

263.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{ln}\left( {1 + \frac{1}{n}} \right)$

264\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{n^{2}}{1 + n^{4}}$

265.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{n^{e}}{1 + n^{\pi}}$

266\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}2^{1\text{/}n}$

267.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n^{1\text{/}n}$

268\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n}\left( {1 - n^{1\text{/}n}} \right)$ (*Hint:* $n^{1\text{/}n} \approx 1 + \text{ln}(n)\text{/}n$ for large $n.)$

$\sum\limits_{n = 1}^{\infty}(-1)^{n}\left( {1 - n^{1\text{/}n}} \right)$ （提示：$n^{1\text{/}n} \approx 1 + \text{ln}(n)\text{/}n$，当 $n$ 较大时）

269.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\left( {1 - \text{cos}\left( \frac{1}{n} \right)} \right)$ (*Hint:* $\text{cos}(1\text{/}n) \approx 1 - 1\text{/}n^{2}$ for large $n.)$

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\left( {1 - \text{cos}\left( \frac{1}{n} \right)} \right)$ （提示：$\text{cos}(1\text{/}n) \approx 1 - 1\text{/}n^{2}$，当 $n$ 较大时）

270\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\sqrt{n + 1} - \sqrt{n}} \right)$ (*Hint:* Rationalize the numerator.)

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\sqrt{n + 1} - \sqrt{n}} \right)$ （提示：有理化分子。）

271.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}} \right)$ (*Hint:* Find common denominator then rationalize numerator.)

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}} \right)$ （提示：先通分，再有理化分子。）

272\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\text{ln}\left( {n + 1} \right) - \text{ln}\ n} \right)$

273.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\left( {\text{tan}^{-1}\left( {n + 1} \right) - \text{tan}^{-1}n} \right)$ (*Hint:* Use Mean Value Theorem.)

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\left( {\text{tan}^{-1}\left( {n + 1} \right) - \text{tan}^{-1}n} \right)$ （提示：使用均值定理。）

274\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\left( {n + 1} \right)^{2} - n^{2}} \right)$

275.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\frac{1}{n} - \frac{1}{n + 1}} \right)$

276\.

$\sum\limits_{n = 1}^{\infty}\frac{\text{cos}\left( {n\pi} \right)}{n}$

277.

$\sum\limits_{n = 1}^{\infty}\frac{\text{cos}\left( {n\pi} \right)}{n^{1\text{/}n}}$

278\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n}\mspace{2mu}\text{sin}\mspace{2mu}\left( \frac{n\pi}{2} \right)$

279.

$\sum\limits_{n = 1}^{\infty}\text{sin}\left( {n\pi\text{/}2} \right)\text{sin}\left( {1\text{/}n} \right)$

In each of the following problems, use the estimate $\left| R_{N} \right| \leq b_{N + 1}$ to find a value of $N$ that guarantees that the sum of the first $N$ terms of the alternating series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$ differs from the infinite sum by at most the given error. Calculate the partial sum $S_{N}$ for this $N.$

在下列各题中，利用估计 $\left| R_{N} \right| \leq b_{N + 1}$ 求出一个 $N$ 值，使得交错级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$ 前 $N$ 项之和与无穷和之差不超过所给误差。对此 $N$ 计算部分和 $S_{N}$。

280\.

\[T\] $b_{n} = 1\text{/}n,$ error $< 10^{-5}$

\[T\] $b_{n} = 1\text{/}n,$ 误差 $< 10^{-5}$

281.

\[T\] $b_{n} = 1\text{/}\text{ln}(n),$ $n \geq 2,$ error $< 10^{-1}$

\[T\] $b_{n} = 1\text{/}\text{ln}(n),$ $n \geq 2,$ 误差 $< 10^{-1}$

282\.

\[T\] $b_{n} = 1\text{/}\sqrt{n},$ error $< 10^{-3}$

\[T\] $b_{n} = 1\text{/}\sqrt{n},$ 误差 $< 10^{-3}$

283.

\[T\] $b_{n} = 1\text{/}2^{n},$ error $< 10^{-6}$

\[T\] $b_{n} = 1\text{/}2^{n},$ 误差 $< 10^{-6}$

284\.

\[T\] $b_{n} = \text{ln}\left( {1 + \frac{1}{n}} \right),$ error $< 10^{-3}$

\[T\] $b_{n} = \text{ln}\left( {1 + \frac{1}{n}} \right),$ 误差 $< 10^{-3}$

285.

\[T\] $b_{n} = 1\text{/}n^{2},$ error $< 10^{-6}$

\[T\] $b_{n} = 1\text{/}n^{2},$ 误差 $< 10^{-6}$

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

对于下列习题，判断下列各陈述为真还是为假。若陈述为假，给出一个使其为假的反例。

286\.

If $b_{n} \geq 0$ is decreasing and $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0,$ then $\sum\limits_{n = 1}^{\infty}\left( {b_{2n - 1} - b_{2n}} \right)$ converges absolutely.

若 $b_{n} \geq 0$ 递减且 $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0$，则 $\sum\limits_{n = 1}^{\infty}\left( {b_{2n - 1} - b_{2n}} \right)$ 绝对收敛。

287.

If $b_{n} \geq 0$ is decreasing, then $\sum\limits_{n = 1}^{\infty}\left( {b_{2n - 1} - b_{2n}} \right)$ converges absolutely.

若 $b_{n} \geq 0$ 递减，则 $\sum\limits_{n = 1}^{\infty}\left( {b_{2n - 1} - b_{2n}} \right)$ 绝对收敛。

288\.

If $b_{n} \geq 0$ and $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0$ then ${\sum\limits_{n = 1}^{\infty}{(\frac{1}{2}(}}b_{3n - 2} + b_{3n - 1})\text{−}b_{3n})$ converges.

若 $b_{n} \geq 0$ 且 $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0$，则 ${\sum\limits_{n = 1}^{\infty}{(\frac{1}{2}(}}b_{3n - 2} + b_{3n - 1})\text{−}b_{3n})$ 收敛。

289.

If $b_{n} \geq 0$ is decreasing and $\sum\limits_{n = 1}^{\infty}{(b_{3n - 2} + b_{3n - 1} - b_{3n})}$ converges then $\sum\limits_{n = 1}^{\infty}b_{3n - 2}$ converges.

若 $b_{n} \geq 0$ 递减且 $\sum\limits_{n = 1}^{\infty}{(b_{3n - 2} + b_{3n - 1} - b_{3n})}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}b_{3n - 2}$ 收敛。

290\.

If $b_{n} \geq 0$ is decreasing and $\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}b_{n}}$ converges conditionally but not absolutely, then $b_{n}$ does not tend to zero.

若 $b_{n} \geq 0$ 递减且 $\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}b_{n}}$ 条件收敛但不绝对收敛，则 $b_{n}$ 不趋于零。

291.

Let $a_{n}^{+} = a_{n}$ if $a_{n} \geq 0$ and $a_{n}^{-} = \text{−}a_{n}$ if $a_{n} < 0.$ (Also, $a_{n}^{+} = 0\ \text{if}\ a_{n} < 0$ and $a_{n}^{-} = 0\ \text{if}\ a_{n} \geq 0.)$ If $\sum\limits_{n = 1}^{\infty}a_{n}$ converges conditionally but not absolutely, then neither $\sum\limits_{n = 1}^{\infty}a_{n}^{+}$ nor $\sum\limits_{n = 1}^{\infty}a_{n}^{-}$ converge.

设当 $a_{n} \geq 0$ 时 $a_{n}^{+} = a_{n}$，当 $a_{n} < 0$ 时 $a_{n}^{-} = \text{−}a_{n}$。（同样，$a_{n}^{+} = 0\ \text{if}\ a_{n} < 0$，$a_{n}^{-} = 0\ \text{if}\ a_{n} \geq 0$。）若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 条件收敛但不绝对收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}^{+}$ 与 $\sum\limits_{n = 1}^{\infty}a_{n}^{-}$ 均不收敛。

292\.

Suppose that $a_{n}$ is a sequence of positive real numbers and that $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

设 $a_{n}$ 是一个正实数的数列，且 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

Suppose that $b_{n}$ is an arbitrary sequence of ones and minus ones. Does ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ necessarily converge?

设 $b_{n}$ 是由 $1$ 与 $-1$ 任意组成的数列。${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ 是否必然收敛？

293.

Suppose that $a_{n}$ is a sequence such that ${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ converges for every possible sequence $b_{n}$ of zeros and ones. Does $\sum\limits_{n = 1}^{\infty}a_{n}$ converge absolutely?

设 $a_{n}$ 是一个数列，使得对任意由 $0$ 和 $1$ 组成的数列 $b_{n}$，${\sum\limits_{n = 1}^{\infty}a_{n}}b_{n}$ 都收敛。$\sum\limits_{n = 1}^{\infty}a_{n}$ 是否绝对收敛？

The following series do not satisfy the hypotheses of the alternating series test as stated.

下列级数不满足所陈述的交错级数判别法的假设条件。

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

在每种情形下，说明哪个假设未被满足。说明该级数是否绝对收敛。

294\.

$\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{\text{sin}^{2}n}{n}}$

295.

$\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{\text{cos}^{2}n}{n}}$

296\.

$1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + \text{⋯}$

297.

$1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + \text{⋯}$

298\.

Show that the alternating series $1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{4} - \frac{1}{8} + \text{⋯}$ does

证明交错级数 $1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} + \frac{1}{4} - \frac{1}{8} + \text{⋯}$

not converge. What hypothesis of the alternating series test is not met?

不收敛。交错级数判别法的哪个假设未被满足？

299.

Suppose that $\sum a_{n}$ converges absolutely. Show that the series consisting of the positive terms $a_{n}$ also converges.

设 $\sum a_{n}$ 绝对收敛。证明由正项 $a_{n}$ 组成的级数也收敛。

300\.

Show that the alternating series $\frac{2}{3} - \frac{3}{5} + \frac{4}{7} - \frac{5}{9} + \text{⋯}$ does not converge. What hypothesis of the alternating series test is not met?

证明交错级数 $\frac{2}{3} - \frac{3}{5} + \frac{4}{7} - \frac{5}{9} + \text{⋯}$ 不收敛。交错级数判别法的哪个假设未被满足？

301.

The formula $\text{cos}\mspace{2mu}\theta = 1 - \frac{\theta^{2}}{2\text{!}} + \frac{\theta^{4}}{4\text{!}} - \frac{\theta^{6}}{6\text{!}} + \text{⋯}$ will be derived in the next chapter. Use the remainder $\left| R_{N} \middle| \leq b_{N + 1} \right.$ to find a bound for the error in estimating $\text{cos}\mspace{2mu}\theta$ by the fifth partial sum $1 - \theta^{2}\text{/}2\text{!} + \theta^{4}\text{/}4\text{!}\text{−}\theta^{6}\text{/}6\text{!} + \theta^{8}\text{/}8\text{!}$ for $\theta = 1,$ $\theta = \pi\text{/}6,$ and $\theta = \pi.$

公式 $\text{cos}\mspace{2mu}\theta = 1 - \frac{\theta^{2}}{2\text{!}} + \frac{\theta^{4}}{4\text{!}} - \frac{\theta^{6}}{6\text{!}} + \text{⋯}$ 将在下一章推导。利用余项 $\left| R_{N} \middle| \leq b_{N + 1} \right.$ 求用第五个部分和 $1 - \theta^{2}\text{/}2\text{!} + \theta^{4}\text{/}4\text{!}\text{−}\theta^{6}\text{/}6\text{!} + \theta^{8}\text{/}8\text{!}$ 估计 $\text{cos}\mspace{2mu}\theta$ 的误差界，分别取 $\theta = 1,$ $\theta = \pi\text{/}6,$ 和 $\theta = \pi.$

302\.

The formula $\text{sin}\mspace{2mu}\theta = \theta - \frac{\theta^{3}}{3\text{!}} + \frac{\theta^{5}}{5\text{!}} - \frac{\theta^{7}}{7\text{!}} + \text{⋯}$ will be derived in the next chapter. Use the remainder $\left| R_{N} \middle| \leq b_{N + 1} \right.$ to find a bound for the error in estimating $\text{sin}\mspace{2mu}\theta$ by the fifth partial sum $\theta - \theta^{3}\text{/}3\text{!}\text{−}\theta^{5}\text{/}5\text{!}\text{−}\theta^{7}\text{/}7\text{!} + \theta^{9}\text{/}9\text{!}$ for $\theta = 1,$ $\theta = \pi\text{/}6,$ and $\theta = \pi.$

公式 $\text{sin}\mspace{2mu}\theta = \theta - \frac{\theta^{3}}{3\text{!}} + \frac{\theta^{5}}{5\text{!}} - \frac{\theta^{7}}{7\text{!}} + \text{⋯}$ 将在下一章推导。利用余项 $\left| R_{N} \middle| \leq b_{N + 1} \right.$ 求用第五个部分和 $\theta - \theta^{3}\text{/}3\text{!}\text{−}\theta^{5}\text{/}5\text{!}\text{−}\theta^{7}\text{/}7\text{!} + \theta^{9}\text{/}9\text{!}$ 估计 $\text{sin}\mspace{2mu}\theta$ 的误差界，分别取 $\theta = 1,$ $\theta = \pi\text{/}6,$ 和 $\theta = \pi.$

303.

How many terms in $\text{cos}\mspace{2mu}\theta = 1 - \frac{\theta^{2}}{2\text{!}} + \frac{\theta^{4}}{4\text{!}} - \frac{\theta^{6}}{6\text{!}} + \text{⋯}$ are needed to approximate $\text{cos}\mspace{2mu} 1$ accurate to an error of at most $0.00001\text{?}$

要使得 $\text{cos}\mspace{2mu} 1$ 的近似值误差至多为 $0.00001$，需要 $\text{cos}\mspace{2mu}\theta = 1 - \frac{\theta^{2}}{2\text{!}} + \frac{\theta^{4}}{4\text{!}} - \frac{\theta^{6}}{6\text{!}} + \text{⋯}$ 中的多少项？

304\.

How many terms in $\text{sin}\mspace{2mu}\theta = \theta - \frac{\theta^{3}}{3\text{!}} + \frac{\theta^{5}}{5\text{!}} - \frac{\theta^{7}}{7\text{!}} + \text{⋯}$ are needed to approximate $\text{sin}\mspace{2mu} 1$ accurate to an error of at most $0.00001\text{?}$

要使得 $\text{sin}\mspace{2mu} 1$ 的近似值误差至多为 $0.00001$，需要 $\text{sin}\mspace{2mu}\theta = \theta - \frac{\theta^{3}}{3\text{!}} + \frac{\theta^{5}}{5\text{!}} - \frac{\theta^{7}}{7\text{!}} + \text{⋯}$ 中的多少项？

305.

Sometimes the alternating series ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n - 1}}b_{n}$ converges to a certain fraction of an absolutely convergent series $\sum\limits_{n = 1}^{\infty}b_{n}$ at a faster rate. Given that ${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}} = \frac{\pi^{2}}{6},$ find $12 = 1 - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \text{⋯}.$ Which of the series $6{\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}}$ and $S{\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n^{2}}}$ gives a better estimation of $\pi^{2}$ using $1000$ terms?

有时交错级数 ${\sum\limits_{n = 1}^{\infty}{(-1)}^{n - 1}}b_{n}$ 会以更快的速率收敛到绝对收敛级数 $\sum\limits_{n = 1}^{\infty}b_{n}$ 的某个分数倍。已知 ${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}} = \frac{\pi^{2}}{6}$，求 $12 = 1 - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \text{⋯}$。在使用 $1000$ 项时，级数 $6{\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}}$ 与 $S{\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n^{2}}}$ 哪个对 $\pi^{2}$ 的估计更好？

The following alternating series converge to given multiples of $\pi.$ Find the value of $N$ predicted by the remainder estimate such that the $N\text{th}$ partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum $N$ for which the error bound holds, and give the desired approximate value in each case. Up to $15$ decimals places, $\pi = 3.141592653589793\text{…}.$

下列交错级数收敛到 $\pi$ 的给定倍数。由余项估计求出使该级数的第 $N$ 个部分和在给定的误差范围内精确逼近左端的 $N$ 值。找出使误差界成立的最小 $N$，并在每种情形下给出所需的近似值。精确到 $15$ 位小数，$\pi = 3.141592653589793\text{…}.$

306\.

\[T\] $\frac{\pi}{4} = {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}},$ error $< 0.0001$

\[T\] $\frac{\pi}{4} = {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}},$ 误差 $< 0.0001$

307.

\[T\] $\frac{\pi}{\sqrt{12}} = {\sum\limits_{k = 0}^{\infty}\frac{(-3)^{\text{−}k}}{2k + 1}},$ error $< 0.0001$

\[T\] $\frac{\pi}{\sqrt{12}} = {\sum\limits_{k = 0}^{\infty}\frac{(-3)^{\text{−}k}}{2k + 1}},$ 误差 $< 0.0001$

308\.

\[T\] The series $\sum\limits_{n = 0}^{\infty}\frac{\text{sin}\left( {x + \pi n} \right)}{x + \pi n}$ plays an important role in signal processing. Show that $\sum\limits_{n = 0}^{\infty}\frac{\text{sin}\left( {x + \pi n} \right)}{x + \pi n}$ converges whenever $0 < x < \pi.$ (*Hint:* Use the formula for the sine of a sum of angles.)

\[T\] 级数 $\sum\limits_{n = 0}^{\infty}\frac{\text{sin}\left( {x + \pi n} \right)}{x + \pi n}$ 在信号处理中起着重要作用。证明：只要 $0 < x < \pi$，$\sum\limits_{n = 0}^{\infty}\frac{\text{sin}\left( {x + \pi n} \right)}{x + \pi n}$ 就收敛。（提示：使用两角和的正弦公式。）

309.

\[T\] If ${\sum\limits_{n = 1}^{N}{(-1)^{n - 1}\frac{1}{n}}}\rightarrow\text{ln}\mspace{2mu} 2,$ what is $1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{8} - \frac{1}{10} - \frac{1}{12} + \text{⋯}\text{?}$

\[T\] 若 ${\sum\limits_{n = 1}^{N}{(-1)^{n - 1}\frac{1}{n}}}\rightarrow\text{ln}\mspace{2mu} 2$，则 $1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{8} - \frac{1}{10} - \frac{1}{12} + \text{⋯}$ 是多少？

310\.

\[T\] Plot the series $\sum\limits_{n = 1}^{100}\frac{\text{cos}\left( {2\pi nx} \right)}{n}$ for $0 \leq x < 1.$ Explain why $\sum\limits_{n = 1}^{100}\frac{\text{cos}\left( {2\pi nx} \right)}{n}$ diverges when $x = 0,1.$ How does the series behave for other $x\text{?}$

\[T\] 对 $0 \leq x < 1$ 画出级数 $\sum\limits_{n = 1}^{100}\frac{\text{cos}\left( {2\pi nx} \right)}{n}$ 的图像。解释为何当 $x = 0,1$ 时 $\sum\limits_{n = 1}^{100}\frac{\text{cos}\left( {2\pi nx} \right)}{n}$ 发散。对于其他 $x$，该级数表现如何？

311.

\[T\] Plot the series $\sum\limits_{n = 1}^{100}\frac{\text{sin}\left( {2\pi nx} \right)}{n}$ for $0 \leq x < 1$ and comment on its behavior

\[T\] 对 $0 \leq x < 1$ 画出级数 $\sum\limits_{n = 1}^{100}\frac{\text{sin}\left( {2\pi nx} \right)}{n}$ 的图像，并说明其行为。

312\.

\[T\] Plot the series $\sum\limits_{n = 1}^{100}\frac{\text{cos}(2\pi nx)}{n^{2}}$ for $0 \leq x < 1$ and describe its graph.

\[T\] 对 $0 \leq x < 1$ 画出级数 $\sum\limits_{n = 1}^{100}\frac{\text{cos}(2\pi nx)}{n^{2}}$ 的图像，并描述其图形。

313.

\[T\] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form ${\sum\limits_{n = 1}^{\infty}\frac{S_{n}}{n}},$ where $s_{n}$ is a randomly generated sequence of $\pm 1\text{'s}$ in which the values $\pm 1$ are equally likely to occur. Use a random number generator to produce $1000$ random $\pm 1\text{s}$ and plot the partial sums $S_{N} = {\sum\limits_{n = 1}^{N}\frac{s_{n}}{n}}$ of your random harmonic sequence for $N = 1$ to $1000.$ Compare to a plot of the first $1000$ partial sums of the harmonic series.

\[T\] 交错调和级数因其各项之间的抵消而收敛。其和已知，是因为这种抵消可以显式描述。随机调和级数是形如 ${\sum\limits_{n = 1}^{\infty}\frac{S_{n}}{n}}$ 的级数，其中 $s_{n}$ 是一个随机生成的由 $\pm 1$ 组成的序列，且每个值 $\pm 1$ 出现的概率相等。使用随机数生成器产生 $1000$ 个随机的 $\pm 1$，并画出你的随机调和序列在 $N = 1$ 到 $1000$ 时的部分和 $S_{N} = {\sum\limits_{n = 1}^{N}\frac{s_{n}}{n}}$。将其与调和级数前 $1000$ 个部分和的图像进行比较。

314\.

\[T\] Estimates of $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ can be *accelerated* by writing its partial sums as ${\sum\limits_{n = 1}^{N}\frac{1}{n^{2}}} = {\sum\limits_{n = 1}^{N}\frac{1}{n(n + 1)}} + {\sum\limits_{n = 1}^{N}\frac{1}{n^{2}(n + 1)}}$ and recalling that ${\sum\limits_{n = 1}^{N}\frac{1}{n(n + 1)}} = 1 - \frac{1}{N + 1}$ converges to one as $N\rightarrow\infty.$ Compare the estimate of $\pi^{2}\text{/}6$ using the sums $\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}}$ with the estimate using $1 + {\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}(n + 1)}}.$

\[T\] 把 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ 的部分和写成 ${\sum\limits_{n = 1}^{N}\frac{1}{n^{2}}} = {\sum\limits_{n = 1}^{N}\frac{1}{n(n + 1)}} + {\sum\limits_{n = 1}^{N}\frac{1}{n^{2}(n + 1)}}$，并注意到 ${\sum\limits_{n = 1}^{N}\frac{1}{n(n + 1)}} = 1 - \frac{1}{N + 1}$ 当 $N\rightarrow\infty$ 时收敛到 $1$，就可以*加速*对 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ 的估计。比较用 $\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}}$ 得到的 $\pi^{2}\text{/}6$ 的估计值与用 $1 + {\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}(n + 1)}}$ 得到的估计值。

315.

\[T\] The *Euler transform* rewrites $S = {\sum\limits_{n = 0}^{\infty}{(-1)}^{n}}b_{n}$ as $S = {\sum\limits_{n = 0}^{\infty}{(-1)}^{n}}2^{\text{−}n - 1}{\sum\limits_{m = 0}^{n}\begin{pmatrix}

\[T\] 欧拉变换将 $S = {\sum\limits_{n = 0}^{\infty}{(-1)}^{n}}b_{n}$ 改写为 $S = {\sum\limits_{n = 0}^{\infty}{(-1)}^{n}}2^{\text{−}n - 1}{\sum\limits_{m = 0}^{n}\begin{pmatrix}

n \\

\end{pmatrix}}b_{n - m}.$ For the alternating harmonic series, it takes the form $\text{ln}(2) = {\sum\limits_{n = 1}^{\infty}\frac{{(-1)}^{n - 1}}{n}} = {\sum\limits_{n = 1}^{\infty}\frac{1}{n2^{n}}}.$ Compute partial sums of $\sum\limits_{n = 1}^{\infty}\frac{1}{n2^{n}}$ until they approximate $\text{ln}(2)$ accurate to within $0.0001.$ How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate $\text{ln}(2).$

\end{pmatrix}}b_{n - m}.$ 对于交错调和级数，它表现为 $\text{ln}(2) = {\sum\limits_{n = 1}^{\infty}\frac{{(-1)}^{n - 1}}{n}} = {\sum\limits_{n = 1}^{\infty}\frac{1}{n2^{n}}}$。计算 $\sum\limits_{n = 1}^{\infty}\frac{1}{n2^{n}}$ 的部分和，直到它们对 $\text{ln}(2)$ 的逼近精确到 $0.0001$ 以内。需要多少项？将此答案与估计 $\text{ln}(2)$ 所需的交错调和级数的项数进行比较。

316\.

\[T\] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If $a_{n} \geq 0$ is such that $a_{n}\rightarrow 0$ as $n\rightarrow\infty$ but $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges, then, given any number $A$ there is a sequence $s_{n}$ of $\pm 1\text{'s}$ such that ${\sum\limits_{n = 1}^{\infty}a_{n}}s_{n}\rightarrow A.$ Show this for $A > 0$ as follows.

\[T\] 正文中曾指出，一个条件收敛的级数可以重排以收敛到任意数值。这里有一个更简单但类似的结论。若 $a_{n} \geq 0$ 满足 $a_{n}\rightarrow 0$（当 $n\rightarrow\infty$）但 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散，则对任意数 $A$，存在一个由 $\pm 1$ 组成的数列 $s_{n}$，使得 ${\sum\limits_{n = 1}^{\infty}a_{n}}s_{n}\rightarrow A$。对 $A > 0$ 按如下方式证明这一结论。

1. Recursively define $s_{n}$ by $s_{n} = 1$ if $S_{n - 1} = {\sum\limits_{k = 1}^{n - 1}a_{k}}s_{k} < A$ and $s_{n} = -1$ otherwise.

1. 递归定义 $s_{n}$：若 $S_{n - 1} = {\sum\limits_{k = 1}^{n - 1}a_{k}}s_{k} < A$，则令 $s_{n} = 1$；否则令 $s_{n} = -1$。

2. Explain why eventually $S_{n} \geq A,$ and for any $m$ larger than this $n,$ $A - a_{m} \leq S_{m} \leq A + a_{m}.$

2. 解释为何最终有 $S_{n} \geq A$，并且对于大于此 $n$ 的任意 $m$，都有 $A - a_{m} \leq S_{m} \leq A + a_{m}.$

3. Explain why this implies that $S_{n}\rightarrow A$ as $n\rightarrow\infty.$

3. 解释为何这意味着当 $n\rightarrow\infty$ 时 $S_{n}\rightarrow A$。

5.6 Ratio and Root Tests 5.6 比值判别法与根值判别法

5.6.1 Use the ratio test to determine absolute convergence of a series.
5.6.2 Use the root test to determine absolute convergence of a series.
5.6.3 Describe a strategy for testing the convergence of a given series.

5.6.1 使用比值判别法判断级数是否绝对收敛。
5.6.2 使用根值判别法判断级数是否绝对收敛。
5.6.3 描述检验给定级数收敛性的策略。

In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because they do not require us to find a comparable series. The ratio test will be especially useful in the discussion of power series in the next chapter.

在本节中，我们证明最后两个级数收敛判别法：比值判别法与根值判别法。这两个判别法特别好用，因为它们不需要我们去找一个可比较的级数。在下一章关于幂级数的讨论中，比值判别法将尤其有用。

Throughout this chapter, we have seen that no single convergence test works for all series. Therefore, at the end of this section we discuss a strategy for choosing which convergence test to use for a given series.

在本章中我们已看到，没有任何单一的收敛判别法适用于所有级数。因此，在本节末尾我们将讨论如何选择用于给定级数的收敛判别法。

Ratio Test 比值判别法

Consider a series ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ From our earlier discussion and examples, we know that $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0$ is not a sufficient condition for the series to converge. Not only do we need $a_{n}\rightarrow 0,$ but we need $a_{n}\rightarrow 0$ quickly enough. For example, consider the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ and the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}.$ We know that $1\text{/}n\rightarrow 0$ and $1\text{/}n^{2}\rightarrow 0.$ However, only the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ converges. The series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ diverges because the terms in the sequence $\left\{ 1\text{/}n \right\}$ do not approach zero fast enough as $n\rightarrow\infty.$ Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

考虑级数 ${\sum\limits_{n = 1}^{\infty}a_{n}}$。根据我们前面的讨论与示例可知，$\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0$ 并不是级数收敛的充分条件。我们不仅需要 $a_{n}\rightarrow 0,$ 而且需要 $a_{n}\rightarrow 0$ 的速度足够快。例如，考虑级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ 与级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$。我们知道 $1\text{/}n\rightarrow 0$ 且 $1\text{/}n^{2}\rightarrow 0$。然而，只有级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{2}$ 收敛。级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ 发散，因为数列 $\left\{ 1\text{/}n \right\}$ 中的项当 $n\rightarrow\infty$ 时趋于零的速度不够快。这里我们引入比值判别法，它提供了一种衡量级数各项趋于零快慢的方法。

Ratio Test 比值判别法

Let $\sum\limits_{n = 1}^{\infty}a_{n}$ be a series with nonzero terms. Let

设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项均非零的级数。令

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right|.$$

1. If $0 \leq \rho < 1,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely.

1. 若 $0 \leq \rho < 1,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛。

2. If $\rho > 1$ or $\rho = \infty,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

2. 若 $\rho > 1$ 或 $\rho = \infty,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

3. If $\rho = 1,$ the test does not provide any information.

3. 若 $\rho = 1,$ 该判别法不能提供任何信息。

Proof 证明

Let $\sum\limits_{n = 1}^{\infty}a_{n}$ be a series with nonzero terms.

设 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项均非零的级数。

We begin with the proof of part i. In this case, $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right| < 1.$ Since $0 \leq \rho < 1,$ there exists $R$ such that $0 \leq \rho < R < 1.$ Let $\varepsilon = R - \rho > 0.$ By the definition of limit of a sequence, there exists some integer $N$ such that

我们先证明第 (i) 部分。此时 $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right| < 1$。因为 $0 \leq \rho < 1,$ 存在 $R$ 使得 $0 \leq \rho < R < 1$。令 $\varepsilon = R - \rho > 0$。根据数列极限的定义，存在某个整数 $N$ 使得

$$\left| {\left| \frac{a_{n + 1}}{a_{n}} \right| - \rho} \right| < \varepsilon\ \text{for all}\ n \geq N.$$

Therefore,

因此，

$$\left| \frac{a_{n + 1}}{a_{n}} \right| < \rho + \varepsilon = R\ \text{for all}\ n \geq N$$

and, thus,

因而，

$$\begin{array}{l} \left| a_{N + 1} \middle| < R \middle| a_{N} \right| \\ \left| a_{N + 2} \middle| < R \middle| a_{N + 1} \middle| < R^{2} \middle| a_{N} \right| \\ \left| a_{N + 3} \middle| < R \middle| a_{N + 2} \middle| < R^{2} \middle| a_{N + 1} \middle| < R^{3} \middle| a_{N} \right| \\ \left| a_{N + 4} \middle| < R \middle| a_{N + 3} \middle| < R^{2} \middle| a_{N + 2} \middle| < R^{3} \middle| a_{N + 1} \middle| < R^{4} \middle| a_{N} \right| \\ {\vdots .} \end{array}$$

Since $R < 1,$ the geometric series

由于 $R < 1,$ 几何级数

$$\left. R \middle| a_{N} \middle| + R^{2} \middle| a_{N} \middle| + R^{3} \middle| a_{N} \middle| + \text{⋯} \right.$$

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

收敛。由上述不等式，我们可以应用比较判别法，并得出级数

$$\left| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \middle| a_{N + 3} \middle| + \middle| a_{N + 4} \middle| + \text{⋯} \right.$$

converges. Therefore, since

收敛。因此，由于

$${\sum\limits_{n = 1}^{\infty}|}a_{n}| = {\sum\limits_{n = 1}^{N}|}a_{n}| + {\sum\limits_{n = N + 1}^{\infty}|}a_{n}|$$

where ${\sum\limits_{n = 1}^{N}|}a_{n}|$ is a finite sum and ${\sum\limits_{n = N + 1}^{\infty}|}a_{n}|$ converges, we conclude that ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ converges.

其中 ${\sum\limits_{n = 1}^{N}|}a_{n}|$ 是有限和，而 ${\sum\limits_{n = N + 1}^{\infty}|}a_{n}|$ 收敛，我们得出 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ 收敛。

For part ii.

对于第 (ii) 部分。

Now suppose $\rho = \left| \frac{a_{n + 1}}{a_{n}} \right| > 1$.

现在假设 $\rho = \left| \frac{a_{n + 1}}{a_{n}} \right| > 1$。

There exists some $N$ such that $\left| \frac{a_{n + 1}}{a_{n}} \right| > 1$ for all $n \geq N$.

存在某个 $N$ 使得对一切 $n \geq N$ 都有 $\left| \frac{a_{n + 1}}{a_{n}} \right| > 1$。

Then $\left| a_{n -} \right| > \left| a_{n} \right|$.

则 $\left| a_{n -} \right| > \left| a_{n} \right|$。

Now because $\left| a_{n -} \right| > \left| a_{n} \right|$ for all $n \geq N$, we know that $\left| a_{n} \right| \neq 0$.

现在因为对一切 $n \geq N$ 都有 $\left| a_{n -} \right| > \left| a_{n} \right|$，我们知道 $\left| a_{n} \right| \neq 0$。

It follows that $\left| a_{n} \right| \neq 0$.

由此可见 $\left| a_{n} \right| \neq 0$。

By the divergence test, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

由发散判别法，级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

$$\left| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \middle| a_{N + 3} \middle| + \text{⋯} \right.$$

diverges, and therefore the series ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ diverges.

发散，因而级数 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ 发散。

For part iii. we show that the test does not provide any information if $\rho = 1$ by considering the $p - \text{series}$ $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}.$ For any real number $p,$

对于第 (iii) 部分，我们通过考察 $p - \text{series}$ $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 来说明当 $\rho = 1$ 时该判别法不能提供任何信息。对任意实数 $p,$

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}\left( {n + 1} \right)^{p}}{1\text{/}n^{p}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{p}}{\left( {n + 1} \right)^{p}} = 1.$$

However, we know that if $p \leq 1,$ the $p - \text{series}$ $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ diverges, whereas $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ converges if $p > 1.$

然而，我们知道若 $p \leq 1,$ 则 $p - \text{series}$ $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 发散；而若 $p > 1,$ $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 收敛。

□

The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.

比值判别法对那些项中含有阶乘或指数的级数特别有用，此时项的比值可以简化表达式。比值判别法很方便，因为它不要求我们找一个比较级数。其缺点是，该判别法有时无法提供关于收敛性的任何信息。

Using the Ratio Test 使用比值判别法

For each of the following series, use the ratio test to determine whether the series converges or diverges.

对于下列各个级数，使用比值判别法判断其收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{n\text{!}}$

2. $\sum\limits_{n = 1}^{\infty}\frac{n^{n}}{n\text{!}}$

3. $\sum\limits_{n = 1}^{\infty}\frac{{(-1)}^{n}{(n\text{!})}^{2}}{(2n)\text{!}}$

Solution 解答

1. From the ratio test, we can see that

1. 由比值判别法，我们可以看出

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n + 1}\text{/}(n + 1)\text{!}}{2^{n}\text{/}n\text{!}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n + 1}}{(n + 1)\text{!}} \cdot \frac{n\text{!}}{2^{n}}.$$

Since $(n + 1)\text{!} = (n + 1) \cdot n\text{!},$

由于 $(n + 1)\text{!} = (n + 1) \cdot n\text{!},$

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{2}{n + 1} = 0.$$

Since $\rho < 1,$ the series converges.

由于 $\rho < 1,$ 该级数收敛。

2. We can see that

2. 我们可以看出

$$\begin{array}{cl} \rho & {= \underset{n\rightarrow\infty}{\text{lim}}\frac{\left( {n + 1} \right)^{n + 1}\text{/}\left( {n + 1} \right)\text{!}}{n^{n}\text{/}n\text{!}}} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\frac{\left( {n + 1} \right)^{n + 1}}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{n^{n}}} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left( \frac{n + 1}{n} \right)^{n} = \underset{n\rightarrow\infty}{\text{lim}}\left( {1 + \frac{1}{n}} \right)^{n} = e.} \end{array}$$

Since $\rho > 1,$ the series diverges.

由于 $\rho > 1,$ 该级数发散。

3. Since

3. 由于

$$\begin{array}{cl} \left| \frac{{(-1)}^{n + 1}{((n + 1)\text{!})}^{2}\text{/}(2(n + 1))\text{!}}{{(-1)}^{n}{(n\text{!})}^{2}\text{/}(2n)\text{!}} \right| & {= \frac{(n + 1)\text{!}(n + 1)\text{!}}{(2n + 2)\text{!}} \cdot \frac{(2n)\text{!}}{n\text{!}n\text{!}}} \\ & {= \frac{(n + 1)(n + 1)}{(2n + 2)(2n + 1)}} \end{array}$$

we see that

我们看到

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{(n + 1)(n + 1)}{(2n + 2)(2n + 1)} = \frac{1}{4}.$$

Since $\rho < 1,$ the series converges.

由于 $\rho < 1,$ 该级数收敛。

Use the ratio test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{n^{3}}{3^{n}}$ converges or diverges.

使用比值判别法判断级数 $\sum\limits_{n = 1}^{\infty}\frac{n^{3}}{3^{n}}$ 收敛还是发散。

Root Test 根值判别法

The approach of the root test is similar to that of the ratio test. Consider a series $\sum\limits_{n = 1}^{\infty}a_{n}$ such that $\underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|} = \rho$ for some real number $\rho.$ Then for $N$ sufficiently large, $\left| a_{N} \middle| \approx \rho^{N}. \right.$ Therefore, we can approximate ${\sum\limits_{n = N}^{\infty}|}a_{n}|$ by writing

根值判别法的思路与比值判别法类似。考虑级数 $\sum\limits_{n = 1}^{\infty}a_{n}$，其中对某个实数 $\rho$ 有 $\underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|} = \rho$。于是当 $N$ 足够大时，$\left| a_{N} \middle| \approx \rho^{N}. \right.$ 因此，我们可以把 ${\sum\limits_{n = N}^{\infty}|}a_{n}|$ 近似写为

$$\left| a_{N} \middle| + \middle| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \text{⋯} \approx \rho^{N} + \rho^{N + 1} + \rho^{N + 2} + \text{⋯}. \right.$$

The expression on the right-hand side is a geometric series. As in the ratio test, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely if $0 \leq \rho < 1$ and the series diverges if $\rho \geq 1.$ If $\rho = 1,$ the test does not provide any information. For example, for any *p*-series, ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{p},$ we see that

右边的表达式是一个几何级数。与比值判别法类似，如果 $0 \leq \rho < 1$，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛；如果 $\rho \geq 1$，则级数发散。若 $\rho = 1,$ 该判别法不能提供任何信息。例如，对任意 *p*-级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{p}$，我们看到

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| \frac{1}{n^{p}} \right|} = \underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{p\text{/}n}}.$$

To evaluate this limit, we use the natural logarithm function. Doing so, we see that

为了计算这个极限，我们使用自然对数函数。这样做之后，我们看到

$$\text{ln}\mspace{2mu}\rho = \text{ln}\left( {\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{p\text{/}n}}} \right) = \underset{n\rightarrow\infty}{\text{lim}}\text{ln}\left( \frac{1}{n} \right)^{p\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{p}{n} \cdot \text{ln}\left( \frac{1}{n} \right) = \underset{n\rightarrow\infty}{\text{lim}}\frac{p\mspace{2mu}\text{ln}(1\text{/}n)}{n}.$$

Using L’Hôpital’s rule, it follows that $\text{ln}\mspace{2mu}\rho = 0,$ and therefore $\rho = 1$ for all $p.$ However, we know that the *p*-series only converges if $p > 1$ and diverges if $p < 1.$

利用洛必达法则，可得 $\text{ln}\mspace{2mu}\rho = 0,$ 因而对一切 $p$ 都有 $\rho = 1$。然而，我们知道 *p*-级数仅当 $p > 1$ 时收敛，而当 $p < 1$ 时发散。

Consider the series $\sum\limits_{n = 1}^{\infty}a_{n}.$ Let

考虑级数 $\sum\limits_{n = 1}^{\infty}a_{n}$。令

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|}.$$

1. If $0 \leq \rho < 1,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely.

1. 若 $0 \leq \rho < 1,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛。

2. If $\rho > 1$ or $\rho = \infty,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

2. 若 $\rho > 1$ 或 $\rho = \infty,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

3. If $\rho = 1,$ the test does not provide any information.

3. 若 $\rho = 1,$ 该判别法不能提供任何信息。

The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms $a_{n}$ satisfy $\left| a_{n} \middle| = b_{n}^{n}, \right.$ then $\sqrt[n]{\left| a_{n} \right|} = b_{n}$ and we need only evaluate $\underset{n\rightarrow\infty}{\text{lim}}b_{n}.$

根值判别法对那些项中含有指数的级数很有用。特别地，对于各项 $a_{n}$ 满足 $\left| a_{n} \middle| = b_{n}^{n}, \right.$ 的级数，有 $\sqrt[n]{\left| a_{n} \right|} = b_{n}$，我们只需要计算 $\underset{n\rightarrow\infty}{\text{lim}}b_{n}$。

Using the Root Test 使用根值判别法

For each of the following series, use the root test to determine whether the series converges or diverges.

对于下列各个级数，使用根值判别法判断其收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}\frac{\left( {n^{2} + 3n} \right)^{n}}{\left( {4n^{2} + 5} \right)^{n}}$

2. $\sum\limits_{n = 2}^{\infty}\frac{n^{n}}{\left( {\text{ln}(n)} \right)^{n}}$

Solution 解答

1. To apply the root test, we compute

1. 为了应用根值判别法，我们计算

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left( {n^{2} + 3n} \right)^{n}\text{/}\left( {4n^{2} + 5} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{2} + 3n}{4n^{2} + 5} = \frac{1}{4}.$$

Since $\rho < 1,$ the series converges absolutely.

由于 $\rho < 1,$ 该级数绝对收敛。

2. We have

2. 我们有

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{n^{n}\text{/}\left( {\text{ln}\ n} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n}{\text{ln}\ n} = \infty\ \text{by L’Hôpital’s rule}.$$

Since $\rho = \infty,$ the series diverges.

由于 $\rho = \infty,$ 该级数发散。

Use the root test to determine whether the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{n}$ converges or diverges.

使用根值判别法判断级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{n}$ 收敛还是发散。

Choosing a Convergence Test 收敛性判别法的选择

At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.

至此，我们已经有了一份很长的收敛性判别法清单。然而，并非所有判别法都适用于所有级数。当给定一个级数时，我们必须确定使用哪种判别法最佳。下面给出一种寻找最优判别法的策略。

Choosing a Convergence Test for a Series 为级数选择收敛性判别法

Consider a series ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ In the steps below, we outline a strategy for determining whether the series converges.

考虑一个级数 ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ 下面的步骤中，我们概述一种判断该级数是否收敛的策略。

1. Is $\sum\limits_{n = 1}^{\infty}a_{n}$ a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a $p - \text{series}$ or geometric series? If so, check the power $p$ or the ratio $r$ to determine if the series converges.

1. $\sum\limits_{n = 1}^{\infty}a_{n}$ 是常见级数吗？例如，它是调和级数（发散）还是交错调和级数（收敛）？它是 $p - \text{series}$（$p$ 级数）或几何级数（等比级数）吗？如果是，检查幂 $p$ 或公比 $r$ 以确定该级数是否收敛。

2. Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. If we are interested in absolute convergence, proceed to step $3,$ considering the series of absolute values ${\sum\limits_{n = 1}^{\infty}|}a_{n}|.$

2. 它是交错级数吗？我们关心的是绝对收敛还是仅仅收敛？如果我们只关心级数是否收敛，则使用交错级数判别法。如果我们关心绝对收敛，则进行第 $3$ 步，考虑绝对值级数 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|.$

3. Is the series similar to a $p - \text{series}$ or geometric series? If so, try the comparison test or limit comparison test.

3. 该级数是否类似于 $p - \text{series}$（$p$ 级数）或几何级数（等比级数）？如果是，尝试比较判别法或极限比较判别法。

4. Do the terms in the series contain a factorial or power? If the terms are powers such that $a_{n} = b_{n}^{n},$ try the root test first. Otherwise, try the ratio test first.

4. 级数中的项是否含有阶乘或幂？如果项是形如 $a_{n} = b_{n}^{n}$ 的幂，先尝试根值判别法。否则，先尝试比值判别法。

5. Use the divergence test. If this test does not provide any information, try the integral test.

5. 使用发散判别法（第 n 项判别法）。如果该判别法不能提供任何信息，则尝试积分判别法。

Visit this website for more information on testing series for convergence, plus general information on sequences and series.

访问该网站以获取更多关于级数收敛性判别的信息，以及关于数列与级数的一般信息。

Using Convergence Tests 使用收敛性判别法

For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.

对下列各个级数，确定使用哪种收敛性判别法最佳并说明理由。然后判断该级数收敛还是发散。如果该级数是交错级数，判断它是绝对收敛、条件收敛还是发散。

1. $\sum\limits_{n = 1}^{\infty}\frac{n^{2} + 2n}{n^{3} + 3n^{2} + 1}$

2. $\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}\left( {3n + 1} \right)}{n\text{!}}$

3. $\sum\limits_{n = 1}^{\infty}\frac{e^{n}}{n^{3}}$

4. $\sum\limits_{n = 1}^{\infty}\frac{3^{n}}{\left( {n + 1} \right)^{n}}$

Solution

解答

1. Step 1. The series is not a $p–\text{series}$ or geometric series.

1. 步骤 1。该级数不是 $p–\text{series}$（$p$ 级数），也不是几何级数（等比级数）。

Step 2. The series is not alternating.

步骤 2。该级数不是交错级数。

Step 3. For large values of $n,$ we approximate the series by the expression

步骤 3。对于较大的 $n,$ 我们用如下表达式近似该级数

$$\frac{n^{2} + 2n}{n^{3} + 3n^{2} + 1} \approx \frac{n^{2}}{n^{3}} = \frac{1}{n}.$$

Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n.$ Using the limit comparison test, we see that

因此，使用级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ 来应用比较判别法或极限比较判别法似乎是合理的。利用极限比较判别法，我们得到

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(n^{2} + 2n)\text{/}(n^{3} + 3n^{2} + 1)}{1\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{3} + 2n^{2}}{n^{3} + 3n^{2} + 1} = 1.$$

Since the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ diverges, this series diverges as well.

由于级数 ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ 发散，因此该级数也发散。

2. Step 1.The series is not a familiar series.

2. 步骤 1。该级数不是常见级数。

Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series

步骤 2。该级数是交错级数。由于我们关心绝对收敛，考虑级数

$${\sum\limits_{n = 1}^{\infty}\frac{3n}{(n + 1)\text{!}}}.$$

Step 3. The series is not similar to a *p*-series or geometric series.

步骤 3。该级数不类似于 *p* 级数或几何级数（等比级数）。

Step 4. Since each term contains a factorial, apply the ratio test. We see that

步骤 4。由于每一项都含有阶乘，应用比值判别法。我们得到

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\left( {3\left( {n + 1} \right)} \right)\text{/}\left( {n + 1} \right)\text{!}}{\left( {3n + 1} \right)\text{/}n\text{!}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3n + 3}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{3n + 1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3n + 3}{\left( {n + 1} \right)\left( {3n + 1} \right)} = 0.$$

Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.

因此，该级数收敛；我们得出结论，原级数绝对收敛，从而收敛。

3. Step 1. The series is not a familiar series.

3. 步骤 1。该级数不是常见级数。

Step 2. It is not an alternating series.

步骤 2。它不是交错级数。

Step 3. There is no obvious series with which to compare this series.

步骤 3。没有可与该级数比较的明显级数。

Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.

步骤 4。没有阶乘。有一个幂，但这并不是适合使用根值判别法的情况。

Step 5. To apply the divergence test, we calculate that

步骤 5。为应用发散判别法（第 n 项判别法），我们计算得到

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{e^{n}}{n^{3}} = \infty.$$

Therefore, by the divergence test, the series diverges.

因此，由发散判别法（第 n 项判别法）可知，该级数发散。

4. Step 1. This series is not a familiar series.

4. 步骤 1。该级数不是常见级数。

Step 2. It is not an alternating series.

步骤 2。它不是交错级数。

Step 3. There is no obvious series with which to compare this series.

步骤 3。没有可与该级数比较的明显级数。

Step 4. Since each term is a power of $n,$ we can apply the root test. Since

步骤 4。由于每一项是 $n$ 的幂，我们可以应用根值判别法。由于

$$\underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left( \frac{3}{n + 1} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3}{n + 1} = 0,$$

by the root test, we conclude that the series converges.

由根值判别法可知，该级数收敛。

For the series $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{3^{n} + n},$ determine which convergence test is the best to use and explain why.

对于级数 $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{3^{n} + n},$ 确定使用哪种收敛性判别法最佳并说明理由。

In Table 5.3, we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series $\sum\limits_{n = 1}^{\infty}a_{n}$ to have nonnegative terms, if $\sum\limits_{n = 1}^{\infty}a_{n}$ has negative terms, these tests can be applied to ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ to test for absolute convergence.

在表 5.3 中，我们总结了各项收敛性判别法及其适用情形。注意，虽然比较判别法、极限比较判别法与积分判别法要求级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 具有非负项，但若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 含有负项，则可将这些判别法应用于绝对值级数 ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ 来检验绝对收敛。

Series or Test	Conclusions	Comments
Divergence Test For any series $\sum\limits_{n = 1}^{\infty}a_{n},$ evaluate $\underset{n\rightarrow\infty}{\text{lim}}a_{n}.$	If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ the test is inconclusive.	This test cannot prove convergence of a series.
	If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ the series diverges.	This test cannot prove convergence of a series.
Geometric Series $\sum\limits_{n = 1}^{\infty}ar^{n - 1}$	If \|r\| < 1, the series converges to a/(1−r).	Any geometric series can be reindexed to be written in the form a + ar + ar² + ⋯, where a is the initial term and r is the ratio.
	If \|r\| ≥ 1, the series diverges.
p-Series $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}$	If p > 1, the series converges.	For p = 1, we have the harmonic series $\sum\limits_{n = 1}^{\infty}1\text{/}n.$
p-Series $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}$	If p ≤ 1, the series diverges.
Comparison Test For $\sum\limits_{n = 1}^{\infty}a_{n}$ with nonnegative terms, compare with a known series $\sum\limits_{n = 1}^{\infty}b_{n}.$	If a_n ≤ b_n for all n ≥ N and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.	Typically used for a series similar to a geometric or p-series. It can sometimes be difficult to find an appropriate series.
	If a_n ≥ b_n for all n ≥ N and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.
Limit Comparison Test For $\sum\limits_{n = 1}^{\infty}a_{n}$ with positive terms, compare with a series $\sum\limits_{n = 1}^{\infty}b_{n}$ by evaluating $L = \underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}}.$	If L is a real number and L ≠ 0, then $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ both converge or both diverge.	Typically used for a series similar to a geometric or p-series. Often easier to apply than the comparison test.
	If L = 0 and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.
	If L = ∞ and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.
Integral Test If there exists a positive, continuous, decreasing function f such that a_n = f(n) for all n ≥ N, evaluate ∫_N^∞f(x)dx.	∫_N^∞f(x)dx and $\sum\limits_{n = 1}^{\infty}a_{n}$ both converge or both diverge.	Limited to those series for which the corresponding function f can be easily integrated.
Alternating Series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}\ \text{or}\ \sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n}$	If b_{n + 1} ≤ b_n for all n ≥ 1 and b_n → 0, then the series converges.	Only applies to alternating series.
Ratio Test For any series $\sum\limits_{n = 1}^{\infty}a_{n}$ with nonzero terms, let $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left\| \frac{a_{n + 1}}{a_{n}} \right\|.$	If 0 ≤ ρ < 1, the series converges absolutely.	Often used for series involving factorials or exponentials.
	If ρ > 1 or ρ = ∞, the series diverges.
	If ρ = 1, the test is inconclusive.
Root Test For any series $\sum\limits_{n = 1}^{\infty}a_{n},$ let $\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left\| a_{n} \right\|}.$	If 0 ≤ ρ < 1, the series converges absolutely.	Often used for series where \|a_n\| = b_nⁿ.
	If ρ > 1 or ρ = ∞, the series diverges.
	If ρ = 1, the test is inconclusive.

级数或判别法	结论	说明
发散判别法（第 n 项判别法）对任意级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 计算 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}.$	若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ 该判别法无法判定。	该判别法不能证明级数收敛。
	若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ 则级数发散。	该判别法不能证明级数收敛。
几何级数（等比级数） $\sum\limits_{n = 1}^{\infty}ar^{n - 1}$	若 \|r\| < 1，级数收敛于 a/(1−r)。	任何几何级数都可以重新索引，写成 a + ar + ar² + ⋯ 的形式，其中 a 为首项，r 为公比。
几何级数（等比级数） $\sum\limits_{n = 1}^{\infty}ar^{n - 1}$	若 \|r\| ≥ 1，级数发散。	任何几何级数都可以重新索引，写成 a + ar + ar² + ⋯ 的形式，其中 a 为首项，r 为公比。
p 级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}$	若 p > 1，级数收敛。	当 p = 1 时，得到调和级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n.$
p 级数 $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}$	若 p ≤ 1，级数发散。	当 p = 1 时，得到调和级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n.$
比较判别法对具有非负项的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 与一个已知级数 $\sum\limits_{n = 1}^{\infty}b_{n}$ 比较。	若对所有 n ≥ N 有 a_n ≤ b_n，且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。	通常用于与几何级数或 p 级数相似的级数。有时难以找到合适的比较级数。
	若对所有 n ≥ N 有 a_n ≥ b_n，且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。	通常用于与几何级数或 p 级数相似的级数。有时难以找到合适的比较级数。
极限比较判别法对具有正项的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 通过与级数 $\sum\limits_{n = 1}^{\infty}b_{n}$ 比较，计算 $L = \underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}}.$	若 L 为实数且 L ≠ 0，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$ 同时收敛或同时发散。	通常用于与几何级数或 p 级数相似的级数。往往比比较判别法更易使用。
	若 L = 0 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。
	若 L = ∞ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。
积分判别法若存在正值、连续、递减的函数 f，使得对所有 n ≥ N 有 a_n = f(n)，则计算 ∫_N^∞f(x)dx。	∫_N^∞f(x)dx 与 $\sum\limits_{n = 1}^{\infty}a_{n}$ 同时收敛或同时发散。	仅限于对应的函数 f 易于积分的级数。
交错级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}\ \text{or}\ \sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n}$	若对所有 n ≥ 1 有 b_{n + 1} ≤ b_n 且 b_n → 0，则级数收敛。	仅适用于交错级数。
比值判别法（达朗贝尔判别法）对任意具有非零项的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 令 $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left\| \frac{a_{n + 1}}{a_{n}} \right\|.$	若 0 ≤ ρ < 1，级数绝对收敛。	常用于涉及阶乘或指数的级数。
	若 ρ > 1 或 ρ = ∞，级数发散。
	若 ρ = 1，判别法无法判定。
根值判别法（柯西根值判别法）对任意级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 令 $\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left\| a_{n} \right\|}.$	若 0 ≤ ρ < 1，级数绝对收敛。	常用于满足 \|a_n\| = b_nⁿ 的级数。
	若 ρ > 1 或 ρ = ∞，级数发散。
	若 ρ = 1，判别法无法判定。

Table 5.3 Summary of Convergence Tests

表 5.3 收敛性判别法总结

Series Converging to $\pi$ and $1\text{/}\pi$ 收敛于 $\pi$ 与 $1\text{/}\pi$ 的级数

Dozens of series exist that converge to $\pi$ or an algebraic expression containing $\pi.$ Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of $\pi$ in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.

存在许多收敛于 $\pi$ 或含有 $\pi$ 的代数表达式的级数。这里我们考察几个例子并比较它们的收敛速度。所谓收敛速度，是指部分和达到与实际值相差某个给定范围以内所需的项数。前两个例子中级数表示 $\pi$ 的方式可以用 Maclaurin 级数解释，这将在下一章讨论。第三个例子涉及超出本书范围的内容。

1. The series

1. 该级数为

$$\pi = 4\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{2n - 1} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \text{⋯}$$

was discovered by Gregory and Leibniz in the late $1600\text{s}\text{.}$ This result follows from the Maclaurin series for $f(x) = \text{tan}^{-1}x.$ We will discuss this series in the next chapter.

由 Gregory 和 Leibniz 在 17 世纪末（$1600\text{s}\text{.}$）发现。这一结果源于 $f(x) = \text{tan}^{-1}x$ 的 Maclaurin 级数。我们将在下一章讨论该级数。

1. Prove that this series converges.

1. 证明该级数收敛。

2. Evaluate the partial sums $S_{n}$ for $n = 10,20,50,100.$

2. 计算部分和 $S_{n}$，其中 $n = 10,20,50,100$。

3. Use the remainder estimate for alternating series to get a bound on the error $R_{n}.$

3. 利用交错级数的余项估计，给出误差 $R_{n}$ 的界。

4. What is the smallest value of $N$ that guarantees $\left| R_{N} \middle| < 0.01\text{?} \right.$ Evaluate $S_{N}.$

4. 保证 $\left| R_{N} \middle| < 0.01\text{?} \right.$ 成立的 $N$ 的最小值是多少？计算 $S_{N}$。

2. The series

2. 该级数为

$$\begin{array}{cl} \pi & {= 6\sum\limits_{n = 0}^{\infty}\frac{\left( {2n} \right)\text{!}}{2^{4n + 1}\left( {n\text{!}} \right)^{2}\left( {2n + 1} \right)}} \\ & {= 6\left( {\frac{1}{2} + \frac{1}{2 \cdot 3}\left( \frac{1}{2} \right)^{3} + \frac{1 \cdot 3}{2 \cdot 4 \cdot 5} \cdot \left( \frac{1}{2} \right)^{5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7}\left( \frac{1}{2} \right)^{7} + \text{⋯}} \right)} \end{array}$$

has been attributed to Newton in the late $1600\text{s}\text{.}$ The proof of this result uses the Maclaurin series for $f(x) = \text{sin}^{-1}x.$

归功于 Newton，于 17 世纪末（$1600\text{s}\text{.}$）。该结果的证明使用了 $f(x) = \text{sin}^{-1}x$ 的 Maclaurin 级数。

1. Prove that the series converges.

1. 证明该级数收敛。

2. Evaluate the partial sums $S_{n}$ for $n = 5,10,20.$

2. 计算部分和 $S_{n}$，其中 $n = 5,10,20$。

3. Compare $S_{n}$ to $\pi$ for $n = 5,10,20$ and discuss the number of correct decimal places.

3. 对 $n = 5,10,20$，将 $S_{n}$ 与 $\pi$ 比较，并讨论正确的小数位数。

3. The series

3. 该级数为

$$\frac{1}{\pi} = \frac{\sqrt{8}}{9801}\sum\limits_{n = 0}^{\infty}\frac{\left( {4n} \right)\text{!}(1103 + 26390n)}{\left( {n\text{!}} \right)^{4}396^{4n}}$$

was discovered by Ramanujan in the early $1900\text{s}\text{.}$ William Gosper, Jr., used this series to calculate $\pi$ to an accuracy of more than $17$ million digits in the $\text{mid-}1980\text{s}\text{.}$ At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for $\pi$ and $1\text{/}\pi.$

由 Ramanujan 在 20 世纪初（$1900\text{s}\text{.}$）发现。William Gosper, Jr. 利用该级数将 $\pi$ 计算到超过 $17$ 百万位的精度，这在当时是世界纪录。此后，该级数以及 Ramanujan 的其他级数促使数学家找到了许多表示 $\pi$ 与 $1\text{/}\pi$ 的其他级数形式。

1. Prove that this series converges.

1. 证明该级数收敛。

2. Evaluate the first term in this series. Compare this number with the value of $\pi$ from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?

2. 计算该级数的第一项。将此数值与计算工具给出的 $\pi$ 值比较。这两个数值在多少位小数上一致？如果加上级数的前两项呢？

3. Investigate the life of Srinivasa Ramanujan $(1887\text{–}1920)$ and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.

3. 研究 Srinivasa Ramanujan $(1887\text{–}1920)$ 的生平并写一份简短概述。Ramanujan 是数学史上最引人入胜的人物之一。他基本上是自学成才，没有接受过正规的数学训练，却以极具原创性的方式对数学的许多高深领域作出了贡献。

Section 5.6 Exercises 5.6 节习题

Use the ratio test to determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

使用比值判别法判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛，其中 $a_{n}$ 由下列各题给出。若比值判别法无法判定，请说明。

317.

$a_{n} = 1\text{/}n\text{!}$

318\.

$a_{n} = 10^{n}\text{/}n\text{!}$

319.

$a_{n} = n^{2}\text{/}2^{n}$

320\.

$a_{n} = n^{10}\text{/}2^{n}$

321.

$\sum\limits_{n = 1}^{\infty}\frac{{(n\text{!})}^{3}}{(3n)\text{!}}$

322\.

$\sum\limits_{n = 1}^{\infty}\frac{2^{3n}{(n\text{!})}^{3}}{(3n)\text{!}}$

323.

$\sum\limits_{n = 1}^{\infty}\frac{(2n)\text{!}}{n^{2n}}$

324\.

$\sum\limits_{n = 1}^{\infty}\frac{(2n)\text{!}}{{(2n)}^{n}}$

325.

$\sum\limits_{n = 1}^{\infty}\frac{n\text{!}}{{(n\text{/}e)}^{n}}$

326\.

$\sum\limits_{n = 1}^{\infty}\frac{(2n)\text{!}}{{(n\text{/}e)}^{2n}}$

327.

$\sum\limits_{n = 1}^{\infty}\frac{{(2^{n}n\text{!})}^{2}}{{(2n)}^{2n}}$

Use the root test to determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, where $a_{n}$ is as follows.

使用根值判别法判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛，其中 $a_{n}$ 如下。

328\.

$a_{k} = \left( \frac{k - 1}{2k + 3} \right)^{k}$

329.

$a_{k} = \left( \frac{2k^{2} - 1}{k^{2} + 3} \right)^{k}$

330\.

$a_{n} = \frac{\left( {\text{ln}\ n} \right)^{2n}}{n^{n}}$

331.

$a_{n} = n\text{/}2^{n}$

332\.

$a_{n} = n\text{/}e^{n}$

333.

$a_{k} = \frac{k^{e}}{e^{k}}$

334\.

$a_{k} = \frac{\pi^{k}}{k^{\pi}}$

335.

$a_{n} = \left( {\frac{1}{e} + \frac{1}{n}} \right)^{n}$

336\.

$a_{k} = \frac{1}{\left( {1 + \text{ln}\mspace{2mu} k} \right)^{k}}$

For this exercise, let n start at 2.

本习题中，令 n 从 2 开始。

337.

$a_{n} = \frac{\left( {\text{ln}\left( {1 + \text{ln}\ n} \right)} \right)^{n}}{\left( {\text{ln}\ n} \right)^{n}}$

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series $\sum\limits_{k = 1}^{\infty}a_{k}$ with given terms $a_{k}$ converges, or state if the test is inconclusive.

在下列各题中，酌情选用比值判别法或根值判别法，判断给定项 $a_{k}$ 的级数 $\sum\limits_{k = 1}^{\infty}a_{k}$ 是否收敛；若无法判定，请说明。

338\.

$a_{k} = \frac{k\text{!}}{1 \cdot 3 \cdot 5\text{⋯}(2k - 1)}$

339.

$a_{k} = \frac{2 \cdot 4 \cdot 6\text{⋯}2k}{(2k)\text{!}}$

340\.

$a_{k} = \frac{1 \cdot 4 \cdot 7\text{⋯}(3k - 2)}{3^{k}k\text{!}}$

341.

$a_{n} = \left( {1 - \frac{1}{n}} \right)^{n^{2}}$

342\.

$a_{k} = \left( {\frac{1}{k + 1} + \frac{1}{k + 2} + \text{⋯} + \frac{1}{2k}} \right)^{k}$ (*Hint:* Compare $a_{k}^{1\text{/}k}$ to ${\int_{k}^{2k}\frac{dt}{t}}. )$

$a_{k} = \left( {\frac{1}{k + 1} + \frac{1}{k + 2} + \text{⋯} + \frac{1}{2k}} \right)^{k}$（*提示：* 将 $a_{k}^{1\text{/}k}$ 与 ${\int_{k}^{2k}\frac{dt}{t}}$ 比较。）

343.

$a_{k} = \left( {\frac{1}{k + 1} + \frac{1}{k + 2} + \text{⋯} + \frac{1}{3k}} \right)^{k}$

344\.

$a_{n} = {(n^{1\text{/}n} - 1)}^{n}$

Use the ratio test to determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, or state if the ratio test is inconclusive.

使用比值判别法判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛，若比值判别法无法判定请说明。

345.

$\sum\limits_{n = 1}^{\infty}\frac{3^{n^{2}}}{2^{n^{3}}}$

346\.

$\sum\limits_{n = 1}^{\infty}\frac{2^{n^{2}}}{n^{n}n\text{!}}$

Use the root and limit comparison tests to determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

使用根值判别法与极限比较判别法判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛。

347.

$a_{n} = 1\text{/}x_{n}^{n}$ where $x_{n + 1} = \frac{1}{2}x_{n} + \frac{1}{x_{n}},$ $x_{1} = 1$ (*Hint:* Find limit of $\left\{ x_{n} \right\}. )$

$a_{n} = 1\text{/}x_{n}^{n}$，其中 $x_{n + 1} = \frac{1}{2}x_{n} + \frac{1}{x_{n}},$ $x_{1} = 1$（*提示：* 求 $\left\{ x_{n} \right\}$ 的极限。）

In the following exercises, use an appropriate test to determine whether the series converges.

在下列各题中，使用适当的判别法判断该级数是否收敛。

348\.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {n + 1} \right)}{n^{3} + n^{2} + n + 1}$

349.

$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}\left( {n + 1} \right)}{n^{3} + 3n^{2} + 3n + 1}$

350\.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {n + 1} \right)^{2}}{n^{3} + (1.1)^{n}}$

351.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {n - 1} \right)^{n}}{\left( {n + 1} \right)^{n}}$

352\.

$a_{n} = \left( {1 + \frac{1}{n^{2}}} \right)^{n}$ (*Hint:* $\left( {1 + \frac{1}{n^{2}}} \right)^{n^{2}} \approx e$.)

$a_{n} = \left( {1 + \frac{1}{n^{2}}} \right)^{n}$（*提示：* $\left( {1 + \frac{1}{n^{2}}} \right)^{n^{2}} \approx e$。）

353.

$a_{k} = 1\text{/}2^{\text{sin}^{2}k}$

354\.

$a_{k} = 2^{\text{−}\text{sin}(1\text{/}k)}$

355.

$a_{n} = 1\text{/}\begin{pmatrix}{n + 2} \\ n \end{pmatrix}$ where $\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n\text{!}}{k\text{!}\left( {n - k} \right)\text{!}}$

356\.

$a_{k} = 1\text{/}\left( \begin{array}{l} {2k} \\ k \end{array} \right)$

357.

$a_{k} = 2^{k}\text{/}\left( \begin{array}{l} {3k} \\ k \end{array} \right)$

358\.

$a_{k} = \left( \frac{k}{k + \text{ln}\mspace{2mu} k} \right)^{k}$ (*Hint:* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\text{ln}\mspace{2mu} k} \approx e^{\text{−}\text{ln}\mspace{2mu} k}$.)

$a_{k} = \left( \frac{k}{k + \text{ln}\mspace{2mu} k} \right)^{k}$（*提示：* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\text{ln}\mspace{2mu} k} \approx e^{\text{−}\text{ln}\mspace{2mu} k}$。）

359.

$a_{k} = \left( \frac{k}{k + \text{ln}\mspace{2mu} k} \right)^{2k}$ (*Hint:* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\mspace{2mu}\text{ln}\mspace{2mu} k^{2}}$.)$

$a_{k} = \left( \frac{k}{k + \text{ln}\mspace{2mu} k} \right)^{2k}$（*提示：* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\mspace{2mu}\text{ln}\mspace{2mu} k^{2}}$。）

The following series converge by the ratio test. Use summation by parts, ${\sum\limits_{k = 1}^{n}a_{k}}\left( {b_{k + 1} - b_{k}} \right) = \left\lbrack {a_{n + 1}b_{n + 1} - a_{1}b_{1}} \right\rbrack - {\sum\limits_{k = 1}^{n}b_{k + 1}}\left( {a_{k + 1} - a_{k}} \right),$ to find the sum of the given series.

下列级数由比值判别法可知收敛。利用分部求和公式 ${\sum\limits_{k = 1}^{n}a_{k}}\left( {b_{k + 1} - b_{k}} \right) = \left\lbrack {a_{n + 1}b_{n + 1} - a_{1}b_{1}} \right\rbrack - {\sum\limits_{k = 1}^{n}b_{k + 1}}\left( {a_{k + 1} - a_{k}} \right),$ 求所给级数的和。

360\.

$\sum\limits_{k = 1}^{\infty}\frac{k}{2^{k}}$ (*Hint:* Take $a_{k} = k$ and $b_{k} = 2^{1 - k}$.)

$\sum\limits_{k = 1}^{\infty}\frac{k}{2^{k}}$（*提示：* 取 $a_{k} = k$，$b_{k} = 2^{1 - k}$。）

361.

${\sum\limits_{k = 1}^{\infty}\frac{k}{c^{k}}},$ where $c > 1$ (*Hint:* Take $a_{k} = k$ and $b_{k} = c^{1 - k}\text{/}(c - 1)$.)$

${\sum\limits_{k = 1}^{\infty}\frac{k}{c^{k}}}，$ 其中 $c > 1$（*提示：* 取 $a_{k} = k$，$b_{k} = c^{1 - k}\text{/}(c - 1)$。）

362\.

$\sum\limits_{n = 1}^{\infty}\frac{n^{2}}{2^{n}}$

363.

$\sum\limits_{n = 1}^{\infty}\frac{{(n + 1)}^{2}}{2^{n}}$

The *k*th term of each of the following series has a factor $x^{k}.$ Find the range of $x$ for which the ratio test implies that the series converges.

下列各级数的第 *k* 项均含有因子 $x^{k}.$ 求使比值判别法推出该级数收敛的 $x$ 的取值范围。

364\.

$\sum\limits_{k = 1}^{\infty}\frac{x^{k}}{k^{2}}$

365.

$\sum\limits_{k = 1}^{\infty}\frac{x^{2k}}{k^{2}}$

366\.

$\sum\limits_{k = 1}^{\infty}\frac{x^{2k}}{3^{k}}$

367.

$\sum\limits_{k = 1}^{\infty}\frac{x^{k}}{k\text{!}}$

368\.

Does there exist a number $p$ such that $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{n^{p}}$ converges?

是否存在一个数 $p$，使得 $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{n^{p}}$ 收敛？

369.

Let $0 < r < 1.$ For which real numbers $p$ does ${\sum\limits_{n = 1}^{\infty}n^{p}}r^{n}$ converge?

设 $0 < r < 1.$ 对于哪些实数 $p$，${\sum\limits_{n = 1}^{\infty}n^{p}}r^{n}$ 收敛？

370\.

Suppose that $\left. \underset{n\rightarrow\infty}{\text{lim}} \middle| \frac{a_{n + 1}}{a_{n}} \middle| = p. \right.$ For which values of $p$ must ${\sum\limits_{n = 1}^{\infty}2^{n}}a_{n}$ converge?

假设 $\left. \underset{n\rightarrow\infty}{\text{lim}} \middle| \frac{a_{n + 1}}{a_{n}} \middle| = p. \right.$ 对于哪些 $p$ 值，必有 ${\sum\limits_{n = 1}^{\infty}2^{n}}a_{n}$ 收敛？

371.

Suppose that $\left. \underset{n\rightarrow\infty}{\text{lim}} \middle| \frac{a_{n + 1}}{a_{n}} \middle| = p. \right.$ For which values of $r > 0$ is ${\sum\limits_{n = 1}^{\infty}r^{n}}a_{n}$ guaranteed to converge?

假设 $\left. \underset{n\rightarrow\infty}{\text{lim}} \middle| \frac{a_{n + 1}}{a_{n}} \middle| = p. \right.$ 对于哪些 $r > 0$，${\sum\limits_{n = 1}^{\infty}r^{n}}a_{n}$ 必然收敛？

372\.

Suppose that $\left| \frac{a_{n + 1}}{a_{n}} \middle| \leq {(n + 1)}^{p} \right.$ for all $n = 1,2\text{,…}$ where $p$ is a fixed real number. For which values of $p$ is ${\sum\limits_{n = 1}^{\infty}n}\text{!}\ a_{n}$ guaranteed to converge?

假设对一切 $n = 1,2\text{,…}$ 有 $\left| \frac{a_{n + 1}}{a_{n}} \middle| \leq {(n + 1)}^{p} \right.$，其中 $p$ 为固定的实数。对于哪些 $p$ 值，${\sum\limits_{n = 1}^{\infty}n}\text{!}\ a_{n}$ 必然收敛？

373.

For which values of $r > 0,$ if any, does $\sum\limits_{n = 1}^{\infty}r^{\sqrt{n}}$ converge? (*Hint:* ${\sum\limits_{n = 1}^{\infty}a_{n}} = {\sum\limits_{k = 1}^{\infty}{\sum\limits_{n = k^{2}}^{{(k + 1)}^{2} - 1}a_{n}}}. )$

对于任意 $r > 0$（若存在），$\sum\limits_{n = 1}^{\infty}r^{\sqrt{n}}$ 何时收敛？（*提示：* ${\sum\limits_{n = 1}^{\infty}a_{n}} = {\sum\limits_{k = 1}^{\infty}{\sum\limits_{n = k^{2}}^{{(k + 1)}^{2} - 1}a_{n}}}$。）

374\.

Suppose that $\left| \frac{a_{n + 2}}{a_{n}} \middle| \leq r < 1 \right.$ for all $n.$ Can you conclude that $\sum\limits_{n = 1}^{\infty}a_{n}$ converges?

假设对一切 $n$ 有 $\left| \frac{a_{n + 2}}{a_{n}} \middle| \leq r < 1 \right.$。能否由此断定 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛？

375.

Let $a_{n} = 2^{\text{−}\lbrack n\text{/}2\rbrack}$ where $\lbrack x\rbrack$ is the greatest integer less than or equal to $x.$ Determine whether $\sum\limits_{n = 1}^{\infty}a_{n}$ converges and justify your answer.

设 $a_{n} = 2^{\text{−}\lbrack n\text{/}2\rbrack}$，其中 $\lbrack x\rbrack$ 表示不超过 $x$ 的最大整数。判断 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是否收敛，并说明理由。

The following *advanced* exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n}}{a_{n}} < 1\text{/}2,$ then $\sum a_{n}$ converges, while if $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n + 1}}{a_{n}} > 1\text{/}2,$ then $\sum a_{n}$ diverges.

下列*进阶*习题使用广义比值判别法来判断某些级数的收敛性；这些级数出现在特定应用中，而本章的判别法（包括比值判别法与根值判别法）不足以判定其收敛性。该判别法陈述为：若 $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n}}{a_{n}} < 1\text{/}2,$ 则 $\sum a_{n}$ 收敛；若 $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n + 1}}{a_{n}} > 1\text{/}2,$ 则 $\sum a_{n}$ 发散。

376\.

Let $a_{n} = \frac{1}{4}\ \frac{3}{6}\ \frac{5}{8}\text{⋯}\frac{2n - 1}{2n + 2} = \frac{1 \cdot 3 \cdot 5\cdots(2n - 1)}{2^{n}(n + 1)\text{!}}.$ Explain why the ratio test cannot determine convergence of ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ Use the fact that $1 - 1\text{/}(4k)$ is increasing $k$ to estimate $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n}}{a_{n}}.$

设 $a_{n} = \frac{1}{4}\ \frac{3}{6}\ \frac{5}{8}\text{⋯}\frac{2n - 1}{2n + 2} = \frac{1 \cdot 3 \cdot 5\cdots(2n - 1)}{2^{n}(n + 1)\text{!}}.$ 解释为何比值判别法不能确定 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 的收敛性。利用 $1 - 1\text{/}(4k)$ 关于 $k$ 递增这一事实，估计 $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n}}{a_{n}}.$

377.

Let $a_{n} = \frac{1}{1 + x}\ \frac{2}{2 + x}\text{⋯}\frac{n}{n + x}\ \frac{1}{n} = \frac{(n - 1)\text{!}}{(1 + x)(2 + x)\text{⋯}(n + x)}.$ Show that $a_{2n}\text{/}a_{n} \leq e^{\text{−}x\text{/}2}\text{/}2.$ For which $x > 0$ does the generalized ratio test imply convergence of ${\sum\limits_{n = 1}^{\infty}a_{n}}\text{?}$ (*Hint:* Write $2a_{2n}\text{/}a_{n}$ as a product of $n$ factors each smaller than $1\text{/}\left( {1 + x\text{/}\left( {2n} \right)} \right]. )$

设 $a_{n} = \frac{1}{1 + x}\ \frac{2}{2 + x}\text{⋯}\frac{n}{n + x}\ \frac{1}{n} = \frac{(n - 1)\text{!}}{(1 + x)(2 + x)\text{⋯}(n + x)}.$ 证明 $a_{2n}\text{/}a_{n} \leq e^{\text{−}x\text{/}2}\text{/}2.$ 对于哪些 $x > 0$，广义比值判别法可推出 ${\sum\limits_{n = 1}^{\infty}a_{n}}$ 收敛？（*提示：* 将 $2a_{2n}\text{/}a_{n}$ 写成 $n$ 个因子的乘积，每个因子都小于 $1\text{/}\left( {1 + x\text{/}\left( {2n} \right)} \right].$）

378\.

Let $a_{n} = \frac{n^{\text{ln}\ n}}{\left( {\text{ln}\ n} \right)^{n}}.$ Show that $\frac{a_{2n}}{a_{n}}\rightarrow 0$ as $n\rightarrow\infty.$

设 $a_{n} = \frac{n^{\text{ln}\ n}}{\left( {\text{ln}\ n} \right)^{n}}.$ 证明 $\frac{a_{2n}}{a_{n}}\rightarrow 0$ 当 $n\rightarrow\infty.$

Key Terms 关键术语

absolute convergence

绝对收敛

if the series $\sum\limits_{n = 1}^{\infty}|a_{n}|$ converges, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ is said to converge absolutely

如果级数 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 收敛，则称级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛

alternating series

交错级数

a series of the form $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$ or $\sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n},$ where $b_{n} \geq 0,$ is called an alternating series

形如 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$ 或 $\sum\limits_{n = 1}^{\infty}(-1)^{n}b_{n},$ 其中 $b_{n} \geq 0,$ 的级数称为交错级数

alternating series test

交错级数判别法

for an alternating series of either form, if $b_{n + 1} \leq b_{n}$ for all integers $n \geq 1$ and $b_{n}\rightarrow 0,$ then an alternating series converges

对任一种形式的交错级数，如果对所有整数 $n \geq 1$ 都有 $b_{n + 1} \leq b_{n}$ 且 $b_{n}\rightarrow 0,$ 则该交错级数收敛

arithmetic sequence

等差数列

a sequence in which the difference between every pair of consecutive terms is the same is called an arithmetic sequence

相邻项之差都相等的数列称为等差数列

bounded above

上有界的

a sequence $\left\{ a_{n} \right\}$ is bounded above if there exists a constant $M$ such that $a_{n} \leq M$ for all positive integers $n$

若存在常数 $M$ 使得对所有正整数 $n$ 都有 $a_{n} \leq M,$ 则数列 $\left\{ a_{n} \right\}$ 是上有界的

bounded below

下有界的

a sequence $\left\{ a_{n} \right\}$ is bounded below if there exists a constant $M$ such that $M \leq a_{n}$ for all positive integers $n$

若存在常数 $M$ 使得对所有正整数 $n$ 都有 $M \leq a_{n},$ 则数列 $\left\{ a_{n} \right\}$ 是下有界的

bounded sequence

有界数列

a sequence $\left\{ a_{n} \right\}$ is bounded if there exists a constant $M$ such that $\left| a_{n} \right| \leq M$ for all positive integers $n$

若存在常数 $M$ 使得对所有正整数 $n$ 都有 $\left| a_{n} \right| \leq M,$ 则数列 $\left\{ a_{n} \right\}$ 是有界的

comparison test

比较判别法

if $0 \leq a_{n} \leq b_{n}$ for all $n \geq N$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges; if $a_{n} \geq b_{n} \geq 0$ for all $n \geq N$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges

如果对所有 $n \geq N$ 有 $0 \leq a_{n} \leq b_{n}$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛；如果对所有 $n \geq N$ 有 $a_{n} \geq b_{n} \geq 0$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散

conditional convergence

条件收敛

if the series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, but the series $\sum\limits_{n = 1}^{\infty}|a_{n}|$ diverges, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ is said to converge conditionally

如果级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，但级数 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 发散，则称级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 条件收敛

convergence of a series

级数的收敛性

a series converges if the sequence of partial sums for that series converges

如果某级数的部分和数列收敛，则该级数收敛

convergent sequence

收敛数列

a convergent sequence is a sequence $\left\{ a_{n} \right\}$ for which there exists a real number $L$ such that $a_{n}$ is arbitrarily close to $L$ as long as $n$ is sufficiently large

收敛数列是指存在实数 $L,$ 使得只要 $n$ 充分大，$a_{n}$ 就可以任意接近 $L$ 的数列 $\left\{ a_{n} \right\}$

divergence of a series

级数的发散性

a series diverges if the sequence of partial sums for that series diverges

如果某级数的部分和数列发散，则该级数发散

divergence test

发散判别法

if $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ then the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges

如果 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ 则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散

divergent sequence

发散数列

a sequence that is not convergent is divergent

不收敛的数列是发散的

explicit formula

显式公式

a sequence may be defined by an explicit formula such that $a_{n} = f(n)$

数列可由显式公式定义，使得 $a_{n} = f(n)$

geometric sequence

几何数列

a sequence $\left\{ a_{n} \right\}$ in which the ratio $a_{n + 1}\text{/}a_{n}$ is the same for all positive integers $n$ is called a geometric sequence

若对所有正整数 $n,$ 比值 $a_{n + 1}\text{/}a_{n}$ 都相同的数列 $\left\{ a_{n} \right\}$ 称为几何数列

geometric series

几何级数（等比级数）

a geometric series is a series that can be written in the form

几何级数（等比级数）是可以写成如下形式的级数

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1} = a + ar + ar^{2} + ar^{3} + \text{⋯}$$

harmonic series

调和级数

the harmonic series takes the form

调和级数形如

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \text{⋯}$$

index variable

下标变量

the subscript used to define the terms in a sequence is called the index

用于定义数列各项的下标称为下标（索引）

infinite series

无穷级数

an infinite series is an expression of the form

无穷级数是形如

$$a_{1} + a_{2} + a_{3} + \text{⋯} = \sum\limits_{n = 1}^{\infty}a_{n}$$

integral test

积分判别法

for a series $\sum\limits_{n = 1}^{\infty}a_{n}$ with positive terms $a_{n},$ if there exists a continuous, decreasing function $f$ such that $f(n) = a_{n}$ for all positive integers $n,$ then

对于具有正项 $a_{n}$ 的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 若存在连续、递减的函数 $f$ 使得对所有正整数 $n$ 都有 $f(n) = a_{n},$ 则

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{1}^{\infty}{f(x)dx}}$$

either both converge or both diverge

两者要么都收敛，要么都发散

limit comparison test

极限比较判别法

suppose $a_{n},\ b_{n} \geq 0$ for all $n \geq 1.$ If $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow L \neq 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ both converge or both diverge; if $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow 0$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges. If $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow\infty,$ and $\sum\limits_{n = 1}^{\infty}b_{n}$ diverges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges

设对所有 $n \geq 1$ 有 $a_{n},\ b_{n} \geq 0.$ 如果 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow L \neq 0,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 与 $\sum\limits_{n = 1}^{\infty}b_{n}$ 同收敛或同发散；如果 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow 0$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。如果 $\underset{n\rightarrow\infty}{\text{lim}}a_{n}\text{/}b_{n}\rightarrow\infty,$ 且 $\sum\limits_{n = 1}^{\infty}b_{n}$ 发散，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散

limit of a sequence

数列的极限

the real number $L$ to which a sequence converges is called the limit of the sequence

数列收敛到的实数 $L$ 称为该数列的极限

monotone sequence

单调数列

an increasing or decreasing sequence

递增或递减的数列

*p*-series

p 级数

a series of the form $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$

形如 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 的级数

partial sum

部分和

the $k\text{th}$ partial sum of the infinite series $\sum\limits_{n = 1}^{\infty}a_{n}$ is the finite sum

无穷级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 的 $k\text{th}$ 部分和是如下有限和

$$S_{k} = \sum\limits_{n = 1}^{k}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k}$$

ratio test

比值判别法

for a series $\sum\limits_{n = 1}^{\infty}a_{n}$ with nonzero terms, let $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| {a_{n + 1}\text{/}a_{n}} \right|;$ if $0 \leq \rho < 1,$ the series converges absolutely; if $\rho > 1,$ the series diverges; if $\rho = 1,$ the test is inconclusive

对于具非零项的级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 令 $\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| {a_{n + 1}\text{/}a_{n}} \right|;$ 若 $0 \leq \rho < 1,$ 则级数绝对收敛；若 $\rho > 1,$ 则级数发散；若 $\rho = 1,$ 则该判别法无法判定

recurrence relation

递推关系

a recurrence relation is a relationship in which a term $a_{n}$ in a sequence is defined in terms of earlier terms in the sequence

递推关系是指数列中的项 $a_{n}$ 用该数列中前面的项来定义的一种关系

remainder estimate

余项估计

for a series $\sum\limits_{n = 1}^{\infty}a_{n}$ with positive terms $a_{n}$ and a continuous, decreasing function $f$ such that $f(n) = a_{n}$ for all positive integers $n,$ the remainder $R_{N} = \sum\limits_{n = 1}^{\infty}a_{n} - \sum\limits_{n = 1}^{N}a_{n}$ satisfies the following estimate:

对于具有正项 $a_{n}$ 的级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 以及连续、递减的函数 $f,$ 使得对所有正整数 $n$ 都有 $f(n) = a_{n},$ 余项 $R_{N} = \sum\limits_{n = 1}^{\infty}a_{n} - \sum\limits_{n = 1}^{N}a_{n}$ 满足如下估计：

$${\int_{N + 1}^{\infty}{f(x)dx < R_{N} <}}{\int_{N}^{\infty}{f(x)dx}}$$

root test

根值判别法

for a series $\sum\limits_{n = 1}^{\infty}a_{n},$ let $\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|};$ if $0 \leq \rho < 1,$ the series converges absolutely; if $\rho > 1,$ the series diverges; if $\rho = 1,$ the test is inconclusive

对于级数 $\sum\limits_{n = 1}^{\infty}a_{n},$ 令 $\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|};$ 若 $0 \leq \rho < 1,$ 则级数绝对收敛；若 $\rho > 1,$ 则级数发散；若 $\rho = 1,$ 则该判别法无法判定

sequence

数列

an ordered list of numbers of the form $a_{1},a_{2},a_{3}\text{,…}$ is a sequence

形如 $a_{1},a_{2},a_{3}\text{,…}$ 的有序数列（数表）是一个数列

telescoping series

裂项级数（叠缩级数）

a telescoping series is one in which most of the terms cancel in each of the partial sums

裂项级数（叠缩级数）是指在每一个部分和中大部分项都相互抵消的级数

term

项

the number $a_{n}$ in the sequence $\left\{ a_{n} \right\}$ is called the $n\text{th}$ term of the sequence

数列 $\left\{ a_{n} \right\}$ 中的数 $a_{n}$ 称为该数列的第 $n\text{th}$ 项

unbounded sequence

无界数列

a sequence that is not bounded is called unbounded

不有界的数列称为无界数列

Key Equations 关键公式

Harmonic series	${\sum\limits_{n = 1}^{\infty}\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \text{⋯}$
Sum of a geometric series	${\sum\limits_{n = 1}^{\infty}{ar^{n - 1}}} = \frac{a}{1 - r}\ \text{for}\ \|r\| < 1$
Divergence test	$\text{If}\ a_{n}\nrightarrow 0\ \text{as}\ n\rightarrow\infty,{\sum\limits_{n = 1}^{\infty}a_{n}}\ \text{diverges}.$
p-series	${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}}\left\{ \begin{array}{l} {\text{converges if}\ p > 1} \\ {\text{diverges if}\ p \leq 1} \end{array} \right.$
Remainder estimate from the integral test	${\int_{N + 1}^{\infty}{f(x)dx}} < R_{N} < {\int_{N}^{\infty}{f(x)dx}}$
Alternating series	${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n + 1}b_{n}}} = b_{1} - b_{2} + b_{3} - b_{4} + \text{⋯}\ \text{or}$ ${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n}b_{n}}} = \text{−}b_{1} + b_{2} - b_{3} + b_{4} - \text{⋯}$

调和级数	${\sum\limits_{n = 1}^{\infty}\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \text{⋯}$
几何级数的和	${\sum\limits_{n = 1}^{\infty}{ar^{n - 1}}} = \frac{a}{1 - r}\ \text{for}\ \|r\| < 1$
发散判别法	$\text{If}\ a_{n}\nrightarrow 0\ \text{as}\ n\rightarrow\infty,{\sum\limits_{n = 1}^{\infty}a_{n}}\ \text{diverges}.$
p 级数	${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}}\left\{ \begin{array}{l} {\text{converges if}\ p > 1} \\ {\text{diverges if}\ p \leq 1} \end{array} \right.$
由积分判别法得到的余项估计	${\int_{N + 1}^{\infty}{f(x)dx}} < R_{N} < {\int_{N}^{\infty}{f(x)dx}}$
交错级数	${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n + 1}b_{n}}} = b_{1} - b_{2} + b_{3} - b_{4} + \text{⋯}\ \text{or}$ ${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n}b_{n}}} = \text{−}b_{1} + b_{2} - b_{3} + b_{4} - \text{⋯}$

Key Concepts 关键概念

5.1 Sequences 5.1 数列

To determine the convergence of a sequence given by an explicit formula $a_{n} = f(n),$ we use the properties of limits for functions.
If $\left\{ a_{n} \right\}$ and $\left\{ b_{n} \right\}$ are convergent sequences that converge to $A$ and $B,$ respectively, and $c$ is any real number, then the sequence $\left\{ ca_{n} \right\}$ converges to $c \cdot A,$ the sequences $\left\{ a_{n} \pm b_{n} \right\}$ converge to $A \pm B,$ the sequence $\left\{ a_{n} \cdot b_{n} \right\}$ converges to $A \cdot B,$ and the sequence $\left\{ a_{n}\text{/}b_{n} \right\}$ converges to $A\text{/}B,$ provided $B \neq 0.$
If a sequence is bounded and monotone, then it converges, but not all convergent sequences are monotone.
If a sequence is unbounded, it diverges, but not all divergent sequences are unbounded.
The geometric sequence $\left\{ r^{n} \right\}$ converges if and only if $|r| < 1$ or $r = 1.$

对于由显式公式 $a_{n} = f(n)$ 给出的数列，我们依据函数极限的性质来判断其收敛性。
若 $\left\{ a_{n} \right\}$ 与 $\left\{ b_{n} \right\}$ 是收敛数列，分别收敛于 $A$ 与 $B$，且 $c$ 为任意实数，则数列 $\left\{ ca_{n} \right\}$ 收敛于 $c \cdot A$，数列 $\left\{ a_{n} \pm b_{n} \right\}$ 收敛于 $A \pm B$，数列 $\left\{ a_{n} \cdot b_{n} \right\}$ 收敛于 $A \cdot B$，而数列 $\left\{ a_{n}\text{/}b_{n} \right\}$ 在 $B \neq 0$ 时收敛于 $A\text{/}B$。
若有界数列单调，则它收敛；但并非所有收敛数列都单调。
若数列无界，则它发散；但并非所有发散数列都无界。
几何数列 $\left\{ r^{n} \right\}$ 收敛当且仅当 $|r| < 1$ 或 $r = 1$。

5.2 Infinite Series 5.2 无穷级数

Given the infinite series

给定如下无穷级数

$$\sum\limits_{n = 1}^{\infty}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯}$$

and the corresponding sequence of partial sums $\left\{ S_{k} \right\}$ where

以及对应的部分和数列 $\left\{ S_{k} \right\}$，其中

$$S_{k} = \sum\limits_{n = 1}^{k}a_{n} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k},$$

the series converges if and only if the sequence $\left\{ S_{k} \right\}$ converges.

该级数收敛当且仅当数列 $\left\{ S_{k} \right\}$ 收敛。

The geometric series $\sum\limits_{n = 1}^{\infty}ar^{n - 1}$ converges if $|r| < 1$ and diverges if $|r| \geq 1.$ For $|r| < 1,$

几何级数 $\sum\limits_{n = 1}^{\infty}ar^{n - 1}$ 在 $|r| < 1$ 时收敛，在 $|r| \geq 1$ 时发散。当 $|r| < 1$ 时，

$$\sum\limits_{n = 1}^{\infty}ar^{n - 1} = \frac{a}{1 - r}.$$

The harmonic series

调和级数

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \text{⋯}$$

diverges.

发散。

A series of the form ${\sum\limits_{n = 1}^{\infty}{\lbrack b_{n} - b_{n + 1}\rbrack}} = \lbrack b_{1} - b_{2}\rbrack + \lbrack b_{2} - b_{3}\rbrack + \lbrack b_{3} - b_{4}\rbrack + \text{⋯} + \lbrack b_{n} - b_{n + 1}\rbrack + \text{⋯}$

形如 ${\sum\limits_{n = 1}^{\infty}{\lbrack b_{n} - b_{n + 1}\rbrack}} = \lbrack b_{1} - b_{2}\rbrack + \lbrack b_{2} - b_{3}\rbrack + \lbrack b_{3} - b_{4}\rbrack + \text{⋯} + \lbrack b_{n} - b_{n + 1}\rbrack + \text{⋯}$ 的级数

is a telescoping series. The $k\text{th}$ partial sum of this series is given by $S_{k} = b_{1} - b_{k + 1}.$ The series will converge if and only if $\underset{k\rightarrow\infty}{\text{lim}}b_{k + 1}$ exists. In that case,

是裂项级数。该级数的第 $k\text{th}$ 个部分和由 $S_{k} = b_{1} - b_{k + 1}$ 给出。该级数收敛当且仅当 $\underset{k\rightarrow\infty}{\text{lim}}b_{k + 1}$ 存在。此时，

$${\sum\limits_{n = 1}^{\infty}{\lbrack b_{n} - b_{n + 1}\rbrack}} = b_{1} - \underset{k\rightarrow\infty}{\text{lim}}{\left( b_{k + 1} \right).}$$

5.3 The Divergence and Integral Tests 5.3 发散判别法与积分判别法

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ then the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.
If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ the series $\sum\limits_{n = 1}^{\infty}a_{n}$ may converge or diverge.
If $\sum\limits_{n = 1}^{\infty}a_{n}$ is a series with positive terms $a_{n}$ and $f$ is a continuous, decreasing function such that $f(n) = a_{n}$ for all positive integers $n,$ then

若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0$，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。
若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0$，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 可能收敛也可能发散。
若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 是一个各项为正的级数 $a_{n}$，且 $f$ 是连续、递减的函数，满足对所有正整数 $n$ 都有 $f(n) = a_{n}$，则

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{1}^{\infty}{f(x)dx}}$$

either both converge or both diverge. Furthermore, if $\sum\limits_{n = 1}^{\infty}a_{n}$ converges, then the $N\text{th}$ partial sum approximation $S_{N}$ is accurate up to an error $R_{N}$ where ${\int_{N + 1}^{\infty}{f(x)dx < R_{N} < {\int_{N}^{\infty}{f(x)dx}}}}.$

二者同时收敛或同时发散。此外，若 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛，则第 $N\text{th}$ 个部分和近似 $S_{N}$ 的误差不超过 $R_{N}$，其中 ${\int_{N + 1}^{\infty}{f(x)dx < R_{N} < {\int_{N}^{\infty}{f(x)dx}}}}.$

The *p*-series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ converges if $p > 1$ and diverges if $p \leq 1.$

p 级数 $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$ 在 $p > 1$ 时收敛，在 $p \leq 1$ 时发散。

5.4 Comparison Tests 5.4 比较判别法

The comparison tests are used to determine convergence or divergence of series with positive terms.
When using the comparison tests, a series $\sum\limits_{n = 1}^{\infty}a_{n}$ is often compared to a geometric or *p*-series.

比较判别法用于判断各项为正的级数的收敛性或发散性。
使用比较判别法时，级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 常与几何级数或 p 级数进行比较。

5.5 Alternating Series 5.5 交错级数

For an alternating series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n},$ if $b_{k + 1} \leq b_{k}$ for all $k$ and $b_{k}\rightarrow 0$ as $k\rightarrow\infty,$ the alternating series converges.
If $\sum\limits_{n = 1}^{\infty}|a_{n}|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

对于交错级数 $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}b_{n}$，若对所有 $k$ 都有 $b_{k + 1} \leq b_{k}$，且当 $k\rightarrow\infty$ 时 $b_{k}\rightarrow 0$，则该交错级数收敛。
若 $\sum\limits_{n = 1}^{\infty}|a_{n}|$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

5.6 Ratio and Root Tests 5.6 比值判别法与根值判别法

For the ratio test, we consider

对于比值判别法，我们考虑

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right|.$$

If $\rho < 1,$ the series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely. If $\rho > 1,$ the series diverges. If $\rho = 1,$ the test does not provide any information. This test is useful for series whose terms involve factorials.

若 $\rho < 1$，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛。若 $\rho > 1$，则级数发散。若 $\rho = 1$，该判别法不能提供任何信息。该判别法适用于通项中含有阶乘的级数。

For the root test, we consider

对于根值判别法，我们考虑

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|}.$$

If $\rho < 1,$ the series $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely. If $\rho > 1,$ the series diverges. If $\rho = 1,$ the test does not provide any information. The root test is useful for series whose terms involve powers.

若 $\rho < 1$，则级数 $\sum\limits_{n = 1}^{\infty}a_{n}$ 绝对收敛。若 $\rho > 1$，则级数发散。若 $\rho = 1$，该判别法不能提供任何信息。根值判别法适用于通项中含有幂的级数。

For a series that is similar to a geometric series or $p - \text{series,}$ consider one of the comparison tests.

对于与几何级数或 $p - \text{series,}$ 类似的级数，可考虑使用某个比较判别法。

Review Exercises 复习习题

*True or False?* Justify your answer with a proof or a counterexample.

*True or False?* 用证明或反例说明你的答案。

379.

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

380\.

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

若 $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ 则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 发散。

381.

If $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

若 $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ 收敛，则 $\sum\limits_{n = 1}^{\infty}a_{n}$ 收敛。

382\.

If $\sum\limits_{n = 1}^{\infty}{2^{n}a_{n}}$ converges, then $\sum\limits_{n = 1}^{\infty}{(-2)^{n}a_{n}}$ converges.

若 $\sum\limits_{n = 1}^{\infty}{2^{n}a_{n}}$ 收敛，则 $\sum\limits_{n = 1}^{\infty}{(-2)^{n}a_{n}}$ 收敛。

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

该数列是否有界、单调，且收敛或发散？若收敛，求其极限。

383.

$a_{n} = \frac{3 + n^{2}}{1 - n}$

384\.

$a_{n} = \text{ln}\left( \frac{1}{n} \right)$

385.

$a_{n} = \frac{\text{ln}\left( {n + 1} \right)}{\sqrt{n + 1}}$

386\.

$a_{n} = \frac{2^{n + 1}}{5^{n}}$

387.

$a_{n} = \frac{\text{ln}\left( {\text{cos}\mspace{2mu} n} \right)}{n}$

Is the series convergent or divergent?

该级数是收敛还是发散？

388\.

$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2} + 5n + 4}$

389.

$\sum\limits_{n = 1}^{\infty}{\text{ln}\left( \frac{n + 1}{n} \right)}$

390\.

$\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{n^{4}}$

391.

$\sum\limits_{n = 1}^{\infty}\frac{e^{n}}{n\text{!}}$

392\.

$\sum\limits_{n = 1}^{\infty}n^{\text{−}{({n + 1\text{/}n})}}$

Is the series convergent or divergent? If convergent, is it absolutely convergent?

该级数是收敛还是发散？若收敛，它是否绝对收敛？

393.

$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}$

394\.

$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n}n\text{!}}{3^{n}}$

395.

$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n}n\text{!}}{n^{n}}$

396\.

$\sum\limits_{n = 1}^{\infty}{\text{sin}\left( \frac{n\pi}{2} \right)}$

397.

$\sum\limits_{n = 1}^{\infty}{\text{cos}\left( {\pi n} \right)e^{\text{−}n}}$

Evaluate

计算

398\.

$\sum\limits_{n = 1}^{\infty}\frac{2^{n + 4}}{7^{n}}$

399.

$\sum\limits_{n = 1}^{\infty}\frac{1}{(n + 1)(n + 2)}$

400\.

A legend from India tells that a mathematician invented chess for a king. The king enjoyed the game so much he allowed the mathematician to demand any payment. The mathematician asked for one grain of rice for the first square on the chessboard, two grains of rice for the second square on the chessboard, four grains of rice for the third square on the chessboard, and so on. Find an exact expression for the total payment (in grains of rice) requested by the mathematician. Assuming there are $30,000$ grains of rice in $1$ pound, and $2000$ pounds in $1$ ton, how many tons of rice did the mathematician attempt to receive?

印度有一则传说：一位数学家为国王发明了国际象棋。国王非常喜爱这个游戏，于是允许数学家提出任何报酬要求。数学家要求：棋盘第一格放 1 粒米，第二格放 2 粒米，第三格放 4 粒米，依此类推。求数学家要求的报酬总量（以粒米计）的精确表达式。假设 1 磅有 $30,000$ 粒米，1 吨有 $2000$ 磅，那么数学家试图得到多少吨米？

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula $x_{n + 1} = bx_{n},$ where $x_{n}$ is the population of houseflies at generation $n,$ and $b$ is the average number of offspring per housefly who survive to the next generation. Assume a starting population $x_{0}.$

以下问题考虑家蝇的一个简单种群模型，它可由递推公式 $x_{n + 1} = bx_{n}$ 表示，其中 $x_{n}$ 是第 $n$ 代的个体数量，$b$ 是每只家蝇存活到下一代的平均后代数。假设初始种群为 $x_{0}.$

401.

Find $\underset{n\rightarrow\infty}{\text{lim}}x_{n}$ if $b > 1,$ $b < 1,$ and $b = 1.$

求 $\underset{n\rightarrow\infty}{\text{lim}}x_{n}$（当 $b > 1,$ $b < 1,$ 以及 $b = 1$ 时）。

402\.

Find an expression for $S_{n} = {\sum\limits_{i = 0}^{n}x_{i}}$ in terms of $b$ and $x_{0}.$ What does it physically represent?

用 $b$ 与 $x_{0}$ 表示 $S_{n} = {\sum\limits_{i = 0}^{n}x_{i}}$ 的表达式。它在物理上代表什么？

403.

If $b = \frac{3}{4}$ and $x_{0} = 100,$ find $S_{10}$ and $\underset{n\rightarrow\infty}{\text{lim}}S_{n}$

若 $b = \frac{3}{4}$ 且 $x_{0} = 100,$ 求 $S_{10}$ 与 $\underset{n\rightarrow\infty}{\text{lim}}S_{n}$。

404\.

For what values of $b$ will the series converge and diverge? What does the series converge to?

对于哪些 $b$ 的值，该级数收敛或发散？该级数收敛于什么？

5 Sequences and Series 数列与级数

5.1 Sequences 5.1 数列

Terminology of Sequences 数列的术语

Finding Explicit Formulas 求显式公式

Solution 解答

Defined by Recurrence Relations 由递推关系定义的数列

Solution 解答

Limit of a Sequence 数列的极限

Limit of a Sequence Defined by a Function 由函数定义的数列的极限

Algebraic Limit Laws 代数极限法则

Proof 证明

Determining Convergence and Finding Limits 判定收敛性与求极限

Solution 解答

Continuous Functions Defined on Convergent Sequences 定义在收敛数列上的连续函数

Proof 证明

Limits Involving Continuous Functions Defined on Convergent Sequences 涉及定义在收敛数列上的连续函数的极限

Solution 解答

Squeeze Theorem for Sequences 数列的夹逼定理

Proof 证明

Using the Squeeze Theorem 应用夹逼定理

Solution 解答

Bounded Sequences 有界数列

Convergent Sequences Are Bounded 收敛数列必有界

Monotone Convergence Theorem 单调收敛定理

Using the Monotone Convergence Theorem 应用单调收敛定理

Solution 解答

Fibonacci Numbers 斐波那契数

Section 5.1 Exercises 5.1 节 习题

5.2 Infinite Series 5.2 无穷级数

Sums and Series 和与级数

Evaluating Limits of Sequences of Partial Sums 求部分和数列的极限

Solution 解答

The Harmonic Series 调和级数

Algebraic Properties of Convergent Series 收敛级数的代数性质

Using Algebraic Properties of Convergent Series 利用收敛级数的代数性质

Solution 解答

Geometric Series 几何级数

Determining Convergence or Divergence of a Geometric Series 判断几何级数的收敛或发散

Solution 解答

Writing Repeating Decimals as Fractions of Integers 把循环小数写成分数

Solution 解答

Chapter Opener: Finding the Area of the Koch Snowflake 章节开篇：求科赫雪花的面积

Solution 解答

Analysis 分析

Telescoping Series 裂项级数

Evaluating a Telescoping Series 求裂项级数

Solution 解答

Euler’s Constant 欧拉常数

Section 5.2 Exercises 5.2 节习题

5.3 The Divergence and Integral Tests 5.3 发散判别法与积分判别法

Divergence Test 发散判别法

Divergence Test 发散判别法

Using the divergence test 使用发散判别法

Solution 解答

Integral Test 积分判别法

Integral Test 积分判别法

Using the Integral Test 使用积分判别法

Solution 解答

The *p*-Series p 级数

Testing for Convergence of *p*-series p 级数收敛性的判别

Solution 解答

Estimating the Value of a Series 估计级数的和

Remainder Estimate from the Integral Test 由积分判别法得到的余项估计

Estimating the Value of a Series 估计级数的和

Solution 解答

Section 5.3 Exercises 5.3 节习题

5.4 Comparison Tests 5.4 比较判别法

Comparison Test 比较判别法

Comparison Test 比较判别法

Proof 证明

Using the Comparison Test 使用比较判别法

Solution 解答

Limit Comparison Test 极限比较判别法

Limit Comparison Test 极限比较判别法

Using the Limit Comparison Test 使用极限比较判别法

Solution 解答

Section 5.4 Exercises 5.4 节习题

5.5 Alternating Series 5.5 交错级数

The Alternating Series Test 交错级数判别法

Proof 证明

Section 5.1 Exercises 5.1 节习题

The p-Series p 级数

Testing for Convergence of p-series p 级数收敛性的判别

Section 5.6 Exercises 5.6 节习题