← 学习库 OpenStax Calculus Vol 2 目录

5 Sequences and Series

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-sequences-and-series

（该页为章节总览/导航页，无独立正文；本章内容请见其下各小节。）

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Introduction

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-introduction

Chapter Outline

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5.1 Sequences

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-1-sequences

5.1 Sequences

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.

Terminology of Sequences

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

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

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

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

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.

Let’s consider the infinite, ordered list

$$2,4,8,16,32\text{,…}\ .$$

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:

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

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

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

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$a_{n} = 2a_{n - 1}.$

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

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

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.

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

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

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

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

$$3,7,11,15,19\text{,…}\ \text{.}$$

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

$$\left\{ \begin{matrix}

{a_{1} = 3} \\

{a_{n} = a_{n - 1} + 4\ \text{for}\ n \geq 2.}

\end{matrix} \right.$$

Note that

$$\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

$$\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 $a_{n} = cn + b.$

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

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

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

$$\begin{array}{l}

\\

{a_{1} = 2} \\

{a_{n} = - \frac{1}{3} \cdot a_{n - 1}\ \text{for}\ n \geq 2.}

\end{array}$$

Alternatively, since

$$\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

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

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 $a_{n} = cr^{n}.$

Finding Explicit Formulas

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

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

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

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

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 $a_{n} = \frac{3^{n}}{3n + 1}.$

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

Defined by Recurrence Relations

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

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

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

Solution

1. Writing out the first few terms, we have

$$\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,

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

2. Write out the first few terms:

$$\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

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

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

Limit of a Sequence

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):

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 $n\rightarrow\infty.$

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 $n\rightarrow\infty.$

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

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 $n\rightarrow\infty.$

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.

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

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

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

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.

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.

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

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

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.

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

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

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 $L.$

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 $\underset{n\rightarrow\infty}{\text{lim}}\left( {-5n + 2} \right) = \rightarrow\text{−}\infty.$

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 $n\rightarrow\infty.$

Limit of a Sequence Defined by a Function

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

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

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

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

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 $r > 1.$

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

$$\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 $r < 0.$

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 $\left\{ b_{n} \right\}.$

Algebraic Limit Laws

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

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

We prove part iii.

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 $n \geq N,$

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

□

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

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

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

Determining Convergence and Finding Limits

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

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

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

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

Therefore,

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

The sequence converges and its limit is $5.$

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

$$\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 $-3\text{/}4.$

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

$$\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.

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

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

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

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

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,

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

Using properties of logarithms, we write

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

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

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

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

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

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 $e^{4}.$

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.

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 $x = 5,$

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

Continuous Functions Defined on Convergent Sequences

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).

Proof

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 $f(L).$

□

Limits Involving Continuous Functions Defined on Convergent Sequences

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

Solution

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

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

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

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

Squeeze Theorem for Sequences

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

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

If there exists a real number $L$ such that

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

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

Proof

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 $n \geq M,$

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

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 $L.$

□

Using the Squeeze Theorem

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

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

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

Solution

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

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

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.

2. Since

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

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

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

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.

$$r^{n}\rightarrow 0\ \text{if}\ |r| < 1$$ (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

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.

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

$$a_{n} \leq M$$

for all positive integers $n.$

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

$$M \leq a_{n}$$

for all positive integers $n.$

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

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

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.

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.

Convergent Sequences Are Bounded

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

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.

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

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

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

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

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.

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

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

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

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

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 $n \geq n_{0}.$

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

Monotone Convergence Theorem

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.

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).

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

Using the Monotone Convergence Theorem

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

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

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

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

Solution

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

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

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.

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

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.

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

$$\left\{ a_{n + 1} \right\} = \left\{ {a_{2,}a_{3},a_{4}\text{,…}} \right\},$$ 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,

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

Combining this fact with the equation

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

and taking the limit of both sides of the equation

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

we can conclude that

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

2. Writing out the first several terms,

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

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

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

To show this, first rewrite

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

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,

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

if and only if

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

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

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

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

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

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

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

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

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

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

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

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

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

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

and therefore

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

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

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

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

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.

Fibonacci Numbers

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

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

Here we look at properties of the Fibonacci numbers.

1. Write out the first twenty Fibonacci numbers.

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

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 $n \geq 0.$

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

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

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

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. Use the answer in 2 c. to show that

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

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

Section 5.1 Exercises

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

1.

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

2\.

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

3.

$a_{1} = 1$ 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 $n \geq 1$

5.

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

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 $n \geq 1.$

7.

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 $n \geq 1.$

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 $n \geq 1.$

9.

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 $n \geq 1.$

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.)

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 $n \geq 2.$

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

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 $a_{n} = f(n).$

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.

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.

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.

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

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.

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.

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

39.

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

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

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

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.

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.

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.

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.

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

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

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.

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

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

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.

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

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

3. Is the amount increasing or decreasing?

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 $\text{\$}1000.$

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

61.

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

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

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

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.

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 $1\text{/}3.$

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 $\pi.$

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.

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.

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?

---

5.2 Infinite Series

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-2-infinite-series

5.2 Infinite Series

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.

Sums and Series

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

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

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

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

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

$$\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

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

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

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

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 $\left\{ S_{k} \right\}.$

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

$$\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.

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

| | | | | |

|---------|----------|---------|------------|------------|

| $k$ | $5$ | $10$ | $15$ | $20$ |

| $S_{k}$ | $1.9375$ | $1.998$ | $1.999939$ | $1.999998$ |

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

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

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.

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.

An infinite series is an expression of the form

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

For each positive integer $k,$ the sum

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

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

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

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

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

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

can also be written as

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

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

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

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

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

Evaluating Limits of Sequences of Partial Sums

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

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

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

$$\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

$$\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.

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

$$\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

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

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

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

$$\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

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

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

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

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

The Harmonic Series

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

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

| | | | | | | |

|---------|-----------|-----------|-----------|-----------|------------|-------------|

| $k$ | $10$ | $100$ | $1000$ | $10,000$ | $100,000$ | $1,000,000$ |

| $S_{k}$ | $2.92897$ | $5.18738$ | $7.48547$ | $9.78761$ | $12.09015$ | $14.39273$ |

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.

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:

$$\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 $S_{4},$

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

Therefore, we conclude that

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

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

$$\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.

Algebraic Properties of Convergent Series

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.

Algebraic Properties of Convergent Series

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.

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)

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)

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)

Using Algebraic Properties of Convergent Series

Evaluate

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

Solution

We showed earlier that

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

and

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

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

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

Using the sum rule, write

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

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

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

Geometric Series

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

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

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

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

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

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

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

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

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

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

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

$$\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.

Therefore,

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

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

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

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

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

A geometric series is a series of the form

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

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

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

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

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

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

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

$$\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

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

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

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

Determining Convergence or Divergence of a Geometric Series

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

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

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

Solution

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

$$\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

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

2. Writing this series as

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

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

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.

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.

Writing Repeating Decimals as Fractions of Integers

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

Solution

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

$$\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

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

Thus,

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

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

Chapter Opener: Finding the Area of the Koch Snowflake

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.

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.

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.

Solution

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

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

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

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

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

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

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

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

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

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

and the length of each side is

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

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

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

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

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

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

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

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

$$\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

$$\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,

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

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

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

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

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

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

$$\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

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

Analysis

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

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

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

Telescoping Series

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

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

Therefore, the series can be written as

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

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

$$\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,

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

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 $1.$

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.

For example, any series of the form

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

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

$$\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

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

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

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

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

Evaluating a Telescoping Series

Determine whether the telescoping series

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

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

Solution

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

$$\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,

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

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

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

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.

Euler’s Constant

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

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

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

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

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

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

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.)

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

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.

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

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

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

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

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

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

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

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

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

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

Section 5.2 Exercises

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

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.

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 $S.$

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.

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

82\.

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

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.

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.

87.

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

88\.

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

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 ${\sum a_{n}}.$

93.

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

94\.

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

95.

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

96\.

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

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.

97.

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

98\.

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

99.

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

100\.

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

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

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.

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.)

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 $\left. 1\text{/}\left( {n2^{n}} \right). \right)$

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

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

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

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.

112\.

Evaluate $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)}$ (*Hint:* $\left. \frac{1}{n\left( {n + 1} \right)\left( {n + 2} \right)} = \frac{1}{2n} - \frac{1}{n + 1} + \frac{1}{2\left( {n + 2} \right)} \right)$

113.

Evaluate ${\sum\limits_{n = 2}^{\infty}\frac{2}{n^{3} - n}}.$

114\.

Find a formula for $\sum\limits_{n = 1}^{\infty}\frac{1}{n\left( {n + N} \right)}$ where $N$ is a positive integer.

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.

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.

Each of the following infinite series converges to the given multiple of $\pi$ or $1\text{/}\pi.$

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, $\pi = 3.141592653589793....$

117.

\[T\] $\pi = -3 + {\sum\limits_{n = 1}^{\infty}\frac{n2^{n}n\text{!}^{2}}{\left( {2n} \right)\text{!}}},$ error $< 0.0001$

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.

1. What is the probability that a fair coin will come up tails $n$ times in a row?

2. Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.

122\.

\[T\] Find the probability that a fair coin is flipped a multiple of three times before coming up heads.

123.

\[T\] Find the probability that a fair coin will come up heads for the second time after an even number of flips.

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.

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.

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).

1. Show that the interest accumulated after $n$ quarters is $\text{\$}10\left( {\frac{1.01^{n + 1} - 1}{0.01} - n} \right).$

2. Find the first eight terms of the sequence.

3. How much interest has accumulated after $2$ years?

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.)

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.

129.

Suppose that $a_{n} \geq 0$ is a sequence of numbers. Explain why the sequence of partial sums of $a_{n}$ is increasing.

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 $a_{n} \geq 0\text{?}$

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.)

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 $k = 1\text{/}4.$

133.

\[T\] Suppose that $\sum\limits_{n = 1}^{\infty}a_{n}$ is a convergent series of positive terms. Explain why $\underset{N\rightarrow\infty}{\text{lim}}{\sum\limits_{n = N + 1}^{\infty}{a_{n} = 0}}.$

134\.

\[T\] Find the length of the dashed zig-zag path in the following figure.

135.

\[T\] Find the total length of the dashed path in the following figure.

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.

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.

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5.3 The Divergence and Integral Tests

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-3-the-divergence-and-integral-tests

5.3 The Divergence and Integral Tests

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.

Divergence Test

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 $k\rightarrow\infty.$

To verify this, notice that from the algebraic properties,

$$\underset{k\rightarrow\infty}{\text{lim}}a_{k} = \underset{k\rightarrow\infty}{\text{lim}}(S_{k} - S_{k - 1}) = \underset{k\rightarrow\infty}{\text{lim}}S_{k} - \underset{k\rightarrow\infty}{\text{lim}}S_{k - 1} = S - S = 0.$$

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:

$$\text{If}\ a_{n}\nrightarrow 0\ \text{as}\ n\rightarrow\infty,{\sum\limits_{n = 1}^{\infty}a_{n}}\ \text{diverges}.$$ (5.8)

This test is known as the divergence test because it provides a way of proving that a series diverges.

Divergence Test

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.

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.

Using the divergence test

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.

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

1. Since $n\text{/}\left( {3n - 1} \right)\rightarrow 1\text{/}3 \neq 0,$ by the divergence test, we can conclude that

$$\sum\limits_{n = 1}^{\infty}\frac{n}{3n - 1}$$

diverges.

2. Since $1\text{/}n^{3}\rightarrow 0,$ the divergence test is inconclusive.

3. Since $e^{1\text{/}n^{2}}\rightarrow 1 \neq 0,$ by the divergence test, the series

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

diverges.

What does the divergence test tell us about the series ${\sum\limits_{n = 1}^{\infty}\text{cos}}(1\text{/}n^{2})\text{?}$

Integral Test

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.

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

$${\sum\limits_{n = 1}^{k}\frac{1}{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \text{⋯} + \frac{1}{k} > {\int_{1}^{k + 1}{\frac{1}{x}dx.}}$$

Therefore, for each $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

$$S_{k} = {\sum\limits_{n = 1}^{k}\frac{1}{n}} > {\int_{1}^{k + 1}{\frac{1}{x}dx}} = \text{ln}\mspace{2mu} x\ |_{1}^{k + 1} = \text{ln}(k + 1) - \text{ln}(1) = \text{ln}(k + 1).$$

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.

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

$${\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}} = 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \text{⋯} + \frac{1}{k^{2}} < 1 + {\int_{1}^{k}{\frac{1}{x^{2}}dx}}.$$

Therefore, for each $k,$ the $k\text{th}$ partial sum $S_{k}$ satisfies

$$S_{k} = {\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}} < 1 + {\int_{1}^{k}{\frac{1}{x^{2}}dx}} = 1 - \left. \frac{1}{x} \right|_{1}^{k} = 1 - \frac{1}{k} + 1 = 2 - \frac{1}{k} < 2.$$

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:

$$S_{k} = S_{k - 1} + \frac{1}{k^{2}}\ \text{for}\ k \geq 2.$$

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.

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

$$S_{k} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k} < a_{1} + {\int_{1}^{k}f}(x)dx < a_{1} + {\int_{1}^{\infty}f}(x)dx.$$

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

$$S_{k} = a_{1} + a_{2} + a_{3} + \text{⋯} + a_{k} > {\int_{1}^{k + 1}f}(x)dx.$$

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.

Integral Test

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:

1. $f$ is continuous,

2. $f$ is decreasing, and

3. $f(n) = a_{n}$ for all integers $n \geq N.$

Then

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{N}^{\infty}{f(x)dx}}$$

both converge or both diverge (see Figure 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,

$${\sum\limits_{n = 1}^{\infty}\left( \frac{1}{e} \right)^{n}} = \frac{1}{e} + \left( \frac{1}{e} \right)^{2} + \left( \frac{1}{e} \right)^{3} + \text{⋯}$$

is a geometric series with initial term $a = 1\text{/}e$ and ratio $r = 1\text{/}e,$ which converges to

$$\frac{1\text{/}e}{1 - (1\text{/}e)} = \frac{1\text{/}e}{(e - 1)\text{/}e} = \frac{1}{e - 1}.$$

However, the related integral ${\int_{1}^{\infty}{(1\text{/}e)}^{x}}dx$ satisfies

$${\int_{1}^{\infty}\left( \frac{1}{e} \right)^{x}}dx = {\int_{1}^{\infty}e^{\text{−}x}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}e^{\text{−}x}}dx = \underset{b\rightarrow\infty}{\text{lim}} - e^{\text{−}x}|_{1}^{b} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {\text{−}e^{\text{−}b} + e^{-1}} \right\rbrack = \frac{1}{e}.$$

Using the Integral Test

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.

1. ${\sum\limits_{n = 1}^{\infty}1}\text{/}n^{3}$

2. $\sum\limits_{n = 1}^{\infty}1\text{/}\sqrt{2n - 1}$

Solution

1. Compare

$${\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3}}}\mspace{9mu}\text{and}\ {\int_{1}^{\infty}\frac{1}{x^{3}}}dx.$$

We have

$${\int_{1}^{\infty}\frac{1}{x^{3}}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}{\frac{1}{x^{3}}dx}} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2x^{2}}|_{1}^{b}} \right\rbrack = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2b^{2}} + \frac{1}{2}} \right\rbrack = \frac{1}{2}.$$

Thus the integral ${\int_{1}^{\infty}1}\text{/}x^{3}dx$ converges, and therefore so does the series

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

2. Compare

$$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{2n - 1}}\mspace{9mu}\text{and}\ {\int_{1}^{\infty}{\frac{1}{\sqrt{2x - 1}}dx}}.$$

Since

$$\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

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

diverges.

Use the integral test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{n}{3n^{2} + 1}$ converges or diverges.

The *p*-Series

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.

For any real number $p,$ the series

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

is called a *p*-series.

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.

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,

$$\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}\ \text{diverges if}\ p \leq 0.$$

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

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n^{p}}\ \text{and}\ {\int_{1}^{\infty}{\frac{1}{x^{p}}dx}}.$$

We have already considered the case when $p = 1.$ Here we consider the case when $p > 0,p \neq 1.$ For this case,

$$\int_{1}^{\infty}{\frac{1}{x^{p}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{1}^{b}{\frac{1}{x^{p}}dx = \underset{b\rightarrow\infty}{\text{lim}}\frac{1}{1 - p}x^{1 - p}|_{1}^{b} = \underset{b\rightarrow\infty}{\text{lim}}\frac{1}{1 - p}\left\lbrack {b^{1 - p} - 1} \right\rbrack.}}}$$

Because

$$b^{1 - p}\rightarrow 0\ \text{if}\ p > 1\ \text{and}\ b^{1 - p}\rightarrow\infty\ \text{if}\ p < 1,$$

we conclude that

$${\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 $0 < p < 1.$

In summary,

$${\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

For each of the following series, determine whether it converges or diverges.

1. $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{4}}$

2. $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2\text{/}3}}$

Solution

1. This is a *p*-series with $p = 4 > 1,$ so the series converges.

2. Since $p = 2\text{/}3 < 1,$ the series diverges.

Does the series $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{5\text{/}4}}$ converge or diverge?

Estimating the Value of a Series

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

$$R_{N} = {\sum\limits_{n = 1}^{\infty}a_{n}} - {\sum\limits_{n = 1}^{N}a_{n}}$$

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 $R_{N}.$

Remainder Estimate from the Integral Test

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:

1. $f$ is continuous,

2. $f$ is decreasing, and

3. $f(n) = a_{n}$ for all integers $n \geq 1.$

Let $S_{N}$ be the *N*th partial sum of $\sum\limits_{n = 1}^{\infty}a_{n}.$ For all positive integers $N,$

$$S_{N} + {\int_{N + 1}^{\infty}{f(x)dx < \sum\limits_{n = 1}^{\infty}a_{n} < S_{N} + {\int_{N}^{\infty}{f(x)dx}}}}.$$

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:

$${\int_{N + 1}^{\infty}{f(x)dx}} < R_{N} < {\int_{N}^{\infty}{f(x)dx.}}$$ (5.10)

This is known as the remainder estimate.

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,

$$R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯} > {\int_{N + 1}^{\infty}{f(x)dx}}$$

and

$$R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + \text{⋯} < {\int_{N}^{\infty}{f(x)dx}}.$$

We conclude that

$${\int_{N + 1}^{\infty}{f(x)dx < R_{N} < {\int_{N}^{\infty}{f(x)dx}}}}.$$

Since

$$\sum\limits_{n = 1}^{\infty}a_{n} = S_{N} + R_{N},$$

where $S_{N}$ is the $N\text{th}$ partial sum, we conclude that

$$S_{N} + {\int_{N + 1}^{\infty}{f(x)dx < \sum\limits_{n = 1}^{\infty}a_{n} < S_{N} + {\int_{N}^{\infty}{f(x)dx}}}}.$$

Estimating the Value of a Series

Consider the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{3}.$

1. Calculate $S_{10} = \sum\limits_{n = 1}^{10}1\text{/}n^{3}$ and estimate the error.

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 $0.001.$

Solution

1. Using a calculating utility, we have

$$S_{10} = 1 + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \frac{1}{4^{3}} + \text{⋯} + \frac{1}{10^{3}} \approx 1.19753.$$

By the remainder estimate, we know

$$R_{N} < {\int_{N}^{\infty}{\frac{1}{x^{3}}dx}}.$$

We have

$$\int_{10}^{\infty}{\frac{1}{x^{3}}dx = \underset{b\rightarrow\infty}{\text{lim}}{\int_{10}^{b}{\frac{1}{x^{3}}dx = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2x^{2}}}\rbrack_{N}^{b} \right.} = \underset{b\rightarrow\infty}{\text{lim}}\left\lbrack {- \frac{1}{2b^{2}} + \frac{1}{2N^{2}}} \right\rbrack = \frac{1}{2N^{2}}.}}$$

Therefore, the error is $R_{10} < 1\text{/}2(10)^{2} = 0.005.$

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 $N = 23.$

For $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{4}},$ calculate $S_{5}$ and estimate the error $R_{5}.$

Section 5.3 Exercises

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.

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.

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.

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.

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 $a_{n} = f(n).$

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.

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.

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 ${\int_{n}^{2n}\frac{1}{t}}dt.\text{)}$

184\.

Find the limit as $n\rightarrow\infty$ of $\frac{1}{n} + \frac{1}{n + 1} + \text{⋯} + \frac{1}{3n}$

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

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:* $\frac{1}{k + 1} < {\int_{k}^{k + 1}\frac{1}{x}}dx$.)

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 $N = 10,20,\text{and}\ 50$.

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.

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 $H_{N} = 1 + 1\text{/}2 + \text{⋯} + 1\text{/}N.$

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 $\lbrack 1,\infty).$

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 $0.0001.$

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 $p\text{?}$

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 $100.$

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 $100.$

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5.4 Comparison Tests

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-4-comparison-tests

5.4 Comparison Tests

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.

Comparison Test

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.

For example, consider the series

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

This series looks similar to the convergent series

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

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

$$0 < \frac{1}{n^{2} + 1} < \frac{1}{n^{2}}$$

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

$$S_{k} = \sum\limits_{n = 1}^{k}\frac{1}{n^{2} + 1} < \sum\limits_{n = 1}^{k}\frac{1}{n^{2}} < \sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}.$$

(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

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

converges.

Similarly, consider the series

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

This series looks similar to the divergent series

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

The sequence of partial sums for each series is monotone increasing and

$$\frac{1}{n - 1\text{/}2} > \frac{1}{n} > 0$$

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

$$S_{k} = \sum\limits_{n = 1}^{k}\frac{1}{n - 1\text{/}2} > \sum\limits_{n = 1}^{k}\frac{1}{n}.$$

(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

$$\sum\limits_{n = 1}^{\infty}\frac{1}{n - 1\text{/}2}\ $$

diverges.

| $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |

|----------------------------------------------|-------|-------------------|----------|----------|----------|----------|----------|----------|

| $\sum\limits_{n = 1}^{k}\frac{1}{n^{2} + 1}$ | $0.5$ | $0.7$ | $0.8$ | $0.8588$ | $0.8973$ | $0.9243$ | $0.9443$ | $0.9597$ |

| $\sum\limits_{n = 1}^{k}\frac{1}{n^{2}}$ | $1$ | $\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)

| $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |

|---------------------------------------------------|-----|----------|-------------------|----------|----------|----------|----------|----------|

| $\sum\limits_{n = 1}^{k}\frac{1}{n - 1\text{/}2}$ | $2$ | $2.6667$ | $3.0667$ | $3.3524$ | $3.5746$ | $3.7564$ | $3.9103$ | $4.0436$ |

| $\sum\limits_{n = 1}^{k}\frac{1}{n}$ | $1$ | $1.5$ | $\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

Comparison Test

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.

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.

Proof

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 $a_{n} \geq 0,$

$$S_{k} = a_{1} + a_{2} + \text{⋯} + a_{k} \leq a_{1} + a_{2} + \text{⋯} + a_{k} + a_{k + 1} = S_{k + 1}.$$

Therefore, the sequence of partial sums is increasing. Further, since $a_{n} \leq b_{n}$ for all $n \geq N,$ then

$${\sum\limits_{n = N}^{k}{a_{n} \leq}}{\sum\limits_{n = N}^{k}{b_{n} \leq}}{\sum\limits_{n = 1}^{\infty}{b_{n} = L.}}$$

Therefore, for all $k \geq 1,$

$$S_{k} = \left( {a_{1} + a_{2} + \text{⋯} + a_{N - 1}} \right) + {\sum\limits_{n = N}^{k}{a_{n} \leq}}\left( {a_{1} + a_{2} + \text{⋯} + a_{N - 1}} \right) + L.$$

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.

□

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.

Using the Comparison Test

For each of the following series, use the comparison test to determine whether the series converges or diverges.

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

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

3. $\sum\limits_{n = 2}^{\infty}\frac{1}{\text{ln}(n)}$

Solution

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,

$$\frac{1}{n^{3} + 3n + 1} < \frac{1}{n^{3}}$$

for every positive integer $n.$ Therefore, we can conclude that $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{3} + 3n + 1}$ converges.

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,

$$\frac{1}{2^{n} + 1} < \frac{1}{2^{n}}$$

for every positive integer $n.$ Therefore, we see that $\sum\limits_{n = 1}^{\infty}\frac{1}{2^{n} + 1}$ converges.

3. Compare to $\sum\limits_{n = 2}^{\infty}\frac{1}{n}.$ Since

$$\frac{1}{\text{ln}(n)} > \frac{1}{n}$$

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.

Use the comparison test to determine if the series $\sum\limits_{n = 1}^{\infty}\frac{n}{n^{3} + n + 1}$ converges or diverges.

Limit Comparison Test

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

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

It is natural to compare this series with the convergent series

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

However, this series does not satisfy the hypothesis necessary to use the comparison test because

$$\frac{1}{n^{2} - 1} > \frac{1}{n^{2}}$$

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

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

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

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

If

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}} = L \neq 0,$$

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

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}(n^{2} - 1)}{1\text{/}n^{2}} = \underset{n\rightarrow\infty}{\text{lim}}\ \frac{n^{2}}{n^{2} - 1} = 1.$$

Since $\sum\limits_{n = 2}^{\infty}1\text{/}n^{2}$ converges, we conclude that

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

converges.

The limit comparison test can be used in two other cases. Suppose

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n}}{b_{n}} = 0.$$

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

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

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.

Limit Comparison Test

Let $a_{n},\ b_{n} \geq 0$ for all $n \geq 1.$

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.

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.

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.

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

$$\frac{1\text{/}\sqrt{n}}{1\text{/}n^{3}} = \frac{n^{3}}{\sqrt{n}} = n^{5\text{/}2}\rightarrow\infty\ \text{as}\ n\rightarrow\infty.$$

Similarly, we see that

$$\frac{1\text{/}n^{2}}{1\text{/}n^{3}} = n\rightarrow\infty\ \text{as}\ n\rightarrow\infty.$$

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

Using the Limit Comparison Test

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.

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

1. Compare this series to $\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}}.$ Calculate

$$\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.

2. Compare this series to $\sum\limits_{n = 1}^{\infty}\left( \frac{2}{3} \right)^{n}.$ We see that

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(2^{n} + 1)\text{/}3^{n}}{2^{n}\text{/}3^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n} + 1}{3^{n}} \cdot \frac{3^{n}}{2^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n} + 1}{2^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\left\lbrack {1 + \left( \frac{1}{2} \right)^{n}} \right\rbrack = 1.$$

Therefore,

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(2^{n} + 1)\text{/}3^{n}}{2^{n}\text{/}3^{n}} = 1.$$

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.

3. Since $\text{ln}\mspace{2mu} n < n,$ compare with $\sum\limits_{n = 1}^{\infty}\frac{1}{n}.$ We see that

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n^{2}} \cdot \frac{n}{1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n}.$$

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

$$\underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} x}{x} = \underset{x\rightarrow\infty}{\text{lim}}\frac{1}{x} = 0.$$

Therefore, $\underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n\text{/}n = 0,$ and, consequently,

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n} = 0.$$

Since the limit is $0$ but $\sum\limits_{n = 1}^{\infty}\frac{1}{n}$ diverges, the limit comparison test does not provide any information.

Compare with $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ instead. In this case,

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\ n\text{/}n^{2}}{1\text{/}n^{2}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\ n}{n^{2}} \cdot \frac{n^{2}}{1} = \underset{n\rightarrow\infty}{\text{lim}}\text{ln}\mspace{2mu} n = \infty.$$

Since the limit is $\infty$ but $\sum\limits_{n = 1}^{\infty}\frac{1}{n^{2}}$ converges, the test still does not provide any information.

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

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n\text{/}n^{2}}{1\text{/}n^{3\text{/}2}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{n^{2}} \cdot \frac{n^{3\text{/}2}}{1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} n}{\sqrt{n}}.$$

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,

$$\underset{x\rightarrow\infty}{\text{lim}}\frac{\text{ln}\mspace{2mu} x}{\sqrt{x}} = \underset{x\rightarrow\infty}{\text{lim}}\frac{2\sqrt{x}}{x} = \underset{x\rightarrow\infty}{\text{lim}}\frac{2}{\sqrt{x}} = 0.$$

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.

Use the limit comparison test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{5^{n}}{3^{n} + 2}$ converges or diverges.

Section 5.4 Exercises

Use the comparison test to determine whether the following series converge.

194\.

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

195.

$\sum\limits_{n = 1}^{\infty}a_{n}$ where $a_{n} = \frac{1}{n\left( {n + {1\text{/}2}} \right)}$

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.

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:*$\left( {1 - \frac{1}{n}} \right)^{n}\rightarrow 1\text{/}e.)$

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 $1 - e^{-1\text{/}n} \approx 1\text{/}n.)$

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 $p\text{?}$

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 $p\text{?}$

224\.

For which $p$ does the series $\sum\limits_{n = 1}^{\infty}2^{pn}\text{/}3^{n}$ converge?

225.

For which $p > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{2^{n}}$ converge?

226\.

For which $r > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{r^{n^{2}}}{2^{n}}$ converge?

227.

For which $r > 0$ does the series $\sum\limits_{n = 1}^{\infty}\frac{2^{n}}{r^{n^{2}}}$ converge?

228\.

Find all values of $p$ and $q$ such that $\sum\limits_{n = 1}^{\infty}\frac{n^{p}}{\left( {n\text{!}} \right)^{q}}$ converges.

229.

Does $\underset{n = 1}{\overset{\infty}{}}\left( \frac{n\pi}{2} \right)$ converge or diverge? Explain.

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 $\sum\limits_{n = 1}^{\infty}\frac{\left| {\text{sin}\mspace{2mu} n} \right|}{\sqrt{n}}.$

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 ${\sum\limits_{n = 1}^{\infty}a_{n}}\ b_{n} \leq \frac{1}{2}\left( {{\sum\limits_{n = 1}^{\infty}a_{n}^{2}} + {\sum\limits_{n = 1}^{\infty}b_{n}^{2}}} \right).$

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 $\text{ln}\mspace{2mu} n.)$

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 $p - \text{series}\text{.})$

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 $1\text{/}n.)$

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?

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?

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?

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?

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.

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.

241.

Let $b_{n}$ be an infinite sequence of zeros and ones. What is the largest possible value of $x = {\sum\limits_{n = 1}^{\infty}{b_{n}\text{/}2^{n}}}\text{?}$

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?

243.

Explain why, if $x > 1\text{/}2,$ then $x$ cannot be written $x = {\sum\limits_{n = 2}^{\infty}\frac{b_{n}}{2^{n}}}\left( {b_{n} = 0\ \text{or}\ 1,\ b_{1} = 0} \right).$

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?

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?

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.

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.

1. How many whole numbers $n$ have $d$ digits?

2. How many $d\text{-digit}$ whole numbers $h(d).$ do not contain $9$ as one or more of their digits?

3. What is the smallest $d\text{-digit}$ number $m(d)\text{?}$

4. Explain why the deleted harmonic series is bounded by ${\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}}.$

5. Show that $\sum\limits_{d = 1}^{\infty}\frac{h(d)}{m(d)}$ converges.

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.)

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 $\text{ln}\left( {1 + t} \right) \leq t,$ $t > 0.)$

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5.5 Alternating Series

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-5-alternating-series

5.5 Alternating Series

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.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

$$\sum\limits_{n = 1}^{\infty}\left( {- \frac{1}{2}} \right)^{n} = - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \text{⋯}$$ (5.11)

and

$$\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \text{⋯}$$ (5.12)

are both alternating series.

Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form

$${\sum\limits_{n = 1}^{\infty}{{(-1)}^{n + 1}b_{n}}} = b_{1} - b_{2} + b_{3} - b_{4} + \text{⋯}$$ (5.13)

or

$$\sum\limits_{n = 1}^{\infty}{(-1)^{n}b_{n} = \text{−}b_{1} + b_{2} - b_{3} + b_{4} - \text{⋯}}$$ (5.14)

Where $b_{n} > 0$ for all positive integers *n*.

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.

To prove this, we look at the sequence of partial sums $\left\{ S_{k} \right\}$ (Figure 5.17).

Proof

Consider the odd terms $S_{2k + 1}$ for $k \geq 0.$ Since $1\text{/}(2k + 1) < 1\text{/}2k,$

$$S_{2k + 1} = S_{2k - 1} - \frac{1}{2k} + \frac{1}{2k + 1} < S_{2k - 1}.$$

Therefore, $\left\{ S_{2k + 1} \right\}$ is a decreasing sequence. Also,

$$S_{2k + 1} = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \text{⋯} + \left( {\frac{1}{2k - 1} - \frac{1}{2k}} \right) + \frac{1}{2k + 1} > 0.$$

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

$$S_{2k} = S_{2k - 2} + \frac{1}{2k - 1} - \frac{1}{2k} > S_{2k - 2}$$

and

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

Therefore, by the Monotone Convergence Theorem, the sequence $\left\{ S_{2k} \right\}$ also converges. Since

$$S_{2k + 1} = S_{2k} + \frac{1}{2k + 1},$$

we know that

$$\underset{k\rightarrow\infty}{\text{lim}}S_{2k + 1} = \underset{k\rightarrow\infty}{\text{lim}}S_{2k} + \underset{k\rightarrow\infty}{\text{lim}}\frac{1}{2k + 1}.$$

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 $S.$

It can also be shown that $S = \text{ln}\mspace{2mu} 2,$ and we can write

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

□

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.

Alternating Series Test

An alternating series of the form

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

converges if

1. $0 < b_{n + 1} \leq b_{n}$ for all $n \geq 1$ and

2. $\underset{n\rightarrow\infty}{\text{lim}}b_{n} = 0.$

This is known as the alternating series test.

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 $n \geq N.$

Convergence of Alternating Series

For each of the following alternating series, determine whether the series converges or diverges.

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

1. Since

$\frac{1}{\left( {n + 1} \right)^{2}} < \frac{1}{n^{2}}\quad\text{and}\quad\frac{1}{n^{2}}\rightarrow 0,$

the series converges.

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

$\underset{n\rightarrow\infty}{\text{lim}}\frac{(-1)^{n + 1}n}{n + 1} \neq 0,$

the series diverges.

Determine whether the series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}2^{n}$ converges or diverges.

Remainder of an Alternating Series

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

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

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

$$\left| R_{N} \right| = \left| {S - S_{N}} \right| \leq \left| {S_{N + 1} - S_{N}} \right| = b_{n + 1}.$$

Remainders in Alternating Series

Consider an alternating series of the form

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

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

$$\left| R_{N} \middle| \leq b_{N + 1}. \right.$$

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 $b_{N + 1}.$

Estimating the Remainder of an Alternating Series

Consider the alternating series

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

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 $S_{10}.$

Solution

From the theorem stated above,

$|R_{10}| \leq b_{11} = \frac{1}{11^{2}} \approx 0.008265.$

Find a bound for $R_{20}$ when approximating $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}n$ by $S_{20}.$

Absolute and Conditional Convergence

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.

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.

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.

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.

Absolute Convergence Implies Convergence

If $\sum\limits_{n = 1}^{\infty}|a_{n}|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

Proof

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

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

converges. By using the algebraic properties for convergent series, we conclude that

$$\sum\limits_{n = 1}^{\infty}a_{n} = \sum\limits_{n = 1}^{\infty}(\left| a_{n} \right| + a_{n})\text{−}\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$$

converges.

□

Absolute versus Conditional Convergence

For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.

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

1. We can see that

$$\sum\limits_{n = 1}^{\infty}\left| \frac{(-1)^{n + 1}}{3n + 1} \right| = \sum\limits_{n = 1}^{\infty}\frac{1}{3n + 1}$$

diverges by using the limit comparison test with the harmonic series.

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}(3n + 1)}{1\text{/}n} = \frac{1}{3}$$

Thus, applying Theorem 5.13, the series cannot converge absolutely. Moreover, because of the alternating series test, we can see that the series converges.

$$\frac{1}{3\left( {n + 1} \right) + 1} < \frac{1}{3n + 1}\ \text{and}\ \frac{1}{3n + 1}\rightarrow 0,$$

We can conclude that $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\text{/}(3n + 1)$ converges conditionally.

2. Noting that $\left| \text{cos}\mspace{2mu} n \middle| \leq 1, \right.$ to determine whether the series converges absolutely, compare

$$\sum\limits_{n = 1}^{\infty}\left| \frac{\text{cos}\mspace{2mu} n}{n^{2}} \right|$$

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.

Determine whether the series $\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}n\text{/}(2n^{3} + 1)$ converges absolutely, converges conditionally, or diverges.

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.

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

$$1 + \frac{1}{3} + \frac{1}{5} + \text{⋯} + \frac{1}{2k - 1} > 10.$$

(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

$$1 + \frac{1}{3} + \text{⋯} + \frac{1}{2k - 1} - \frac{1}{2} + \frac{1}{2k + 1} + \text{⋯} + \frac{1}{2j + 1} > 100.$$

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.

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.

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.

Rearranging Series

Use the fact that

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \text{⋯} = \text{ln}\mspace{2mu} 2$$

to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is $3\mspace{2mu}\text{ln}(2)\text{/}2.$

Solution

Let

$${\sum\limits_{n = 1}^{\infty}a_{n}} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \text{⋯}.$$

Since ${\sum\limits_{n = 1}^{\infty}a_{n}} = \text{ln}(2),$ by the algebraic properties of convergent series,

$${\sum\limits_{n = 1}^{\infty}\frac{1}{2}}a_{n} = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \text{⋯} = \frac{1}{2}{\sum\limits_{n = 1}^{\infty}a_{n}} = \frac{\text{ln}\mspace{2mu} 2}{2}.$$

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

$${\sum\limits_{n = 1}^{\infty}b_{n}} = 0 + \frac{1}{2} + 0 - \frac{1}{4} + 0 + \frac{1}{6} + 0 - \frac{1}{8} + \text{⋯} = \frac{\text{ln}\mspace{2mu} 2}{2}.$$

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

$${\sum\limits_{n = 1}^{\infty}{(a_{n} + b_{n})}} = {\sum\limits_{n = 1}^{\infty}a_{n}} + {\sum\limits_{n = 1}^{\infty}b_{n}} = \text{ln}\mspace{2mu} 2 + \frac{\text{ln}\mspace{2mu} 2}{2} = \frac{3\mspace{2mu}\text{ln}\mspace{2mu} 2}{2}.$$

Now adding the corresponding terms, $a_{n}$ and $b_{n},$ we see that

$$\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

$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \text{⋯} = \frac{3\mspace{2mu}\text{ln}(2)}{2}.$$

Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

Section 5.5 Exercises

State whether each of the following series converges absolutely, conditionally, or not at all.

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 $n.)$

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 $n.)$

270\.

$\sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\left( {\sqrt{n + 1} - \sqrt{n}} \right)$ (*Hint:* Rationalize the numerator.)

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.)

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.)

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 $N.$

280\.

\[T\] $b_{n} = 1\text{/}n,$ error $< 10^{-5}$

281.

\[T\] $b_{n} = 1\text{/}\text{ln}(n),$ $n \geq 2,$ error $< 10^{-1}$

282\.

\[T\] $b_{n} = 1\text{/}\sqrt{n},$ error $< 10^{-3}$

283.

\[T\] $b_{n} = 1\text{/}2^{n},$ error $< 10^{-6}$

284\.

\[T\] $b_{n} = \text{ln}\left( {1 + \frac{1}{n}} \right),$ error $< 10^{-3}$

285.

\[T\] $b_{n} = 1\text{/}n^{2},$ error $< 10^{-6}$

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.

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.

287.

If $b_{n} \geq 0$ is decreasing, then $\sum\limits_{n = 1}^{\infty}\left( {b_{2n - 1} - b_{2n}} \right)$ converges absolutely.

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.

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.

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.

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.

292\.

Suppose that $a_{n}$ is a sequence of positive real numbers and that $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

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?

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?

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

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

not converge. What hypothesis of the alternating series test is not met?

299.

Suppose that $\sum a_{n}$ converges absolutely. Show that the series consisting of the positive terms $a_{n}$ also converges.

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?

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 $\theta = \pi.$

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 $\theta - \theta^{3}\text{/}3\text{!} + \theta^{5}\text{/}5\text{!}\text{−}\theta^{7}\text{/}7\text{!} + \theta^{9}\text{/}9\text{!}$ for $\theta = 1,$ $\theta = \pi\text{/}6,$ and $\theta = \pi.$

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 $0.00001\text{?}$

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 $0.00001\text{?}$

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?

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, $\pi = 3.141592653589793\text{…}.$

306\.

\[T\] $\frac{\pi}{4} = {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}},$ error $< 0.0001$

307.

\[T\] $\frac{\pi}{\sqrt{12}} = {\sum\limits_{k = 0}^{\infty}\frac{(-3)^{\text{−}k}}{2k + 1}},$ error $< 0.0001$

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.)

309.

\[T\] If ${\sum\limits_{n = 1}^{N}{(-1)^{n - 1}\frac{1}{n}}}\rightarrow\text{ln}\mspace{2mu} 2,$ what is $1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{8} - \frac{1}{10} - \frac{1}{12} + \text{⋯}\text{?}$

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 $x\text{?}$

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

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.

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.

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 $1 + {\sum\limits_{n = 1}^{1000}\frac{1}{n^{2}(n + 1)}}.$

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}

n \\

m

\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 $\text{ln}(2).$

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.

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.

2. Explain why eventually $S_{n} \geq A,$ and for any $m$ larger than this $n,$ $A - a_{m} \leq S_{m} \leq A + a_{m}.$

3. Explain why this implies that $S_{n}\rightarrow A$ as $n\rightarrow\infty.$

---

5.6 Ratio and Root Tests

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-6-ratio-and-root-tests

5.6 Ratio and Root Tests

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.

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.

Ratio Test

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.

Ratio Test

Let $\sum\limits_{n = 1}^{\infty}a_{n}$ be a series with nonzero terms. Let

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right|.$$

1. If $0 \leq \rho < 1,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely.

2. If $\rho > 1$ or $\rho = \infty,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

3. If $\rho = 1,$ the test does not provide any information.

Proof

Let $\sum\limits_{n = 1}^{\infty}a_{n}$ be a series with nonzero terms.

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

$$\left| {\left| \frac{a_{n + 1}}{a_{n}} \right| - \rho} \right| < \varepsilon\ \text{for all}\ n \geq N.$$

Therefore,

$$\left| \frac{a_{n + 1}}{a_{n}} \right| < \rho + \varepsilon = R\ \text{for all}\ n \geq N$$

and, thus,

$$\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

$$\left. R \middle| a_{N} \middle| + R^{2} \middle| a_{N} \middle| + R^{3} \middle| a_{N} \middle| + \text{⋯} \right.$$

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

$$\left| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \middle| a_{N + 3} \middle| + \middle| a_{N + 4} \middle| + \text{⋯} \right.$$

converges. Therefore, since

$${\sum\limits_{n = 1}^{\infty}|}a_{n}| = {\sum\limits_{n = 1}^{N}|}a_{n}| + {\sum\limits_{n = N + 1}^{\infty}|}a_{n}|$$

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.

For part ii.

Now suppose $\rho = \left| \frac{a_{n + 1}}{a_{n}} \right| > 1$.

There exists some $N$ such that $\left| \frac{a_{n + 1}}{a_{n}} \right| > 1$ for all $n \geq N$.

Then $\left| a_{n -} \right| > \left| a_{n} \right|$.

Now because $\left| a_{n -} \right| > \left| a_{n} \right|$ for all $n \geq N$, we know that $\left| a_{n} \right| \neq 0$.

It follows that $\left| a_{n} \right| \neq 0$.

By the divergence test, the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

$$\left| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \middle| a_{N + 3} \middle| + \text{⋯} \right.$$

diverges, and therefore the series ${\sum\limits_{n = 1}^{\infty}|}a_{n}|$ diverges.

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 $p,$

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{1\text{/}\left( {n + 1} \right)^{p}}{1\text{/}n^{p}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{p}}{\left( {n + 1} \right)^{p}} = 1.$$

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 $p > 1.$

□

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.

Using the Ratio Test

For each of the following series, use the ratio test to determine whether the series converges or diverges.

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

1. From the ratio test, we can see that

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n + 1}\text{/}(n + 1)\text{!}}{2^{n}\text{/}n\text{!}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{2^{n + 1}}{(n + 1)\text{!}} \cdot \frac{n\text{!}}{2^{n}}.$$

Since $(n + 1)\text{!} = (n + 1) \cdot n\text{!},$

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{2}{n + 1} = 0.$$

Since $\rho < 1,$ the series converges.

2. We can see that

$$\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.

3. Since

$$\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

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\frac{(n + 1)(n + 1)}{(2n + 2)(2n + 1)} = \frac{1}{4}.$$

Since $\rho < 1,$ the series converges.

Use the ratio test to determine whether the series $\sum\limits_{n = 1}^{\infty}\frac{n^{3}}{3^{n}}$ converges or diverges.

Root Test

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

$$\left| a_{N} \middle| + \middle| a_{N + 1} \middle| + \middle| a_{N + 2} \middle| + \text{⋯} \approx \rho^{N} + \rho^{N + 1} + \rho^{N + 2} + \text{⋯}. \right.$$

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

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| \frac{1}{n^{p}} \right|} = \underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{p\text{/}n}}.$$

To evaluate this limit, we use the natural logarithm function. Doing so, we see that

$$\text{ln}\mspace{2mu}\rho = \text{ln}\left( {\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n^{p\text{/}n}}} \right) = \underset{n\rightarrow\infty}{\text{lim}}\text{ln}\left( \frac{1}{n} \right)^{p\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{p}{n} \cdot \text{ln}\left( \frac{1}{n} \right) = \underset{n\rightarrow\infty}{\text{lim}}\frac{p\mspace{2mu}\text{ln}(1\text{/}n)}{n}.$$

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 $p < 1.$

Root Test

Consider the series $\sum\limits_{n = 1}^{\infty}a_{n}.$ Let

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|}.$$

1. If $0 \leq \rho < 1,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges absolutely.

2. If $\rho > 1$ or $\rho = \infty,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

3. If $\rho = 1,$ the test does not provide any information.

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 $\underset{n\rightarrow\infty}{\text{lim}}b_{n}.$

Using the Root Test

For each of the following series, use the root test to determine whether the series converges or diverges.

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

1. To apply the root test, we compute

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left( {n^{2} + 3n} \right)^{n}\text{/}\left( {4n^{2} + 5} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{2} + 3n}{4n^{2} + 5} = \frac{1}{4}.$$

Since $\rho < 1,$ the series converges absolutely.

2. We have

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{n^{n}\text{/}\left( {\text{ln}\ n} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n}{\text{ln}\ n} = \infty\ \text{by L’Hôpital’s rule}.$$

Since $\rho = \infty,$ the series diverges.

Use the root test to determine whether the series $\sum\limits_{n = 1}^{\infty}1\text{/}n^{n}$ converges or diverges.

Choosing a Convergence Test

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.

Choosing a Convergence Test for a Series

Consider a series ${\sum\limits_{n = 1}^{\infty}a_{n}}.$ In the steps below, we outline a strategy for determining whether the series converges.

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.

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

3. Is the series similar to a $p - \text{series}$ or geometric series? If so, try the comparison test or limit comparison test.

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.

5. Use the divergence test. If this test does not provide any information, try the integral test.

Visit this website for more information on testing series for convergence, plus general information on sequences and series.

Using Convergence Tests

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.

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

1. Step 1. The series is not a $p–\text{series}$ or geometric series.

Step 2. The series is not alternating.

Step 3. For large values of $n,$ we approximate the series by the expression

$$\frac{n^{2} + 2n}{n^{3} + 3n^{2} + 1} \approx \frac{n^{2}}{n^{3}} = \frac{1}{n}.$$

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

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{(n^{2} + 2n)\text{/}(n^{3} + 3n^{2} + 1)}{1\text{/}n} = \underset{n\rightarrow\infty}{\text{lim}}\frac{n^{3} + 2n^{2}}{n^{3} + 3n^{2} + 1} = 1.$$

Since the series ${\sum\limits_{n = 1}^{\infty}1}\text{/}n$ diverges, this series diverges as well.

2. Step 1.The series is not a familiar series.

Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series

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

Step 3. The series is not similar to a *p*-series or geometric series.

Step 4. Since each term contains a factorial, apply the ratio test. We see that

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\left( {3\left( {n + 1} \right)} \right)\text{/}\left( {n + 1} \right)\text{!}}{\left( {3n + 1} \right)\text{/}n\text{!}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3n + 3}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{3n + 1} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3n + 3}{\left( {n + 1} \right)\left( {3n + 1} \right)} = 0.$$

Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.

3. Step 1. The series is not a familiar series.

Step 2. It is not an alternating series.

Step 3. There is no obvious series with which to compare this series.

Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.

Step 5. To apply the divergence test, we calculate that

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{e^{n}}{n^{3}} = \infty.$$

Therefore, by the divergence test, the series diverges.

4. Step 1. This series is not a familiar series.

Step 2. It is not an alternating series.

Step 3. There is no obvious series with which to compare this series.

Step 4. Since each term is a power of $n,$ we can apply the root test. Since

$$\underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left( \frac{3}{n + 1} \right)^{n}} = \underset{n\rightarrow\infty}{\text{lim}}\frac{3}{n + 1} = 0,$$

by the root test, we conclude that the series converges.

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.

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.

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.
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 $\sum\limits_{n = 1}^{\infty}1\text{/}n.$
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 an ≤ bn 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 an ≥ bn 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 an = 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 bn + 1 ≤ bn for all n ≥ 1 and bn → 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 |an|=bnn.
If ρ > 1 or ρ = ∞, the series diverges.
If ρ = 1, the test is inconclusive.

Table 5.3 Summary of Convergence Tests

Series Converging to $\pi$ and $1\text{/}\pi$

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.

1. The series

$$\pi = 4\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{2n - 1} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \text{⋯}$$

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.

1. Prove that this series converges.

2. Evaluate the partial sums $S_{n}$ for $n = 10,20,50,100.$

3. Use the remainder estimate for alternating series to get a bound on the error $R_{n}.$

4. What is the smallest value of $N$ that guarantees $\left| R_{N} \middle| < 0.01\text{?} \right.$ Evaluate $S_{N}.$

2. The series

$$\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 $f(x) = \text{sin}^{-1}x.$

1. Prove that the series converges.

2. Evaluate the partial sums $S_{n}$ for $n = 5,10,20.$

3. Compare $S_{n}$ to $\pi$ for $n = 5,10,20$ and discuss the number of correct decimal places.

3. The series

$$\frac{1}{\pi} = \frac{\sqrt{8}}{9801}\sum\limits_{n = 0}^{\infty}\frac{\left( {4n} \right)\text{!}(1103 + 26390n)}{\left( {n\text{!}} \right)^{4}396^{4n}}$$

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 $1\text{/}\pi.$

1. Prove that this series converges.

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?

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.

Section 5.6 Exercises

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.

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.

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.

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.

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 ${\int_{k}^{2k}\frac{dt}{t}}.)$

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.

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.

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 $\left\{ x_{n} \right\}.)$

In the following exercises, use an appropriate test to determine whether the series converges.

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:* $\left( {1 + \frac{1}{n^{2}}} \right)^{n^{2}} \approx e.)$

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:* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\text{ln}\mspace{2mu} k} \approx e^{\text{−}\text{ln}\mspace{2mu} k}.)$

359.

$a_{k} = \left( \frac{k}{k + \text{ln}\mspace{2mu} k} \right)^{2k}$ (*Hint:* $a_{k} = \left( {1 + \frac{\text{ln}\mspace{2mu} k}{k}} \right)^{\text{−}(k\text{/}\text{ln}\mspace{2mu} k)\mspace{2mu}\text{ln}\mspace{2mu} k^{2}}.)$

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.

360\.

$\sum\limits_{k = 1}^{\infty}\frac{k}{2^{k}}$ (*Hint:* Take $a_{k} = k$ and $b_{k} = 2^{1 - k}.)$

361.

${\sum\limits_{k = 1}^{\infty}\frac{k}{c^{k}}},$ where $c > 1$ (*Hint:* Take $a_{k} = k$ and $b_{k} = c^{1 - k}\text{/}(c - 1).)$

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.

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?

369.

Let $0 < r < 1.$ For which real numbers $p$ does ${\sum\limits_{n = 1}^{\infty}n^{p}}r^{n}$ converge?

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?

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?

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?

373.

For which values of $r > 0,$ if any, does $\sum\limits_{n = 1}^{\infty}r^{\sqrt{n}}$ converge? (*Hint:* ${\sum\limits_{n = 1}^{\infty}a_{n}} = {\sum\limits_{k = 1}^{\infty}{\sum\limits_{n = k^{2}}^{{(k + 1)}^{2} - 1}a_{n}}}.)$

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?

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.

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.

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 $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{2n}}{a_{n}}.$

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 $1\text{/}\left( {1 + x\text{/}\left( {2n} \right)} \right).)$

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 $n\rightarrow\infty.$

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Chapter Review

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-chapter-review

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Key Terms

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Key Terms

absolute convergence

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

alternating series

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

alternating series test

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

arithmetic sequence

a sequence in which the difference between every pair of consecutive terms is the same is called an arithmetic sequence

bounded above

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 $n$

bounded below

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 $n$

bounded sequence

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 $n$

comparison test

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

conditional convergence

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

convergence of a series

a series converges if the sequence of partial sums for that series converges

convergent sequence

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

divergence of a series

a series diverges if the sequence of partial sums for that series diverges

divergence test

if $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ then the series $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges

divergent sequence

a sequence that is not convergent is divergent

explicit formula

a sequence may be defined by an explicit formula such that $a_{n} = f(n)$

geometric sequence

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

geometric series

a geometric series is a series that can be written in the form

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

harmonic series

the harmonic series takes the form

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

index variable

the subscript used to define the terms in a sequence is called the index

infinite series

an infinite series is an expression of the form

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

integral test

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

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{1}^{\infty}{f(x)dx}}$$

either both converge or both diverge

limit comparison test

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

limit of a sequence

the real number $L$ to which a sequence converges is called the limit of the sequence

monotone sequence

an increasing or decreasing sequence

*p*-series

a series of the form $\sum\limits_{n = 1}^{\infty}1\text{/}n^{p}$

partial sum

the $k\text{th}$ partial sum of the infinite series $\sum\limits_{n = 1}^{\infty}a_{n}$ is the finite sum

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

ratio test

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

recurrence relation

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

remainder estimate

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:

$${\int_{N + 1}^{\infty}{f(x)dx < R_{N} <}}{\int_{N}^{\infty}{f(x)dx}}$$

root test

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

sequence

an ordered list of numbers of the form $a_{1},a_{2},a_{3}\text{,…}$ is a sequence

telescoping series

a telescoping series is one in which most of the terms cancel in each of the partial sums

term

the number $a_{n}$ in the sequence $\left\{ a_{n} \right\}$ is called the $n\text{th}$ term of the sequence

unbounded sequence

a sequence that is not bounded is called unbounded

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Key Equations

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Key Equations

| | |

|-------------------------------|------------------------------------------------------------------------------------------------------|

| 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 | ${\int_{N + 1}^{\infty}{f(x)dx}} < R_{N} < {\int_{N}^{\infty}{f(x)dx}}$ |

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Key Concepts

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Key Concepts

5.1 Sequences

5.2 Infinite Series

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

and the corresponding sequence of partial sums $\left\{ S_{k} \right\}$ where

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

the series converges if and only if the sequence $\left\{ S_{k} \right\}$ converges.

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

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

diverges.

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,

$${\sum\limits_{n = 1}^{\infty}{\lbrack b_{n} - b_{n + 1}\rbrack}} = b_{1} - \underset{k\rightarrow\infty}{\text{lim}}{\left( b_{k + 1} \right).}$$

5.3 The Divergence and Integral Tests

$$\sum\limits_{n = 1}^{\infty}a_{n}\text{and}{\int_{1}^{\infty}{f(x)dx}}$$

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 ${\int_{N + 1}^{\infty}{f(x)dx < R_{N} < {\int_{N}^{\infty}{f(x)dx}}}}.$

5.4 Comparison Tests

5.5 Alternating Series

5.6 Ratio and Root Tests

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{a_{n + 1}}{a_{n}} \right|.$$

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.

$$\rho = \underset{n\rightarrow\infty}{\text{lim}}\sqrt[n]{\left| a_{n} \right|}.$$

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.

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Review Exercises

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/5-review-exercises

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Review Exercises

Calculus Volume 2Review Exercises

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Review Exercises

*True or False?* Justify your answer with a proof or a counterexample.

379.

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} = 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

380\.

If $\underset{n\rightarrow\infty}{\text{lim}}a_{n} \neq 0,$ then $\sum\limits_{n = 1}^{\infty}a_{n}$ diverges.

381.

If $\sum\limits_{n = 1}^{\infty}\left| a_{n} \right|$ converges, then $\sum\limits_{n = 1}^{\infty}a_{n}$ converges.

382\.

If $\sum\limits_{n = 1}^{\infty}{2^{n}a_{n}}$ converges, then $\sum\limits_{n = 1}^{\infty}{(-2)^{n}a_{n}}$ converges.

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

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?

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?

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

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?

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 $x_{0}.$

401.

Find $\underset{n\rightarrow\infty}{\text{lim}}x_{n}$ if $b > 1,$ $b < 1,$ and $b = 1.$

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?

403.

If $b = \frac{3}{4}$ and $x_{0} = 100,$ find $S_{10}$ and $\underset{n\rightarrow\infty}{\text{lim}}S_{n}$

404\.

For what values of $b$ will the series converge and diverge? What does the series converge to?

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This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Attribution information

Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction

Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction

Citation information

© Jul 15, 2026 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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