## Divergence and Integral Tests

### Learning Outcomes

• Use the divergence test to determine whether a series converges or diverges
• Use the integral test to determine the convergence of a series

## Divergence Test

For a series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ to converge, the $n\text{th}$ term ${a}_{n}$ must satisfy ${a}_{n}\to 0$ as $n\to \infty$.

Therefore, from the algebraic limit properties of sequences,

$\underset{k\to \infty }{\text{lim}}{a}_{k}=\underset{k\to \infty }{\text{lim}}\left({S}_{k}-{S}_{k - 1}\right)=\underset{k\to \infty }{\text{lim}}{S}_{k}-\underset{k\to \infty }{\text{lim}}{S}_{k - 1}=S-S=0$.

Therefore, if $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges, the $n\text{th}$ term ${a}_{n}\to 0$ as $n\to \infty$. An important consequence of this fact is the following statement:

$\text{If }{a}_{n}\nrightarrow 0\text{ as }n\to \infty ,\displaystyle\sum _{n=1}^{\infty }{a}_{n}\text{ diverges}$.

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

### Theorem: Divergence Test

If $\underset{n\to \infty }{\text{lim}}{a}_{n}=c\ne 0$ or $\underset{n\to \infty }{\text{lim}}{a}_{n}$ does not exist, then the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.

It is important to note that the converse of this theorem is not true. That is, if $\underset{n\to \infty }{\text{lim}}{a}_{n}=0$, we cannot make any conclusion about the convergence of $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$. For example, $\underset{n\to 0}{\text{lim}}\left(\frac{1}{n}\right)=0$, but the harmonic series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ 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 ${a}_{n}\to 0$, the divergence test is inconclusive.

### Example: 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. $\displaystyle\sum _{n=1}^{\infty }\frac{n}{3n - 1}$
2. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}$
3. $\displaystyle\sum _{n=1}^{\infty }{e}^{\frac{1}{{n}^{2}}}$

### try it

What does the divergence test tell us about the series $\displaystyle\sum _{n=1}^{\infty }\cos\left(\frac{1}{{n}^{2}}\right)\text{?}$

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

## 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 ${S}_{{2}^{k}}>1+\frac{k}{2}$ 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 1, we depict the harmonic series by sketching a sequence of rectangles with areas $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ along with the function $f\left(x\right)=\frac{1}{x}$. From the graph, we see that

$\displaystyle\sum _{n=1}^{k}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{k}>{\displaystyle\int }_{1}^{k+1}\frac{1}{x}dx$.

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

${S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{n}>{\displaystyle\int }_{1}^{k+1}\frac{1}{x}dx=\text{ln}x{|}_{1}^{k+1}=\text{ln}\left(k+1\right)-\text{ln}\left(1\right)=\text{ln}\left(k+1\right)$.

Since $\underset{k\to \infty }{\text{lim}}\text{ln}\left(k+1\right)=\infty$, we see that the sequence of partial sums $\left\{{S}_{k}\right\}$ is unbounded. Therefore, $\left\{{S}_{k}\right\}$ diverges, and, consequently, the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ also diverges.

Now consider the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$. We show how an integral can be used to prove that this series converges. In Figure 2, we sketch a sequence of rectangles with areas $1,\frac{1}{{2}^{2}},\frac{1}{{3}^{2}},\ldots$ along with the function $f\left(x\right)=\frac{1}{{x}^{2}}$. From the graph we see that

$\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}=1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\cdots +\frac{1}{{k}^{2}}<1+{\displaystyle\int }_{1}^{k}\frac{1}{{x}^{2}}dx$.

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

${S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}<1+{\displaystyle\int }_{1}^{k}\frac{1}{{x}^{2}}dx=1-{\frac{1}{x}|}_{1}^{k}=1-\frac{1}{k}+1=2-\frac{1}{k}<2$.

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:

${S}_{k}={S}_{k - 1}+\frac{1}{{k}^{2}}\text{for}k\ge 2$.

Since $\left\{{S}_{k}\right\}$ is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}$ converges.

We can extend this idea to prove convergence or divergence for many different series. Suppose $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ is a series with positive terms ${a}_{n}$ such that there exists a continuous, positive, decreasing function $f$ where $f\left(n\right)={a}_{n}$ for all positive integers. Then, as in Figure 3 (a), for any integer $k$, the $k\text{th}$ partial sum ${S}_{k}$ satisfies

${S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{k}<{a}_{1}+{\displaystyle\int }_{1}^{k}f\left(x\right)dx<1+{\displaystyle\int }_{1}^{\infty }f\left(x\right)dx$.

Therefore, if ${\displaystyle\int }_{1}^{\infty }f\left(x\right)dx$ 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 ${\displaystyle\int }_{1}^{\infty }f\left(x\right)dx$ converges, then the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ also converges. On the other hand, from Figure 3 (b), for any integer $k$, the $k\text{th}$ partial sum ${S}_{k}$ satisfies

${S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{k}>{\displaystyle\int }_{1}^{k+1}f\left(x\right)dx$.

If $\underset{k\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{k+1}f\left(x\right)dx=\infty$, then $\left\{{S}_{k}\right\}$ is an unbounded sequence and therefore diverges. As a result, the series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ also diverges. Since $f$ is a positive function, if ${\displaystyle\int }_{1}^{\infty }f\left(x\right)dx$ diverges, then $\underset{k\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{k+1}f\left(x\right)dx=\infty$. We conclude that if ${\displaystyle\int }_{1}^{\infty }f\left(x\right)dx$ diverges, then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ diverges.

### Theorem: Integral Test

Suppose $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ 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. $f\left(n\right)={a}_{n}$ for all integers $n\ge N$.

Then $\displaystyle\sum _{n=1}^{\infty }{a}_{n}\text{ and }{\displaystyle\int }_{N}^{\infty }f\left(x\right)dx$both converge or both diverge (see Figure 3).

Although convergence of ${\displaystyle\int }_{N}^{\infty }f\left(x\right)dx$ implies convergence of the related series $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$, 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,

$\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{e}\right)}^{n}=\frac{1}{e}+{\left(\frac{1}{e}\right)}^{2}+{\left(\frac{1}{e}\right)}^{3}+\cdots$

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

$\frac{\frac{1}{e}}{1-\left(\frac{1}{e}\right)}=\frac{\frac{1}{e}}{\frac{\left(e - 1\right)}{e}}=\frac{1}{e - 1}$.

However, the related integral ${\displaystyle\int }_{1}^{\infty }{\left(\frac{1}{e}\right)}^{x}dx$ satisfies

${\displaystyle\int }_{1}^{\infty }{\left(\frac{1}{e}\right)}^{x}dx={\displaystyle\int }_{1}^{\infty }{e}^{\text{-}x}dx=\underset{b\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{b}{e}^{\text{-}x}dx=\underset{b\to \infty }{\text{lim}}-{e}^{\text{-}x}{|}_{1}^{b}=\underset{b\to \infty }{\text{lim}}\left[\text{-}{e}^{\text{-}b}+{e}^{-1}\right]=\frac{1}{e}$.

In the following examples, we explore how to use the integral test.  Before doing so, we should note that since one of the conditions of the test is that the function is decreasing, we can use calculus to verify that this condition is met.

### Recall: Showing that a function is decreasing

A differentiable function $f(x)$ is decreasing on an interval $\left( a,b \right)$ if $f'(x) < 0$ for all $x\in\left( a,b \right)$.

### Example: Using the Integral Test

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

1. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}$
2. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{\sqrt{2n - 1}}$

### try it

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

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.