# 8.3 : The Divergence and Integral Tests

- Page ID
- 14716

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums \(\displaystyle {S_k}.\) 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

For a series \(\displaystyle \sum^∞_{n=1}a_n\) to converge, the \(\displaystyle n^{th}\) term \(\displaystyle a_n\) must satisfy \(\displaystyle a_n→0\) as \(\displaystyle n→∞.\) Therefore, from the algebraic limit properties of sequences,

\[\begin{align*} \lim_{k→∞}a_k &= \lim_{k→∞}(S_k−S_{k−1}) \\[5pt] &=\lim_{k→∞}S_k−\lim_{k→∞}S_{k−1} \\[5pt] &=S−S=0. \end{align*}\]

Therefore, if \(\displaystyle \sum_{n=1}^∞a_n\) converges, the \(\displaystyle n^{th}\) term \(\displaystyle a_n→0\) as \(\displaystyle n→∞.\) An important consequence of this fact is the following statement:

If \(\displaystyle a_n↛0\) as \(\displaystyle n→∞,\sum_{n=1}^∞a_n\) diverges.

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

Definition: The Divergence Test

If \(\displaystyle \lim_{n→∞}a_n=c≠0\) or \(\displaystyle \lim_{n→∞}a_n\) does not exist, then the series \(\displaystyle \sum_{n=1}^∞a_n\) diverges.

It is important to note that the converse of this theorem is not true. That is, if \(\displaystyle \lim_{n→∞}a_n=0\), we cannot make any conclusion about the convergence of \(\displaystyle \sum_{n=1}^∞a_n\). For example, \(\displaystyle \lim_{n→0}(1/n)=0\), but the harmonic series \(\displaystyle \sum^∞_{n=1}1/n\) 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 \(\displaystyle a_n→0\), the divergence test is inconclusive.

Example \(\displaystyle \PageIndex{1}\): 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.

- \(\displaystyle \sum^∞_{n=1}\dfrac{n}{3n−1}\)
- \(\displaystyle \sum^∞_{n=1}\dfrac{1}{n^3}\)
- \(\displaystyle \sum^∞_{n=1}e^{1/n^2}\)

**Solution**

- Since \(\displaystyle n/(3n−1)→1/3≠0\), by the divergence test, we can conclude that \(\displaystyle \sum_{n=1}^∞n\dfrac{n}{3n−1}\) diverges.
- Since \(\displaystyle 1/n^3→0\), the divergence test is inconclusive.
- Since \(\displaystyle e^{1/n^2}→1≠0\), by the divergence test, the series \(\displaystyle \sum_{n=1}^∞e^{1/n^2}\) diverges.

Exercise \(\displaystyle \PageIndex{1}\)

What does the divergence test tell us about the series \(\displaystyle \sum_{n=1}^∞cos(1/n^2)\)?

**Hint**-
Look at \(\displaystyle \lim_{n→∞}\cos(1/n^2)\).

**Answer**-
The series diverges.

## Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums \(\displaystyle {S_k}\) and showing that \(\displaystyle S_{2^k}>1+k/2\) for all positive integers \(\displaystyle k\). 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 \(\PageIndex{1}\), we depict the harmonic series by sketching a sequence of rectangles with areas \(\displaystyle 1,1/2,1/3,1/4,…\) along with the function \(\displaystyle f(x)=1/x.\) From the graph, we see that

\[\sum_{n=1}^k\dfrac{1}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+⋯+\dfrac{1}{k}>∫^{k+1}_1\dfrac{1}{x}dx.\]

Therefore, for each \(\displaystyle k\), the \(\displaystyle kth\) partial sum \(\displaystyle S_k\) satisfies

\[\begin{align*} S_k =\sum_{n=1}^k\dfrac{1}{n} >∫^{k+1}_1\dfrac{1}{x}dx &= \left. \ln x \right| ^{k+1}_1 \\[5pt] &= \ln (k+1)−\ln (1) \\[5pt] &=\ln (k+1).\end{align*}\]

Since \(\displaystyle \lim_{k→∞}\ln(k+1)=∞,\) we see that the sequence of partial sums \(\displaystyle {S_k}\) is unbounded. Therefore, \(\displaystyle {S_k}\) diverges, and, consequently, the series \(\displaystyle \sum_{n=1}^∞\dfrac{1}{n}\) also diverges.

Now consider the series \(\displaystyle \sum_{n=1}^∞1/n^2\). We show how an integral can be used to prove that this series converges. In Figure \(\PageIndex{2}\), we sketch a sequence of rectangles with areas \(\displaystyle 1,1/2^2,1/3^2,…\) along with the function \(\displaystyle f(x)=1/x^2\). From the graph we see that

\[\sum_{n=1}^k\dfrac{1}{n^2}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+⋯+\dfrac{1}{k^2}<1+∫^k_1\dfrac{1}{x^2}dx.\]

Therefore, for each \(\displaystyle k\), the \(\displaystyle kth\) partial sum \(\displaystyle S_k\) satisfies

\[\begin{align*} S_k=\sum_{n=1}^k\dfrac{1}{n^2}<1+∫^k_1\dfrac{1}{x^2}dx &=1−\left. \dfrac{1}{x} \right|^k_1 \\[5pt] &=1−\dfrac{1}{k}+1 \\[5pt] &=2−\dfrac{1}{k}<2. \end{align*}\]

We conclude that the sequence of partial sums \(\displaystyle {S_k}\) is bounded. We also see that \(\displaystyle {S_k}\) is an increasing sequence:

\[S_k=S_{k−1}+\dfrac{1}{k^2}\]

for \(\displaystyle k≥2\).

Since \(\displaystyle {S_k}\) is increasing and bounded, by the **Monotone Convergence Theorem**, it converges. Therefore, the series \(\displaystyle \sum_{n=1}^∞1/n^2\) converges.

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

\[S_k=a_1+a_2+a_3+⋯+a_k<a_1+∫^k_1f(x)dx<1+∫^∞_1f(x)dx.\]

Therefore, if \(\displaystyle ∫^∞_1f(x)dx\) converges, then the sequence of partial sums \(\displaystyle {S_k}\) is bounded. Since \(\displaystyle {S_k}\) is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if \(\displaystyle ∫^∞_1f(x)dx\) converges, then the series \(\displaystyle \sum^∞_{n=1}a_n\) also converges. On the other hand, from Figure \(\PageIndex{3b}\), for any integer \(\displaystyle k\), the \(\displaystyle kth\) partial sum \(\displaystyle S_k\) satisfies

\[S_k=a_1+a_2+a_3+⋯+a_k>∫^{k+1}_1f(x)dx.\]

If

\[\displaystyle \lim_{k→∞}∫^{k+1}_1f(x)dx=∞,\]

then \(\displaystyle {S_k}\) is an unbounded sequence and therefore diverges. As a result, the series \(\displaystyle \sum_{n=1}^∞a_n\) also diverges. Since \(\displaystyle f\) is a positive function, if \(\displaystyle ∫^∞_1f(x)dx\) diverges, then

\[\displaystyle \lim_{k→∞}∫^{k+1}_1f(x)dx=∞.\]

We conclude that if \(\displaystyle ∫^∞_1f(x)dx\) diverges, then \(\displaystyle \sum_{n=1}^∞a_n\) diverges.

Definition: The Integral Test

Suppose \(\displaystyle \sum_{n=1}^∞a_n\) is a series with positive terms \(\displaystyle a_n\). Suppose there exists a function \(\displaystyle f\) and a positive integer \(\displaystyle N\) such that the following three conditions are satisfied:

- \(\displaystyle f\) is continuous,
- \(\displaystyle f\) is decreasing, and
- \(\displaystyle f(n)=a_n\) for all integers \(\displaystyle n≥N.\)

Then

\[\sum_{n=1}^∞a_n \nonumber\]

and

\[∫^∞_Nf(x)dx \nonumber\]

both converge or both diverge (Figure \(\PageIndex{3}\)).

Although convergence of \(\displaystyle ∫^∞_Nf(x)dx\) implies convergence of the related series \(\displaystyle \sum_{n=1}^∞a_n\), 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,

\[\sum_{n=1}^∞\left(\dfrac{1}{e}\right)^n=\dfrac{1}{e}+\left(\dfrac{1}{e}\right)^2+\left(\dfrac{1}{e}\right)^3+⋯\]

is a geometric series with initial term \(\displaystyle a=1/e\) and ratio \(\displaystyle r=1/e,\) which converges to

\[\dfrac{1/e}{1−(1/e)}=\dfrac{1/e}{(e−1)/e}=\dfrac{1}{e−1}.\]

However, the related integral \(\displaystyle ∫^∞_1(1/e)^xdx\) satisfies

\[∫^∞_1(\dfrac{1}{e})^xdx=∫^∞_1e^{−x}dx=\lim_{b→∞}∫^b_1e^{−x}dx=\lim_{b→∞}−e^{−x}∣^b_1=\lim_{b→∞}[−e^{−b}+e^{−1}]=\dfrac{1}{e}.\]

Example \(\displaystyle \PageIndex{2}\): Using the Integral Test

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

- \(\displaystyle sum_{n=1}^∞1/n^3\)
- \(\displaystyle \sum^∞_{n=1}1/\sqrt{2n−1}\)

**Solution**

a. Compare

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{n^3}\) and \(\displaystyle ∫^∞_1\dfrac{1}{x^3}dx.\)

We have

\(\displaystyle ∫^∞_1\dfrac{1}{x^3}dx=\lim_{b→∞}∫^b_1\dfrac{1}{x^3}dx=\lim_{b→∞}[−\dfrac{1}{2x^2}∣^b_1]=\lim_{b→∞}[−\dfrac{1}{2b^2}+\dfrac{1}{2}]=\dfrac{1}{2}.\)

Thus the integral \(\displaystyle ∫^∞_11/x^3dx\) converges, and therefore so does the series

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{n^3}\).

b. Compare

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{\sqrt{2n−1}}\) and \(\displaystyle ∫^∞_1\dfrac{1}{\sqrt{2x−1}}dx\).

Since

\(\displaystyle ∫^∞_1\dfrac{1}{\sqrt{2x−1}}dx=\lim_{b→∞}∫^b_1\dfrac{1}{\sqrt{2x−1}}dx=\lim_{b→∞}\sqrt{2x−1}∣^b_1=\lim_{b→∞}[\sqrt{2b−1}−1]=∞,\)

the integral \(\displaystyle ∫^∞_11/\sqrt{2x−1}dx\) diverges, and therefore

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{\sqrt{2n−1}}\)

diverges.

Exercise \(\displaystyle \PageIndex{2}\)

Use the integral test to determine whether the series \(\displaystyle \sum^∞_{n=1}\dfrac{n}{3n^2+1}\) converges or diverges.

**Hint**-
Compare to the integral \(\displaystyle ∫^∞_1\dfrac{x}{3x^2+1}dx.\)

**Answer**-
The series diverges.

## The p-Series

The harmonic series \(\displaystyle \sum^∞_{n=1}1/n\) and the series \(\displaystyle \sum^∞_{n=1}1/n^2\) are both examples of a type of series called a p-series.

Definition: p-series

For any real number \(\displaystyle p\), the series

\[\sum_{n=1}^∞\dfrac{1}{n^p}\]

is called a *p-series***.**

We know the p-series converges if \(\displaystyle p=2\) and diverges if \(\displaystyle p=1\). What about other values of \(\displaystyle p\)? In general, it is difficult, if not impossible, to compute the exact value of most \(\displaystyle p\)-series. However, we can use the tests presented thus far to prove whether a \(\displaystyle p\)-series converges or diverges.

If \(\displaystyle p<0,\) then \(\displaystyle 1/n^p→∞,\) and if \(\displaystyle p=0\), then \(\displaystyle 1/n^p→1.\) Therefore, by the divergence test,

\[\sum_{n=1}^∞1/n^p\]

diverges if \(p≤0\).

If \(\displaystyle p>0,\) then \(\displaystyle f(x)=1/x^p\) is a positive, continuous, decreasing function. Therefore, for \(\displaystyle p>0,\) we use the integral test, comparing

\[\sum_{n=1}^∞\dfrac{1}{n^p}\] and \[∫^∞_1\dfrac{1}{x^p}dx.\]

We have already considered the case when \(\displaystyle p=1.\) Here we consider the case when \(\displaystyle p>0,p≠1.\) For this case,

\[∫^∞_1\dfrac{1}{x^p}dx=\lim_{b→∞}∫^b_1\dfrac{1}{x^p}dx=\lim_{b→∞}\dfrac{1}{1−p}x^{1−p}∣^b_1=\lim_{b→∞}\dfrac{1}{1−p}[b^{1−p}−1].\]

Because

\(\displaystyle b^{1−p}→0\) if \(\displaystyle p>1\) and \(\displaystyle b^{1−p}→∞\) if \(\displaystyle p<1,\)

we conclude that

\[∫^∞_1\dfrac{1}{x^p}dx=\begin{cases}\dfrac{1}{p−1}&ifp>1//∞&ifp<1.\end{cases}\]

Therefore, \(\displaystyle \sum^∞_{n=1}1/n^p\) converges if \(\displaystyle p>1\) and diverges if \(\displaystyle 0<p<1.\)

In summary,

\[\sum_{n=1}^∞\dfrac{1}{n^p}\begin{cases}converges & if p>1//diverges & ifp≤1\end{cases}\].

Example \(\displaystyle \PageIndex{3}\): Testing for Convergence of p-series

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

- \(\displaystyle \sum^∞_{n=1}\dfrac{1}{n^4}\)
- \(\displaystyle \sum^∞_{n=1}\dfrac{1}{n^{2/3}}\)

**Solution**

- This is a p-series with \(\displaystyle p=4>1\),so the series converges.
- Since \(\displaystyle p=2/3<1,\) the series diverges.

Exercise \(\displaystyle \PageIndex{3}\)

Does the series \(\displaystyle \sum^∞_{n=1}\dfrac{1}{n^{5/4}}\) converge or diverge?

**Hint**-
\(\displaystyle p=5/4\)

**Answer**-
The series converges.

## Estimating the Value of a Series

Suppose we know that a series \(\displaystyle \sum_{n=1}^∞a_n\) converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum \(\displaystyle \sum_{n=1}^Na_n\) where \(\displaystyle N\) is any positive integer. The question we address here is, for a convergent series \(\displaystyle \sum^∞_{n=1}a_n\), how good is the approximation \(\displaystyle \sum^N_{n=1}a_n\)? More specifically, if we let

\[R_N=\sum_{n=1}^∞a_n−\sum_{n=1}^Na_n\]

be the remainder when the sum of an infinite series is approximated by the \(\displaystyle Nth\) partial sum, how large is \(\displaystyle R_N\)? For some types of series, we are able to use the ideas from the integral test to estimate \(\displaystyle R_N\).

Remainder Estimate from the Integral Test

Suppose \(\displaystyle \sum^∞_{n=1}a_n\) is a convergent series with positive terms. Suppose there exists a function \(\displaystyle f\) satisfying the following three conditions:

- \(\displaystyle f\) is continuous,
- \(\displaystyle f\) is decreasing, and
- \(\displaystyle f(n)=a_n\) for all integers \(\displaystyle n≥1.\)

Let \(\displaystyle S_N\) be the Nth partial sum of \(\displaystyle \sum^∞_{n=1}a_n\). For all positive integers \(\displaystyle N\),

\[S_N+∫^∞_{N+1}f(x)dx<\sum_{n=1}^∞a_n<S_N+∫^∞_Nf(x)dx.\]

In other words, the remainder \(\displaystyle R_N=\sum^∞_{n=1}a_n−S_N=\sum^∞_{n=N+1}a_n\) satisfies the following estimate:

\[∫^∞_{N+1}f(x)dx<R_N<∫^∞_Nf(x)dx.\]

This is known as the **remainder estimate.**

We illustrate Note in Figure \(\PageIndex{4}\). In particular, by representing the remainder \(\displaystyle R_N=a_{N+1}+a_{N+2}+a_{N+3}+⋯\) as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by \(\displaystyle ∫^∞_Nf(x)dx\) and bounded below by \(\displaystyle ∫^∞_{N+1}f(x)dx.\) In other words,

\[R_N=a_{N+1}+a_{N+2}+a_{N+3}+⋯>∫^∞_{N+1}f(x)dx\]

and

\[R_N=a_{N+1}+a_{N+2}+a_{N+3}+⋯<∫^∞_Nf(x)dx.\]

We conclude that

\[∫^∞_{N+1}f(x)dx<R_N<∫^∞_Nf(x)dx.\]

Since

\[\sum_{n=1}^∞a_n=S_N+R_N,\]

where \(\displaystyle S_N\) is the \(\displaystyle Nth\) partial sum, we conclude that

\[S_N+∫^∞_{N+1}f(x)dx<\sum_{n=1}^∞a_n<S_N+∫^∞_Nf(x)dx.\]

Example \(\displaystyle \PageIndex{4}\): Estimating the Value of a Series

Consider the series \(\displaystyle \sum^∞_{n=1}1/n^3\).

- Calculate \(\displaystyle S_{10}=\sum^{10}_{n=1}1/n^3\) and estimate the error.
- Determine the least value of \(\displaystyle N\) necessary such that \(\displaystyle S_N\) will estimate \(\displaystyle \sum^∞_{n=1}1/n^3\) to within \(\displaystyle 0.001\).

**Solution**

a. Using a calculating utility, we have

\[\displaystyle S_{10}=1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+⋯+\dfrac{1}{10^3}≈1.19753.\]

By the remainder estimate, we know

\[\displaystyle R_N<∫^∞_N\dfrac{1}{x^3}dx.\]

We have

\[\displaystyle ∫^∞_{10}\dfrac{1}{x^3}dx=\lim_{b→∞}∫^b_{10}\dfrac{1}{x^3}dx=\lim_{b→∞}[−\dfrac{1}{2x^2}]^b_N=\lim_{b→∞}[−\dfrac{1}{2b^2}+\dfrac{1}{2N^2}]=\dfrac{1}{2N^2}.\]

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

b. Find \(\displaystyle N\) such that \(\displaystyle R_N<0.001\). In part a. we showed that \(\displaystyle R_N<1/2N^2\). Therefore, the remainder \(\displaystyle R_N<0.001\) as long as \(\displaystyle 1/2N^2<0.001\). That is, we need \(\displaystyle 2N^2>1000\). Solving this inequality for \(\displaystyle N\), we see that we need \(\displaystyle N>22.36\). 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 \(\displaystyle N=23\).

Exercise \(\displaystyle \PageIndex{4}\)

For \(\displaystyle ^∞_{n=1}\dfrac{1}{n^4}\), calculate \(\displaystyle S_5\) and estimate the error \(\displaystyle R_5\).

**Hint**-
Use the remainder estimate \(\displaystyle R_N<∫^∞_N1/x^4dx.\)

**Answer**-
\(\displaystyle S_5≈1.09035, R_5<0.00267\)

## Key Concepts

- If \(\displaystyle \lim_{n→∞}a_n≠0,\) then the series \(\displaystyle \sum^∞_{n=1}a_n\) diverges.
- If \(\displaystyle \lim_{n→∞}a_n=0,\) the series \(\displaystyle \sum^∞_{n=1}a_n\) may converge or diverge.
- If \(\displaystyle \sum^∞_{n=1}a_n\) is a series with positive terms \(\displaystyle a_n\) and \(\displaystyle f\) is a continuous, decreasing function such that \(\displaystyle f(n)=a_n\) for all positive integers \(\displaystyle n\), then

\[\sum_{n=1}^∞a_n\] and \[∫^∞_1f(x)dx\]

either both converge or both diverge. Furthermore, if \(\displaystyle \sum^∞_{n=1}a_n\) converges, then the \(\displaystyle Nth\) partial sum approximation \(\displaystyle S_N\) is accurate up to an error \(\displaystyle R_N\) where \(\displaystyle ∫^∞_{N+1}f(x)dx<R_N<∫^∞_Nf(x)dx\).

- The
*p*-series \(\displaystyle \sum∞^n_{=1}1/n^p\) converges if \(\displaystyle p>1\) and diverges if \(\displaystyle p≤1.\)

## Key Equations

**Divergence test**

If \(\displaystyle a_n↛0\) as \(\displaystyle n→∞,\sum_{n=1}^∞a_n\) diverges.

**p-series**

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{n^p}\begin{cases}converges &ifp>1//diverges &if p≤1\end{cases}\)

**Remainder estimate from the integral test**

\(\displaystyle ∫^∞_{N+1}f(x)dx<R_N<∫^∞_Nf(x)dx\)

## Glossary

**divergence test**- if \(\displaystyle \lim_{n→∞}a_n≠0,\) then the series \(\displaystyle \sum^∞_{n=1}a_n\) diverges

**integral test**-
for a series \(\displaystyle \sum^∞_{n=1}a_n\) with positive terms \(\displaystyle a_n\), if there exists a continuous, decreasing function \(\displaystyle f\) such that \(\displaystyle f(n)=a_n\) for all positive integers \(\displaystyle n\), then

\[\sum_{n=1}^∞a_n\] and \[∫^∞_1f(x)dx\]

either both converge or both diverge

*p*-series- a series of the form \(\displaystyle \sum^∞_{n=1}1/n^p\)

**remainder estimate**-
for a series \(\displaystyle \sum^∞_{n=}1a_n\) with positive terms \(\displaystyle a_n\) and a continuous, decreasing function \(\displaystyle f\) such that \(\displaystyle f(n)=a_n\) for all positive integers \(\displaystyle n\), the remainder \(\displaystyle R_N=\sum^∞_{n=1}a_n−\sum^N_{n=1}a_n\) satisfies the following estimate:

\[∫^∞_{N+1}f(x)dx<R_N<∫^∞_Nf(x)dx\]

## Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.