9.3: The Divergence and Integral Tests (2024)

Divergence Test

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

\[\begin{align*} \lim_{k→∞}a_k = \lim_{k→∞}(S_k−S_{k−1}) \\[4pt] =\lim_{k→∞}S_k−\lim_{k→∞}S_{k−1} \\[4pt] =S−S=0. \end{align*}\]

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

If \( a_n↛0\) 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.

It is important to note that the inverse statementof 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, \(\displaystyle \lim_{n→0}\tfrac{1}{n}=0\), but the harmonic series \(\displaystyle \sum^∞_{n=1}\frac{1}{n}\) 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 \( a_n→0\), the divergence test is inconclusive.

Integral Test

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

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

Therefore, for each \( k\), the \( k^{\text{th}}\) partial sum \( S_k\) satisfies

\[\begin{align*} S_k =\sum_{n=1}^k\dfrac{1}{n} >∫^{k+1}_1\dfrac{1}{x}\,dx = \ln x \big| ^{k+1}_1 \\[4pt] = \ln (k+1)−\ln (1) \\[4pt] =\ln (k+1).\end{align*}\]

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

Now consider the series \(\displaystyle \sum_{n=1}^∞\frac{1}{n^2}\). 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 \( 1,1/2^2,1/3^2,…\) along with the function \( f(x)=\frac{1}{x^2}\). From the graph we see that

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

Therefore, for each \( k\), the \( k^{\text{th}}\) partial sum \( S_k\) satisfies

\[\begin{align*} S_k=\sum_{n=1}^k\dfrac{1}{n^2}<1+∫^k_1\dfrac{1}{x^2}\,dx =1−\left. \dfrac{1}{x} \right|^k_1 \\[4pt] =1−\dfrac{1}{k}+1 \\[4pt] =2−\dfrac{1}{k}<2. \end{align*}\]

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

\[S_k=S_{k−1}+\dfrac{1}{k^2} \nonumber \]

for \( k≥2\).

Since \( {S_k}\) is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series \(\displaystyle \sum_{n=1}^∞\frac{1}{n^2}\) 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 \( a_n\) such that there exists a continuous, positive, decreasing function \( f\) where \( f(n)=a_n\) for all positive integers. Then, as in Figure \(\PageIndex{3a}\), for any integer \( k\), the \( k^{\text{th}}\) partial sum \( S_k\) satisfies

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

Therefore, if \(\displaystyle ∫^∞_1f(x)\,dx\) converges, then the sequence of partial sums \( {S_k}\) is bounded. Since \( {S_k}\) 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 \( k\), the \( k^{\text{th}}\) partial sum \( S_k\) satisfies

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

If

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

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

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

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

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,

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

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

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

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

\[∫^∞_1\left(\frac{1}{e}\right)^x\,dx=∫^∞_1e^{−x}\,dx=\lim_{b→∞}∫^b_1e^{−x}\,dx=\lim_{b→∞}−e^{−x}\big|^b_1=\lim_{b→∞}[−e^{−b}+e^{−1}]=\dfrac{1}{e}. \nonumber \]

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.

We know the \(p\)-series converges if \( p=2\) and diverges if \( p=1\). What about other values of \( p\)? 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 \( 1/n^p→∞,\) and if \( p=0\), then \( 1/n^p→1.\) Therefore, by the divergence test,

\[\sum_{n=1}^∞\dfrac{1}{n^p} \nonumber \]

diverges if \(p≤0\).

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

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

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

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

Because

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

we conclude that

\[∫^∞_1\dfrac{1}{x^p}\,dx=\begin{cases}\dfrac{1}{p−1}, \text{if}\;p>1\\ ∞, \text{if}\;p<1.\end{cases} \nonumber \]

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

In summary,

\[\sum_{n=1}^∞\dfrac{1}{n^p}\begin{cases}\text{converges if}\; p>1\\ \text{diverges if}\;p≤1\end{cases} \nonumber \].

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 \( N\) 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

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

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

We illustrate Note \(\PageIndex{1}\) in Figure \(\PageIndex{4}\). In particular, by representing the remainder \( R_N=a_{N+1}+a_{N+2}+a_{N+3}+⋯\) 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,

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

and

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

We conclude that

\[∫^∞_{N+1}f(x)\,dx<R_N<∫^∞_Nf(x)\,dx. \nonumber \]

Since

\[\sum_{n=1}^∞a_n=S_N+R_N, \nonumber \]

where \( S_N\) is the \(N^{\text{th}}\) partial sum, we conclude that

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

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 \( a_n\) and \( f\) is a continuous, decreasing function such that \( f(n)=a_n\) for all positive integers \( n\), then

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

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

• The \(p\)-series \(\displaystyle \sum_{n=1}^∞\frac{1}{n^p}\) converges if \( p>1\) and diverges if \( p≤1.\)

Key Equations

• Divergence test

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

• p-series

\(\displaystyle \sum_{n=1}^∞\dfrac{1}{n^p}\quad \begin{cases}\text{converges}, \text{if}\;p>1\\\text{diverges}, \text{if}\; p≤1\end{cases}\)

• 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 \( a_n\), if there exists a continuous, decreasing function \( f\) such that \( f(n)=a_n\) for all positive integers \( n\), then

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

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

\[∫^∞_{N+1}f(x)\,dx<R_N<∫^∞_Nf(x)\,dx \nonumber \]