9 Sequences and Series 9.2 Infinite Series 9.4 Comparison Tests

9.3 The Integral Test

Knowing whether or not a series converges is very important, especially when we discuss Power Series in Section 9.8. Theorem 9.2.1 gives criteria for when Geometric series converge and Theorem 9.2.4 gives a quick test to determine if a series diverges. There are many important series whose convergence cannot be determined by these theorems, though, so we introduce a set of tests that allow us to handle a broad range of series. We start with the Integral Test.

Integral Test

We stated in Section 9.1 that a sequence $\{a_{n}\}$ is a function $a(n)$ whose domain is $\mathbb{N}$ , the set of natural numbers. If we can extend $a(n)$ to have the domain of $\mathbb{R}$ , the real numbers, and it is both positive and decreasing on $[1,\infty)$ , then the convergence of $\displaystyle\sum_{n=1}^{\infty}a_{n}$ is the same as $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$ .

Theorem 9.3.1 Integral Test

Let a sequence $\{a_{n}\}$ be defined by $a_{n}=a(n)$ , where $a(n)$ is continuous, positive, and decreasing on $[1,\infty)$ . Then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges, if, and only if, $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$ converges. In other words:

(a)

If $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$ is convergent, then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ is convergent.
(b)

If $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$ is divergent, then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ is divergent.

^†^†margin: Note: Theorem 9.3.1 does not state that the integral and the summation have the same value. Λ

Note that it is not necessary to start the series or the integral at $n=1$ . We may use any interval $[n,\infty]$ on which $a(n)$ is continuous, positive and decreasing. Also the sequence $\{a_{n}\}$ does not have to be strictly decreasing. It must be ultimately decreasing which means it is decreasing for all $n$ larger than some number $N$ .

We can demonstrate the truth of the Integral Test with two simple graphs. In Figure 9.3.1(a), the height of each rectangle is $a(n)=a_{n}$ for $n=1,2,\dotsc$ , and clearly the rectangles enclose more area than the area under $y=a(x)$ . Therefore we can conclude that

\int_{1}^{\infty}a(x)\operatorname{d}\!x<\sum_{n=1}^{\infty}a_{n}.

(9.3.1)

^†^†margin: (a) (b) Figure 9.3.1: Illustrating the truth of the Integral Test. Λ

In Figure 9.3.1(b), we draw rectangles under $y=a(x)$ with the Right-Hand rule, starting with $n=2$ . This time, the area of the rectangles is less than the area under $y=a(x)$ , so $\displaystyle\sum_{n=2}^{\infty}a_{n}<\int_{1}^{\infty}a(x)\operatorname{d}\!x$ . Note how this summation starts with $n=2$ ; adding $a_{1}$ to both sides lets us rewrite the summation starting with $n=1$ :

\sum_{n=1}^{\infty}a_{n}<a_{1}+\int_{1}^{\infty}a(x)\operatorname{d}\!x.

(9.3.2)

Combining Equations (9.3.1) and (9.3.2), we have

\sum_{n=1}^{\infty}a_{n}<a_{1}+\int_{1}^{\infty}a(x)\operatorname{d}\!x<a_{1}+% \sum_{n=1}^{\infty}a_{n}.

(9.3.3)

From Equation (9.3.3) we can make the following two statements:

(a)

If $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges, so does $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$
(because $\displaystyle\sum_{n=1}^{\infty}a_{n}<a_{1}+\int_{1}^{\infty}a(x)\operatorname% {d}\!x$ )
(b)

If $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges, so does $\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x$
(because $\displaystyle\displaystyle\int_{1}^{\infty}a(x)\operatorname{d}\!x<\sum_{n=1}^% {\infty}a_{n}$ .)

Therefore the series and integral either both converge or both diverge. Theorem 9.2.5 allows us to extend this theorem to series where $a(n)$ is positive and decreasing on $[b,\infty)$ for some $b>1$ .

Watch the video:
Integral Test for Series: Why It Works from https://youtu.be/ObiRjUFHJHo

Example 9.3.1 Using the Integral Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^{2}}$ . (The terms of the sequence $\{a_{n}\}=\{(\ln n)/n^{2}\}$ and the n ${}^{\text{th}}$ partial sums are given in Figure 9.3.2.)

SolutionFigure 9.3.2 ^†^†margin: Figure 9.3.2: Plotting the sequence and series in Example 9.3.1. Λ implies that $a(n)=(\ln n)/n^{2}$ is positive and decreasing on $[2,\infty)$ . We can determine this analytically, too. We know $a(n)$ is positive as both $\ln n$ and $n^{2}$ are positive on $[2,\infty)$ . To determine that $a(n)$ is decreasing, consider $a^{\prime}(n)=(1-2\ln n)/n^{3}$ , which is negative for $n\geq 2$ . Since $a^{\prime}(n)$ is negative, $a(n)$ is decreasing.

Applying the Integral Test, we test the convergence of $\displaystyle\int_{1}^{\infty}\frac{\ln x}{x^{2}}\operatorname{d}\!x$ . Integrating this improper integral requires the use of Integration by Parts, with $u=\ln x$ and $dv=1/x^{2}\operatorname{d}\!x$ .

	$\displaystyle\int_{1}^{\infty}\frac{\ln x}{x^{2}}\operatorname{d}\!x$	$\displaystyle=\lim_{t\to\infty}\int_{1}^{t}\frac{\ln x}{x^{2}}\operatorname{d}\!x$
		$\displaystyle=\lim_{t\to\infty}\biggl{(}-\frac{1}{x}\ln x\Big{\|}_{1}^{t}+\int_% {1}^{t}\frac{1}{x^{2}}\operatorname{d}\!x\biggr{)}$
		$\displaystyle=\lim_{t\to\infty}\biggl{(}\Bigl{[}-\frac{1}{x}\ln x-\frac{1}{x}% \Bigr{]}_{1}^{t}\biggr{)}$
		$\displaystyle=\lim_{t\to\infty}\biggl{(}1-\frac{1}{t}-\frac{\ln t}{t}\biggr{)}% .\quad\text{Apply L'Hôpital's Rule:}$
		$\displaystyle=1-0-\lim_{t\to\infty}\frac{1}{t}$
		$\displaystyle=1$

Since $\displaystyle\int_{1}^{\infty}\frac{\ln x}{x^{2}}\operatorname{d}\!x$ converges, so does $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^{2}}$ .

$p$ -Series

Another important type of series is the p-series.

Definition 9.3.1 $p$ -Series, General $p$ -Series

(a) A $p$ -series is a series of the form $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ . (b) A general $p$ -series is a series of the form $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(an+b)^{p}}$ , where $a>0$ , $b$ is a real number, and $an+b\neq 0$ for all $n$ .

Like geometric series, one of the nice things about $p$ -series is that they have easy to determine convergence properties.

Theorem 9.3.2 Convergence of General $p$ -Series

Assume $a$ and $b$ are real numbers and $an+b\neq 0$ for all $n$ .
A general $p$ -series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(an+b)^{p}}$ will converge if, and only if, $p>1$ .

Proof

Consider the integral $\displaystyle\int_{1}^{\infty}\frac{1}{(ax+b)^{p}}\operatorname{d}\!x$ ; assuming $p\neq 1$ ,

$\displaystyle\int_{1}^{\infty}\frac{1}{(ax+b)^{p}}\operatorname{d}\!x$ $\displaystyle=\lim_{t\to\infty}\int_{1}^{t}\frac{1}{(ax+b)^{p}}\operatorname{d% }\!x$

$\displaystyle=\lim_{t\to\infty}\frac{1}{a(1-p)}(ax+b)^{1-p}\Big{|}_{1}^{t}$

$\displaystyle=\lim_{t\to\infty}\frac{1}{a(1-p)}\bigl{(}(at+b)^{1-p}-(a+b)^{1-p% }\bigr{)}.$

This limit converges if and only if, $p>1$ . It is easy to show that the integral also diverges in the case of $p=1$ . (This result is similar to the work preceding Key Idea 8.6.1.)

Therefore $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(an+b)^{p}}$ converges if, and only if, $p>1$ . ∎

Example 9.3.2 Determining convergence of series

Determine the convergence of the following series.

(a)

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$
(b)

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$
(c)

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$
(d)

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$
(e)

$\displaystyle\sum_{n=11}^{\infty}\frac{1}{(\frac{1}{2}n-5)^{3}}$
(f)

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}$

Solution

(a)

This is a $p$ -series with $p=1$ . By Theorem 9.3.2, this series diverges.

This series is a famous series, called the Harmonic Series, so named because of its relationship to harmonics in the study of music and sound.
(b)

This is a $p$ -series with $p=2$ . By Theorem 9.3.2, it converges. Note that the theorem does not give a formula by which we can determine what the series converges to; we just know it converges. A famous, unexpected result is that this series converges to $\displaystyle{\pi^{2}}/{6}$ .
(c)

This is a $p$ -series with $p=1/2$ ; the theorem states that it diverges.
(d)

This is not a $p$ -series; the definition does not allow for alternating signs. Therefore we cannot apply Theorem 9.3.2. We will consider this series again in Section 9.5. (Another famous result states that this series, the Alternating Harmonic Series, converges to $-\ln 2$ .)
(e)

This is a general $p$ -series with $p=3$ , therefore it converges.
(f)

This is not a $p$ -series, but a geometric series with $r=1/2$ . It converges.

In the next section we consider two more convergence tests, both comparison tests. That is, we determine the convergence of one series by comparing it to another series with known convergence.

Exercises 9.3

Terms and Concepts

1.

In order to apply the Integral Test to a sequence $\{a_{n}\}$ , the function $a(n)=a_{n}$ must be xxxxxx, xxxxxx and xxxxxx.
2.

T/F: The Integral Test can be used to determine the sum of a convergent series.

Problems

In Exercises 3–10., use the Integral Test to determine the convergence of the given series.

3.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}$
4.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{4}}$
5.

$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^{2}+1}$
6.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$
7.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}$
8.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^{2}}$
9.

$\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^{n}}$
10.

$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^{3}}$

In Exercises 11–14., find the value(s) of $p$ for which the series is convergent.

11.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^{p}}$
12.

$\displaystyle\sum_{n=1}^{\infty}n(1+n^{2})^{p}$
13.

$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}$
14.

$\displaystyle\sum_{n=3}^{\infty}\frac{1}{n\ln n[\ln(\ln n)]^{p}}$

15.
It can be shown that $\int_{0}^{1}x^{-x}\operatorname{d}\!x=\sum_{n=1}^{\infty}n^{-n}$ . Use the Integral Test to show that the series is convergent, and hence conclude that the integral is convergent. Hint: remember, you only need to show that the integral in the Integral Test is convergent–you do not need to be able to evaluate it.

	$\displaystyle\int_{1}^{\infty}\frac{1}{(ax+b)^{p}}\operatorname{d}\!x$	$\displaystyle=\lim_{t\to\infty}\int_{1}^{t}\frac{1}{(ax+b)^{p}}\operatorname{d% }\!x$
		$\displaystyle=\lim_{t\to\infty}\frac{1}{a(1-p)}(ax+b)^{1-p}\Big{\|}_{1}^{t}$
		$\displaystyle=\lim_{t\to\infty}\frac{1}{a(1-p)}\bigl{(}(at+b)^{1-p}-(a+b)^{1-p% }\bigr{)}.$