3.5 Integral Test

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We determine the convergence or divergence of an infinite series using a related improper integral.

1 Integral Test

Integral Test Suppose that the function $f(x)$ is positive, continuous, and decreasing on the interval $(1, \infty )$ , and suppose that the terms of an infinite series are given by the corresponding function values, i.e., $a_n = f(n)$ .
Then the improper integral and the associated infinite series $\int _1^\infty f(x) \, dx\;\; \mbox {and} \;\;\sum _{n=1}^\infty a_n$ both converge or both diverge. In other words, the behavior of the improper integral determines the behavior of the associated series.

It is sufficient for the function to be decreasing eventually, i.e., on an interval of the form $(a, \infty )$ for some number $a$ .

The rationale for the integral test is that both the improper integral and the the sum of an infinite series can be interpreted as the area under a graph. Moreover, these areas will be comparable. Hence, we can use the logic of the comparison theorem of improper integrals: “less than convergent is convergent and greater than divergent is divergent”. The two figures below show that if the function $f(x)$ satisfies the hypotheses of the integral test and if $a_n = f(n)$ , then we can use the behavior of the integral to make conclusions about the behavior of the series.

example 1 Determine whether the series $\sum _{n=0}^\infty \frac {1}{1+n^2}$ converges or diverges.
Consider the function, $f(x) = \frac {1}{1+x^2}$ Note that $a_n = f(n)$ . This function $f(x)$ is positive and continuous on the interval $(0, \infty )$ . To check that $f(x)$ is decreasing, we need to verify that $f'(x) < 0$ on the interval $(0, \infty )$ . Recall that if the derivative is negative then the function is decreasing. By the quotient rule, $f'(x) = -\frac {2x}{(1+x^2)^2}.$ which is negative on the interval $(0, \infty )$ . Thus $f(x)$ is decreasing on this interval. Having verified that the conditions of the Integral Test are met, we compute the improper integral: \begin{align*} \int _0^\infty \frac{1}{1+x^2} \; dx &= \lim _{b \to \infty } \int _0^b \frac{1}{1+x^2} \; dx\\ &= \lim _{b \to \infty } \tan ^{-1}(x) \bigg |_0^b \\ &= \lim _{b \to \infty } \left [\tan ^{-1}(b) - \tan ^{-1}(0)\right ]\\ &= \lim _{b \to \infty } \left [\tan ^{-1}(b)\right ]\\ &= \frac{\pi }{2}. \end{align*}

The improper integral converges, so by the Integral Test, the associated infinite series $\sum _{n=0}^\infty \frac {1}{1+n^2}$ also converges.

(problem 1a) Use the integral test to determine whether the infinite series $\sum _{n=1}^\infty \frac {1}{n^2}$ converges or diverges.

Let $f(x) = \answer { 1/x^2 }$

Is $f(x)$ positive and continuous on the interval $(1, \infty )$ ? Yes No

The derivative, $f'(x) = \answer {-2/x^3}$

Is $f(x)$ decreasing on the interval $(1, \infty )$ ? Yes No

Evaluate the improper integral: $\int _1^\infty \frac {1}{x^2} \, dx = \answer {1}$

Does the infinite series $\sum _{n=1}^\infty \frac {1}{n^2}$ converge or diverge? Converge Diverge

(problem 1b) Use the integral test to determine whether the infinite series $\sum _{n=0}^\infty \frac {1}{\sqrt {n+1}}$ converges or diverges.

Let $f(x) = \answer { 1/\sqrt {x+1} }$

Is $f(x)$ positive and continuous on the interval $(0, \infty )$ ?Yes No

The derivative, $f'(x) = \answer {-\frac 12 (x+1)^{-3/2}}$

Is $f(x)$ decreasing on the interval $(0, \infty )$ ? Yes No

Evaluate the improper integral: $\int _0^\infty \frac {1}{\sqrt {x+1}} \, dx = \answer {\infty }$

Does the infinite series $\sum _{n=0}^\infty \frac {1}{\sqrt {n+1}}$ converge or diverge?Converge Diverge

example 2 Determine whether the series $\sum _{n=2}^\infty \frac {\ln (n)}{n}$ converges or diverges.
Consider the function $f(x) = \frac {\ln (x)}{x},$ so that $a_n = f(n)$ . This function is positive and continuous on the interval $(2, \infty )$ . To see that the function is decreasing, compute the derivative: $f'(x) = \frac {x\cdot \frac {1}{x}- \ln (x)}{x^2} = \frac {1-\ln (x)}{x^2},$ which is negative if $x > e$ . Hence $f(x)$ is eventually decreasing and we can use the integral test:

\begin{align*} \int _2^\infty \frac{\ln (x)}{x} \; dx &= \lim _{t \to \infty } \int _2^t \frac{\ln (x)}{x} \; dx\\ &= \lim _{t \to \infty } \int _{\ln (2)}^{\ln (t)} u \, du \;\; \left (\text{u-sub with $u = \ln (x), du = \frac{1}{x} dx$}\right )\\ &= \lim _{t \to \infty } \frac{u^2}{2} \bigg |_{\ln (2)}^{\ln (t)}\\ &= \frac 12\lim _{t \to \infty } \left (\ln ^2(t) - \ln ^2(2) \right )\\ &= \infty . \end{align*}

The improper integral diverges, so by the Integral Test, the associated infinite series $\sum _{n=2}^\infty \frac {\ln (n)}{n}$ also diverges.

(problem 2a) Use the integral test to determine whether the infinite series $\sum _{n=2}^\infty \frac {1}{n\ln (n)}$ converges or diverges.

Let $f(x) = \answer {1/(x\ln (x))}$

Is $f(x)$ positive and continuous on the interval $(2, \infty )$ ?Yes No

The derivative, $f'(x) = \answer {-\frac {1+\ln (x)}{x^2 \ln ^2(x)}}$

Is $f(x)$ decreasing on the interval $(2, \infty )$ ? Yes No

Evaluate the improper integral: $\int _2^\infty \frac {1}{x\ln (x)} = \answer {\infty }$

Does the infinite series $\sum _{n=2}^\infty \frac {1}{n\ln (n)}$ converge or diverge? Converge Diverge

(problem 2b) Use the integral test to determine whether the infinite series $\sum _{n=1}^\infty \frac {n}{e^n}$ converges or diverges.

Let $f(x) = \answer {xe^{-x}}$

Is $f(x)$ positive and continuous on the interval $(1, \infty )$ ?Yes No

The derivative, $f'(x) = \answer {e^{-x}(1-x)}$

Is $f(x)$ decreasing on the interval $(1, \infty )$ ? Yes No

Evaluate the improper integral (hint, use integration by parts): $\int _1^\infty \frac {x}{e^x} = \answer {2/e}$

Does the infinite series $\sum _{n=1}^\infty \frac {n}{e^n}$ converge or diverge? Converge Diverge

p-series Let $p > 0$ . The infinite series $\sum _{n=1}^\infty \frac {1}{n^p}$ is called a $p$ -series. In the special case $p = 1$ , the series is known as the harmonic series.

We have seen in the section on improper integrals that the $p$ -integral

$\int _1^\infty \frac {1}{x^p} \; dx.$ converges if $p > 1$ and diverges if $p \leq 1$ . Since the function $f(x) = 1/x^p \; (p > 0)$ is continuous, positive, and decreasing on the interval $(1, \infty )$ , the integral test can be applied to conclude that the associated $p$ -series has the same behavior. This is stated in the following theorem.

$p$ -series The $p$ -series $\sum _{n=1}^\infty \frac {1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$ .

If the series starts at some number $n >1$ , then we shall still refer to the series as a $p$ -series, and the behavior is the same as if it started at $n=1$ .

A non-zero multiple of a $p$ -series has the same behavior as the $p$ -series.

2 Video Lesson

Here is a detailed, lecture style video on the Integral Test:

2024-09-27 13:58:57