Introduction

Welcome to this guide on the Integral Test for Convergence, this article will guide you through what is the Integral Test for Convergence and how to apply it!

The Integral Test

If $f(x)$ is a positive, continuous, and decreasing function for $x\ge m$, where $m$ is a positive integer and the $n$th term expression can be expressed as $a_n=f(n)$, then:

$\sum _{n=m}^{\infty }{a}_n\text{ and }{\int }_m^{\infty }f(x)dx$

both converge or both diverge. Note that in many cases, $m=1$.

It is extremely important to note that the Integral Test does not provide the actual sum, but simply confirms the series' status of convergence or divergence.

Applying the Integral Test

Example 1:

$\sum _{n=1}^{\infty }\frac{n}{n^2+1}$

First, check for the three criteria: positive, continuous, decreasing

Now evaluate the improper integral:

${\int }_1^{\infty }\frac{x}{x^2+1}dx$

$=\underset{b\to \infty }{\lim }{\int }_1^b\frac{x}{x^2+1}dx$

U-substitution: $u=x^2+1⟹du=2xdx$

$=\frac{1}{2}\underset{b\to \infty }{\lim }{\int }_2^{b^2+1}\frac{1}{u}du$

$=\frac{1}{2}\underset{b\to \infty }{\lim }(\ln |b^2+1|-\ln |1^2+1|)$

$=\infty$

The integral diverges, meaning that the series also diverges.

Example 2:

$\sum _{n=1}^{\infty }\frac{1}{n^2+1}$

First, check for the three criteria: positive, continuous, decreasing

Let $a_n=f(n)=\frac{1}{x^2+1}$

Then we can construct and solve this improper integral:

${\int }_1^{\infty }\frac{1}{x^2+1}dx$

$=\underset{b\to \infty }{\lim }{\int }_1^b\frac{1}{x^2+1}dx$

$=\underset{b\to \infty }{\lim }\arctan b-\arctan 1$

$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

Meaning the series converges, but the series doesn't necessarily converge to $\frac{\pi }{4}$

From the examples shown, you should notice that it is extremely important to check if the series meet the 3 requirements of integral test.

Practice

Determine whether the following series converge or diverge using the integral test.