AP®MathsCollege BoardCalculus BCStudy GuidesUnit 10: Infinite Sequences and SeriesTests for Divergence & ConvergenceIntegral Test for Convergence

Integral Test for Convergence (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 19 May 2026

Integral test

What is the integral test?

The integral test is a method of determining whether an infinite series converges or diverges
The integral test states that:
- Given that $a_{n} = f (n)$ , where $f$ is a continuous, positive, decreasing function on $[c, \infty)$
- If the improper integral $\int_{c}^{\infty} f (x) d x$ exists
- Then the series $\sum_{n = c}^{\infty} a_{n}$ converges
  - If the improper integral doesn't exist then the series diverges
For example, you can use the function $f (x) = \frac{1}{x^{3}}$ to test whether $\sum_{n = 1}^{\infty} \frac{1}{n^{3}}$ converges
- $f$ is continuous, positive, and decreasing for $x \geq 1$

Examiner Tips and Tricks

For an FRQ, be sure to state that your function is continuous, positive, and decreasing.

How does the integral test work?

Each term in the infinite series $a_{1} + a_{2} + a_{3} + . . .$ can be represented by the area of a rectangle of width 1 and height $a_{n}$
The integral $\int_{1}^{\infty} f (x) d x = A$ represents the area $A$ under the curve $f (x)$ from $x = 1$ to $\infty$
In the first image, $A$ is an underestimate of the rectangles $a_{1} + a_{2} + a_{3} + . . .$
- so $A < \sum_{n = 1}^{\infty} a_{n}$

Graph showing a curve y=f(x) with marked points at intervals. Grey rectangles overestimate the area under the graph and the areas a1 to a5 represent the infinite series.

In the next image, $A$ is an overestimate of the rectangles $a_{2} + a_{3} + . . .$
- Adding $a_{1}$ to both sides gives $a_{1} + A > \sum_{n = 1}^{\infty} a_{n}$

Graph showing a curve y=f(x) with decreasing step-like shaded rectangular areas under it representing the infinite series. Points (1,a1) to (5,a5) are marked. Axes labelled x and y.

So $A < \sum_{n = 1}^{\infty} a_{n} < A + a_{1}$ which means
- if $A$ is finite, then $\sum_{n = 1}^{\infty} a_{n}$ ⁣must have a finite value (the series converges)
- if $A$ is infinite, then $\sum_{n = 1}^{\infty} a_{n}$ must also be infinite (the series diverges)

Worked Example

Use the integral test to determine whether each of the following series converges or diverges.

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

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

Answer:

(a)

Note that this series is the harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . .$

$f (x) = \frac{1}{x}$ is continuous, positive and decreasing for $x \geq 1$

Evaluate the improper integral $\int_{1}^{\infty} \frac{1}{x} d x$

$\begin{array}{rcl} \int_{1}^{\infty} \frac{1}{x} d x & = & \lim_{p \to \infty} \int_{1}^{p} \frac{1}{x} d x \\ = & \lim_{p \to \infty} {[\ln |x|]}_{1}^{p} \\ = & \lim_{p \to \infty} (\ln p - \ln 1) \\ = & \lim_{p \to \infty} \ln p \\ = & \infty \end{array}$

The improper integral diverges to infinity (as logarithmic growth tends to infinity, $\ln x \to \infty$ as $x \to \infty$ ), so the series diverges

The integral $\int_{1}^{\infty} \frac{1}{x} d x$ diverges to infinity, so by the integral test the series is divergent

(b)

Note that this is a p-series with $p = 2$

$f (x) = \frac{1}{x^{2}}$ is continuous, positive and decreasing for $x \geq 1$

Evaluate the improper integral $\int_{1}^{\infty} \frac{1}{x^{2}} d x$

$\begin{array}{rcl} \int_{1}^{\infty} \frac{1}{x^{2}} d x & = & \lim_{p \to \infty} \int_{1}^{p} x^{- 2} d x \\ = & \lim_{p \to \infty} {[- x^{- 1}]}_{1}^{p} \\ = & \lim_{p \to \infty} {[- \frac{1}{x}]}_{1}^{p} \\ = & \lim_{p \to \infty} (- \frac{1}{p} - (- \frac{1}{1})) \\ = & \lim_{p \to \infty} (1 - \frac{1}{p}) \\ = & 1 - 0 \\ = & 1 \end{array}$

The improper integral converges to a finite value, so the series converges

The integral $\int_{1}^{\infty} \frac{1}{x^{2}} d x$ exists with a finite value, so by the integral test the series is convergent

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