sum of_n=2^infinity (1)/(n(ln n)^2)

f(x) = (1)/(x(ln x)^2) is positive, continuous and decreasing for x ≥ 2; [4pt] integral of_2^infinity (1)/(x(ln x)^2) dx = (1)/(ln 2); The series converges

sum of_n=1^infinity 6ne^-n^2

f(x) = 6xe^-x^2 is positive, continuous and decreasing for x ≥ 1; [4pt] integral of_1^infinity 6xe^-x^2 dx = 3/e; The series converges

sum of_n=1^infinity (3)/((n + 5)^2)

f(x) = (3)/((x + 5)^2) is positive, continuous and decreasing for x ≥ 1; [4pt] integral of_1^infinity (3)/((x + 5)^2) dx = 1/2; The series converges

sum of_n=2^infinity (6)/(nln n)

f(x) = (6)/(xln x) is positive, continuous and decreasing for x ≥ 2; [4pt] integral of_2^infinity (6)/(xln x) dx = infinity; The series diverges

sum of_n=1^infinity (n)/(n^2 + 5)

f(x) = (x)/(x^2 + 5) is positive, continuous and decreasing for x ≥ 3; [4pt] (the terms before that do not affect convergence); [4pt] integral of_1^infinity (x)/(x^2 + 5) dx = infinity; The series diverges

sum of_n=1^infinity (ln n)/(n)

f(x) = (ln x)/(x) is positive, continuous and decreasing for x ≥ 3; [4pt] (the terms before that do not affect convergence); [4pt] integral of_1^infinity (ln x)/(x) dx = infinity; The series diverges

sum of_n=1^infinity (4n)/(n^2 + 9)

f(x) = (4x)/(x^2 + 9) is positive, continuous and decreasing for x ≥ 3; [4pt] (the terms before that do not affect convergence); [4pt] integral of_1^infinity (4x)/(x^2 + 9) dx = infinity; The series diverges

sum of_n=2^infinity (3)/(nln n)

f(x) = (3)/(xln x) is positive, continuous and decreasing for x ≥ 2; [4pt] integral of_2^infinity (3)/(xln x) dx = infinity; The series diverges

sum of_n=1^infinity 1/n^2

f(x) = 1/x^2 is positive, continuous and decreasing for x ≥ 1; [4pt] integral of_1^infinity 1/x^2 dx = 1; The series converges

sum of_n=2^infinity (2)/(n(ln n)^2)

f(x) = (2)/(x(ln x)^2) is positive, continuous and decreasing for x ≥ 2; [4pt] integral of_2^infinity (2)/(x(ln x)^2) dx = (2)/(ln 2); The series converges

sum of_n=1^infinity (4n)/(n^2 + 7)

integral of_1^infinity (4x)/(x^2 + 7) dx = infinity; The series diverges

Worksheets · Calculus

Integral test worksheet

When the terms come from a function that is positive, continuous and decreasing, the series and the integral of that function either both converge or both diverge. The same integral, started at N instead of 1, bounds the error left after N terms.

Select your difficulty

Pick more than one for a sheet that mixes them.

The preview can look cramped on a phone. The PDF and printed sheet come out normal.



Sigma Prep · sigmaprep.io/worksheets Name The Integral Test Date Period Use the integral test to decide whether each series converges, or bound the remainder as asked. $\sum_{n = 2}^{\infty} \frac{1}{n {(\ln n)}^{2}}$ Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} 6 e^{- n}$ $\sum_{n = 1}^{\infty} 6 n e^{- n^{2}}$ Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{3}}$ $\sum_{n = 1}^{\infty} \frac{3}{{(n + 5)}^{2}}$ Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} 5 e^{- 3 n}$ $\sum_{n = 2}^{\infty} \frac{6}{n \ln n}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{2}{n^{4}}$ $\sum_{n = 1}^{\infty} \frac{n}{n^{2} + 5}$ Bound the remainder $R_{5}$ for $\sum_{n = 1}^{\infty} e^{- n}$ $\sum_{n = 1}^{\infty} \frac{\ln n}{n}$ Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} \frac{4}{n^{3}}$ $\sum_{n = 1}^{\infty} \frac{4 n}{n^{2} + 9}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} e^{- 3 n}$ $\sum_{n = 2}^{\infty} \frac{3}{n \ln n}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} 3 e^{- 3 n}$ $\sum_{n = 2}^{\infty} \frac{2}{n {(\ln n)}^{2}}$ Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The Integral Test

Date Period

Use the integral test to decide whether each series converges, or bound the remainder as asked.

$\sum_{n = 2}^{\infty} \frac{1}{n {(\ln n)}^{2}}$
Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} 6 e^{- n}$
$\sum_{n = 1}^{\infty} 6 n e^{- n^{2}}$
Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{3}}$
$\sum_{n = 1}^{\infty} \frac{3}{{(n + 5)}^{2}}$
Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} 5 e^{- 3 n}$
$\sum_{n = 2}^{\infty} \frac{6}{n \ln n}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{2}{n^{4}}$
$\sum_{n = 1}^{\infty} \frac{n}{n^{2} + 5}$
Bound the remainder $R_{5}$ for $\sum_{n = 1}^{\infty} e^{- n}$
$\sum_{n = 1}^{\infty} \frac{\ln n}{n}$
Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} \frac{4}{n^{3}}$
$\sum_{n = 1}^{\infty} \frac{4 n}{n^{2} + 9}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} e^{- 3 n}$
$\sum_{n = 2}^{\infty} \frac{3}{n \ln n}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$
$\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$
Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} 3 e^{- 3 n}$
$\sum_{n = 2}^{\infty} \frac{2}{n {(\ln n)}^{2}}$
Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$



Sigma Prep · sigmaprep.io/worksheets Name The Integral TestAnswer keyVersion 1 Date Period Use the integral test to decide whether each series converges, or bound the remainder as asked. $\sum_{n = 2}^{\infty} \frac{1}{n {(\ln n)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{1}{x {(\ln x)}^{2}} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{1}{x {(\ln x)}^{2}} d x = \frac{1}{\ln 2} \\ The series converges \end{aligned} \end{array}$ Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} 6 e^{- n}$ $R_{20} \leq \frac{6}{e^{20}}$ $\sum_{n = 1}^{\infty} 6 n e^{- n^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = 6 x e^{- x^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} 6 x e^{- x^{2}} d x = \frac{3}{e} \\ The series converges \end{aligned} \end{array}$ Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{3}}$ $R_{6} \leq \frac{1}{72}$ $\sum_{n = 1}^{\infty} \frac{3}{{(n + 5)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{3}{{(x + 5)}^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} \frac{3}{{(x + 5)}^{2}} d x = \frac{1}{2} \\ The series converges \end{aligned} \end{array}$ Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} 5 e^{- 3 n}$ $R_{6} \leq \frac{5}{3 e^{18}}$ $\sum_{n = 2}^{\infty} \frac{6}{n \ln n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{6}{x \ln x} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{6}{x \ln x} d x = \infty \\ The series diverges \end{aligned} \end{array}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{2}{n^{4}}$ $R_{4} \leq \frac{1}{96}$ $\sum_{n = 1}^{\infty} \frac{n}{n^{2} + 5}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{x}{x^{2} + 5} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{x}{x^{2} + 5} d x = \infty \\ The series diverges \end{aligned} \end{array}$ Bound the remainder $R_{5}$ for $\sum_{n = 1}^{\infty} e^{- n}$ $R_{5} \leq \frac{1}{e^{5}}$ $\sum_{n = 1}^{\infty} \frac{\ln n}{n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{\ln x}{x} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{\ln x}{x} d x = \infty \\ The series diverges \end{aligned} \end{array}$ Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} \frac{4}{n^{3}}$ $R_{20} \leq \frac{1}{200}$ $\sum_{n = 1}^{\infty} \frac{4 n}{n^{2} + 9}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{4 x}{x^{2} + 9} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{4 x}{x^{2} + 9} d x = \infty \\ The series diverges \end{aligned} \end{array}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} e^{- 3 n}$ $R_{4} \leq \frac{1}{3 e^{12}}$ $\sum_{n = 2}^{\infty} \frac{3}{n \ln n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{3}{x \ln x} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{3}{x \ln x} d x = \infty \\ The series diverges \end{aligned} \end{array}$ Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$ $R_{4} \leq \frac{1}{192}$ $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{1}{x^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} \frac{1}{x^{2}} d x = 1 \\ The series converges \end{aligned} \end{array}$ Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} 3 e^{- 3 n}$ $R_{8} \leq \frac{1}{e^{24}}$ $\sum_{n = 2}^{\infty} \frac{2}{n {(\ln n)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{2}{x {(\ln x)}^{2}} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{2}{x {(\ln x)}^{2}} d x = \frac{2}{\ln 2} \\ The series converges \end{aligned} \end{array}$ Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ $R_{8} \leq \frac{1}{8}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The Integral TestAnswer keyVersion 1

Date Period

Use the integral test to decide whether each series converges, or bound the remainder as asked.

$\sum_{n = 2}^{\infty} \frac{1}{n {(\ln n)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{1}{x {(\ln x)}^{2}} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{1}{x {(\ln x)}^{2}} d x = \frac{1}{\ln 2} \\ The series converges \end{aligned} \end{array}$
Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} 6 e^{- n}$ $R_{20} \leq \frac{6}{e^{20}}$
$\sum_{n = 1}^{\infty} 6 n e^{- n^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = 6 x e^{- x^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} 6 x e^{- x^{2}} d x = \frac{3}{e} \\ The series converges \end{aligned} \end{array}$
Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{3}}$ $R_{6} \leq \frac{1}{72}$
$\sum_{n = 1}^{\infty} \frac{3}{{(n + 5)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{3}{{(x + 5)}^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} \frac{3}{{(x + 5)}^{2}} d x = \frac{1}{2} \\ The series converges \end{aligned} \end{array}$
Bound the remainder $R_{6}$ for $\sum_{n = 1}^{\infty} 5 e^{- 3 n}$ $R_{6} \leq \frac{5}{3 e^{18}}$
$\sum_{n = 2}^{\infty} \frac{6}{n \ln n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{6}{x \ln x} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{6}{x \ln x} d x = \infty \\ The series diverges \end{aligned} \end{array}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{2}{n^{4}}$ $R_{4} \leq \frac{1}{96}$
$\sum_{n = 1}^{\infty} \frac{n}{n^{2} + 5}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{x}{x^{2} + 5} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{x}{x^{2} + 5} d x = \infty \\ The series diverges \end{aligned} \end{array}$
Bound the remainder $R_{5}$ for $\sum_{n = 1}^{\infty} e^{- n}$ $R_{5} \leq \frac{1}{e^{5}}$
$\sum_{n = 1}^{\infty} \frac{\ln n}{n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{\ln x}{x} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{\ln x}{x} d x = \infty \\ The series diverges \end{aligned} \end{array}$
Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} \frac{4}{n^{3}}$ $R_{20} \leq \frac{1}{200}$
$\sum_{n = 1}^{\infty} \frac{4 n}{n^{2} + 9}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{4 x}{x^{2} + 9} is positive, continuous and decreasing for x \geq 3 \\ (the terms before that do not affect convergence) \\ \int_{1}^{\infty} \frac{4 x}{x^{2} + 9} d x = \infty \\ The series diverges \end{aligned} \end{array}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} e^{- 3 n}$ $R_{4} \leq \frac{1}{3 e^{12}}$
$\sum_{n = 2}^{\infty} \frac{3}{n \ln n}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{3}{x \ln x} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{3}{x \ln x} d x = \infty \\ The series diverges \end{aligned} \end{array}$
Bound the remainder $R_{4}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$ $R_{4} \leq \frac{1}{192}$
$\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{1}{x^{2}} is positive, continuous and decreasing for x \geq 1 \\ \int_{1}^{\infty} \frac{1}{x^{2}} d x = 1 \\ The series converges \end{aligned} \end{array}$
Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} 3 e^{- 3 n}$ $R_{8} \leq \frac{1}{e^{24}}$
$\sum_{n = 2}^{\infty} \frac{2}{n {(\ln n)}^{2}}$ $\begin{array}{l} \begin{aligned} f (x) = \frac{2}{x {(\ln x)}^{2}} is positive, continuous and decreasing for x \geq 2 \\ \int_{2}^{\infty} \frac{2}{x {(\ln x)}^{2}} d x = \frac{2}{\ln 2} \\ The series converges \end{aligned} \end{array}$
Bound the remainder $R_{8}$ for $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ $R_{8} \leq \frac{1}{8}$

How to do these

$\sum_{n = 1}^{\infty} n e^{- n^{2}}$

f(x) = xe^(-x^2) is positive and continuous, and decreasing for x ≥ 1.
Substitute u = x^2: the integral from 1 to infinity is (1/2)e^(-1).
The integral is finite, so the series converges.

The answer is $\int_{1}^{\infty} x e^{- x^{2}} d x = \frac{1}{2 e}$ ; converges.

Where students go wrong

Giving the value of the integral as the sum of the series. The test only says whether the series converges; the integral and the sum are different numbers.

Also called integral test for convergence, Cauchy integral test or integral test remainder estimate.

What you can put on this worksheet

Evaluate the integral and decide: $\sum_{n = 1}^{\infty} \frac{4 n}{n^{2} + 7}$ → $\begin{array}{l} \int_{1}^{\infty} \frac{4 x}{x^{2} + 7} d x = \infty \\ The series diverges \end{array}$
Bound the remainder with an integral: Bound the remainder $R_{20}$ for $\sum_{n = 1}^{\infty} \frac{5}{n^{3}}$ → $R_{20} \leq \frac{1}{160}$

Questions about these worksheets

Yes. Every integral test sheet is free to print and free to download as a PDF. There is no account to make, no email to hand over and no limit on how many you take.

Yes. The answer key prints on a second page, and you can choose whether it shows each question beside its answer or just the answers in a list.

Yes. Print straight from the page, or download the PDF and print that. The sheet is laid out for paper rather than squeezed off a screen, so there is room to work under each question, and the answer key comes out on its own page.

Calculus is usually taken around 12th grade or a first college course, though schools vary and plenty of students meet it earlier or later. Pick the difficulty rather than the grade: the easy level suits a first lesson on it, the hard level suits review before a test.

Yes, as many as you like. The questions are built fresh each time rather than picked from a fixed set of files, so pressing Generate gives a new sheet. That is what you want for a second class, a retake, or two students sitting next to each other.

That is the closest comparison, and Kuta Software is good software. It is also paid software you install on a computer. These worksheets run in a browser tab for free, with no account and nothing to install. Like Kuta, the integral test questions are generated when you ask for them rather than pulled from a fixed set of files, so two classes never get the same sheet, and the answer key comes with it.

Yes. There are three levels, and you can tick more than one for a sheet that mixes them, in which case the questions come out easiest first. The harder levels are not just larger numbers: they ask for something the easy ones do not.

You pick from 2: evaluate the integral and decide and bound the remainder with an integral. Tick as many as you want and set how many of each, or let it spread them evenly.

Giving the value of the integral as the sum of the series. The test only says whether the series converges; the integral and the sum are different numbers.

Integral test worksheet

Select your difficulty

The Integral Test

The Integral TestAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

Are these integral test worksheets free?+

Do the integral test worksheets come with an answer key?+

Are these integral test worksheets printable?+

What grade are these integral test worksheets for?+

Can I get a different version of the same integral test worksheet?+

Is this a free alternative to Kuta Software for integral test?+

Can I choose how hard the integral test questions are?+

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What do students get wrong with integral test?+