lim_n to infinity a_n = π/2; sum of a_n diverges

sum of_n=1^infinity ((n + 2)!)/(n!(6n^2 + 3))

Diverges: the terms approach 1/6

lim_n to infinity a_n = 18; sum of a_n diverges

lim_n to infinity a_n = e^1/3; sum of a_n diverges

sum of_n=1^infinity (4n^2 + 2n + 9)/(3n^2 + 9)

Diverges: the terms approach 4/3

a_n = sqrt(25n^2 + 9)6n + 1

lim_n to infinity a_n = 5/6; sum of a_n diverges

sum of_n=1^infinity sqrt(n^2 + 9)2n + 7

Diverges: the terms approach 1/2

a_n = (4^n + 9)/(6× 4^n + 2)

lim_n to infinity a_n = 1/6; sum of a_n diverges

sum of_n=1^infinity (-8n^2 - 5n - 2)/(3n^2 + 6)

Diverges: the terms approach -8/3

a_n = ((n)/(n + 5))^n

lim_n to infinity a_n = 1/e^5; sum of a_n diverges

sum of_n=1^infinity 3tan ^-1 n

Diverges: the terms approach 3π/2

a_n = ((n + 2)!)/(n!(3n^2 + 7))

lim_n to infinity a_n = 1/3; sum of a_n diverges

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

Diverges: the terms approach 36

lim_n to infinity a_n = e^6; sum of a_n diverges

sum of_n=1^infinity (-5n^2 + 2n - 5)/(7n^2 + 8)

Diverges: the terms approach -5/7

a_n = (6× 3^n + 2)/(7× 3^n + 8)

lim_n to infinity a_n = 6/7; sum of a_n diverges

sum of_n=1^infinity sqrt(16n^2 + 8)6n + 3

Diverges: the terms approach 2/3

a_n = ((n + 2)!)/(n!(4n^2 + 2))

lim_n to infinity a_n = 1/4; sum of a_n diverges

a_n = ((n + 4)/(n))^n

lim_n to infinity a_n = e^4; sum of a_n diverges

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

Diverges: the terms approach e^4

Worksheets · Calculus

nth-term test for divergence worksheet

If the terms of a series do not shrink to zero, adding infinitely many of them cannot settle on a number. That is the whole test, and it only ever proves divergence: terms that do go to zero leave the question open.

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 nth-Term Test Date Period Find the limit of the terms, and say what the nth-term test tells you about the series. $a_{n} = \tan^{- 1} n$ $\sum_{n = 1}^{\infty} \frac{(n + 2)!}{n! (6 n^{2} + 3)}$ $a_{n} = 2 n \sin ⁣ (\frac{9}{n})$ $\sum_{n = 1}^{\infty} \frac{2 \cdot 3^{n} + 5}{2 \cdot 3^{n} + 1}$ $a_{n} = {(1 + \frac{1}{3 n})}^{n}$ $\sum_{n = 1}^{\infty} \frac{4 n^{2} + 2 n + 9}{3 n^{2} + 9}$ $a_{n} = \frac{\sqrt{25 n^{2} + 9}}{6 n + 1}$ $\sum_{n = 1}^{\infty} \frac{\sqrt{n^{2} + 9}}{2 n + 7}$ $a_{n} = \frac{4^{n} + 9}{6 \cdot 4^{n} + 2}$ $\sum_{n = 1}^{\infty} \frac{- 8 n^{2} - 5 n - 2}{3 n^{2} + 6}$ $a_{n} = {(\frac{n}{n + 5})}^{n}$ $\sum_{n = 1}^{\infty} 3 \tan^{- 1} n$ $a_{n} = \frac{(n + 2)!}{n! (3 n^{2} + 7)}$ $\sum_{n = 1}^{\infty} 4 n \tan ⁣ (\frac{9}{n})$ $a_{n} = {(1 + \frac{6}{n})}^{n}$ $\sum_{n = 1}^{\infty} \frac{- 5 n^{2} + 2 n - 5}{7 n^{2} + 8}$ $a_{n} = \frac{6 \cdot 3^{n} + 2}{7 \cdot 3^{n} + 8}$ $\sum_{n = 1}^{\infty} \frac{\sqrt{16 n^{2} + 8}}{6 n + 3}$ $a_{n} = \frac{(n + 2)!}{n! (4 n^{2} + 2)}$ $\sum_{n = 1}^{\infty} {(\frac{n + 1}{n})}^{n}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The nth-Term Test

Date Period

Find the limit of the terms, and say what the nth-term test tells you about the series.

$a_{n} = \tan^{- 1} n$
$\sum_{n = 1}^{\infty} \frac{(n + 2)!}{n! (6 n^{2} + 3)}$
$a_{n} = 2 n \sin ⁣ (\frac{9}{n})$
$\sum_{n = 1}^{\infty} \frac{2 \cdot 3^{n} + 5}{2 \cdot 3^{n} + 1}$
$a_{n} = {(1 + \frac{1}{3 n})}^{n}$
$\sum_{n = 1}^{\infty} \frac{4 n^{2} + 2 n + 9}{3 n^{2} + 9}$
$a_{n} = \frac{\sqrt{25 n^{2} + 9}}{6 n + 1}$
$\sum_{n = 1}^{\infty} \frac{\sqrt{n^{2} + 9}}{2 n + 7}$
$a_{n} = \frac{4^{n} + 9}{6 \cdot 4^{n} + 2}$
$\sum_{n = 1}^{\infty} \frac{- 8 n^{2} - 5 n - 2}{3 n^{2} + 6}$
$a_{n} = {(\frac{n}{n + 5})}^{n}$
$\sum_{n = 1}^{\infty} 3 \tan^{- 1} n$
$a_{n} = \frac{(n + 2)!}{n! (3 n^{2} + 7)}$
$\sum_{n = 1}^{\infty} 4 n \tan ⁣ (\frac{9}{n})$
$a_{n} = {(1 + \frac{6}{n})}^{n}$
$\sum_{n = 1}^{\infty} \frac{- 5 n^{2} + 2 n - 5}{7 n^{2} + 8}$
$a_{n} = \frac{6 \cdot 3^{n} + 2}{7 \cdot 3^{n} + 8}$
$\sum_{n = 1}^{\infty} \frac{\sqrt{16 n^{2} + 8}}{6 n + 3}$
$a_{n} = \frac{(n + 2)!}{n! (4 n^{2} + 2)}$
$\sum_{n = 1}^{\infty} {(\frac{n + 1}{n})}^{n}$



Sigma Prep · sigmaprep.io/worksheets Name The nth-Term TestAnswer keyVersion 1 Date Period Find the limit of the terms, and say what the nth-term test tells you about the series. $a_{n} = \tan^{- 1} n$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{π}{2} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{(n + 2)!}{n! (6 n^{2} + 3)}$ Diverges: the terms approach $\frac{1}{6}$ $a_{n} = 2 n \sin ⁣ (\frac{9}{n})$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = 18 \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{2 \cdot 3^{n} + 5}{2 \cdot 3^{n} + 1}$ Diverges: the terms approach $1$ $a_{n} = {(1 + \frac{1}{3 n})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = e^{1 / 3} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{4 n^{2} + 2 n + 9}{3 n^{2} + 9}$ Diverges: the terms approach $\frac{4}{3}$ $a_{n} = \frac{\sqrt{25 n^{2} + 9}}{6 n + 1}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{5}{6} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{\sqrt{n^{2} + 9}}{2 n + 7}$ Diverges: the terms approach $\frac{1}{2}$ $a_{n} = \frac{4^{n} + 9}{6 \cdot 4^{n} + 2}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{6} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{- 8 n^{2} - 5 n - 2}{3 n^{2} + 6}$ Diverges: the terms approach $- \frac{8}{3}$ $a_{n} = {(\frac{n}{n + 5})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{e^{5}} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} 3 \tan^{- 1} n$ Diverges: the terms approach $\frac{3 π}{2}$ $a_{n} = \frac{(n + 2)!}{n! (3 n^{2} + 7)}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{3} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} 4 n \tan ⁣ (\frac{9}{n})$ Diverges: the terms approach $36$ $a_{n} = {(1 + \frac{6}{n})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = e^{6} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{- 5 n^{2} + 2 n - 5}{7 n^{2} + 8}$ Diverges: the terms approach $- \frac{5}{7}$ $a_{n} = \frac{6 \cdot 3^{n} + 2}{7 \cdot 3^{n} + 8}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{6}{7} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} \frac{\sqrt{16 n^{2} + 8}}{6 n + 3}$ Diverges: the terms approach $\frac{2}{3}$ $a_{n} = \frac{(n + 2)!}{n! (4 n^{2} + 2)}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{4} \\ \sum a_{n} diverges \end{array}$ $\sum_{n = 1}^{\infty} {(\frac{n + 1}{n})}^{n}$ Diverges: the terms approach $e$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

The nth-Term TestAnswer keyVersion 1

Date Period

Find the limit of the terms, and say what the nth-term test tells you about the series.

$a_{n} = \tan^{- 1} n$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{π}{2} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{(n + 2)!}{n! (6 n^{2} + 3)}$ Diverges: the terms approach $\frac{1}{6}$
$a_{n} = 2 n \sin ⁣ (\frac{9}{n})$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = 18 \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{2 \cdot 3^{n} + 5}{2 \cdot 3^{n} + 1}$ Diverges: the terms approach $1$
$a_{n} = {(1 + \frac{1}{3 n})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = e^{1 / 3} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{4 n^{2} + 2 n + 9}{3 n^{2} + 9}$ Diverges: the terms approach $\frac{4}{3}$
$a_{n} = \frac{\sqrt{25 n^{2} + 9}}{6 n + 1}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{5}{6} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{\sqrt{n^{2} + 9}}{2 n + 7}$ Diverges: the terms approach $\frac{1}{2}$
$a_{n} = \frac{4^{n} + 9}{6 \cdot 4^{n} + 2}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{6} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{- 8 n^{2} - 5 n - 2}{3 n^{2} + 6}$ Diverges: the terms approach $- \frac{8}{3}$
$a_{n} = {(\frac{n}{n + 5})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{e^{5}} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} 3 \tan^{- 1} n$ Diverges: the terms approach $\frac{3 π}{2}$
$a_{n} = \frac{(n + 2)!}{n! (3 n^{2} + 7)}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{3} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} 4 n \tan ⁣ (\frac{9}{n})$ Diverges: the terms approach $36$
$a_{n} = {(1 + \frac{6}{n})}^{n}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = e^{6} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{- 5 n^{2} + 2 n - 5}{7 n^{2} + 8}$ Diverges: the terms approach $- \frac{5}{7}$
$a_{n} = \frac{6 \cdot 3^{n} + 2}{7 \cdot 3^{n} + 8}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{6}{7} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} \frac{\sqrt{16 n^{2} + 8}}{6 n + 3}$ Diverges: the terms approach $\frac{2}{3}$
$a_{n} = \frac{(n + 2)!}{n! (4 n^{2} + 2)}$ $\begin{array}{l} \lim_{n \to \infty} a_{n} = \frac{1}{4} \\ \sum a_{n} diverges \end{array}$
$\sum_{n = 1}^{\infty} {(\frac{n + 1}{n})}^{n}$ Diverges: the terms approach $e$

How to do these

$\sum_{n = 1}^{\infty} \frac{3 n + 1}{5 n - 2}$

Divide the top and bottom by n: the terms approach 3/5.
The limit is not 0, so the terms never get small enough for the sum to settle.
The series diverges by the nth-term test.

The answer is Diverges: the terms approach $\frac{3}{5}$ .

Where students go wrong

Reading a limit of 0 as "converges". The harmonic series has terms that go to 0 and still diverges. When the terms go to 0, the nth-term test says nothing, and another test has to decide.

Also called divergence test, test for divergence, nth term test or term test.

What you can put on this worksheet

Find the limit of the terms, then decide: $a_{n} = {(\frac{n + 4}{n})}^{n}$ → $\begin{array}{l} \lim_{n \to \infty} a_{n} = e^{4} \\ \sum a_{n} diverges \end{array}$
Diverges or inconclusive, from the series: $\sum_{n = 1}^{\infty} {(\frac{n + 4}{n})}^{n}$ → Diverges: the terms approach $e^{4}$

Questions about these worksheets

Yes. Every nth-term test for divergence 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 nth-term test for divergence 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: find the limit of the terms, then decide and diverges or inconclusive, from the series. Tick as many as you want and set how many of each, or let it spread them evenly.

Reading a limit of 0 as "converges". The harmonic series has terms that go to 0 and still diverges. When the terms go to 0, the nth-term test says nothing, and another test has to decide.

nth-term test for divergence worksheet

Select your difficulty

The nth-Term Test

The nth-Term TestAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

Are these nth-term test for divergence worksheets free?+

Do the nth-term test for divergence worksheets come with an answer key?+

Are these nth-term test for divergence worksheets printable?+

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