S_N and the sum: sum of_n=1^infinity [ln (n + 2) - ln (n + 1)]

S_N = ln (N + 2) - ln 2; Diverges

S_N and the sum: sum of_n=1^infinity (3)/((2n - 1)(2n + 1))

S_N = 3/2 - (3)/(2(2N + 1)); Sum: 3/2

S_N and the sum: sum of_n=1^infinity (2)/(n(n + 2))

S_N = 3/2 - (1)/(N + 1) - (1)/(N + 2); Sum: 3/2

S_N and the sum: sum of_n=1^infinity (2)/((3n)(3n + 3))

S_N = 2/9 - (2)/(3(3N + 3)); Sum: 2/9

S_N and the sum: sum of_n=1^infinity 5(e^1/n - e^1/(n + 1))

S_N = 5e - 5e^1/(N + 1); Sum: 5e - 5

S_N and the sum: sum of_n=1^infinity (4)/(n(n + 2))

S_N = 3 - (2)/(N + 1) - (2)/(N + 2); Sum: 3

S_N and the sum: sum of_n=1^infinity (2)/((3n + 1)(3n + 4))

S_N = 1/6 - (2)/(3(3N + 4)); Sum: 1/6

S_N and the sum: sum of_n=1^infinity 4((1)/(n + 4) - (1)/(n + 5))

S_N = 4/5 - (4)/(N + 5); Sum: 4/5

Worksheets · Calculus

Telescoping series worksheet

A telescoping series is a difference of shifted terms. Written out, almost everything cancels: the partial sum keeps the first few pieces and the last few, and the sum is whatever the first pieces leave once the last ones die away.

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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 Telescoping Series Date Period Find what each question asks for: the sum, or the Nth partial sum and then the sum. Say so if the series diverges. Sum: $\sum_{n = 1}^{\infty} \ln ⁣ (1 + \frac{1}{n})$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} [\ln (n + 2) - \ln (n + 1)]$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{(n + 2) (n + 3)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{3}{(2 n - 1) (2 n + 1)}$ Sum: $\sum_{n = 1}^{\infty} (e^{1 / n} - e^{1 / (n + 1)})$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} 2 ⁣ (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{n (n + 3)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{n (n + 2)}$ Sum: $\sum_{n = 1}^{\infty} \frac{5}{n (n + 4)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{(3 n) (3 n + 3)}$ Sum: $\sum_{n = 1}^{\infty} \ln ⁣ (\frac{n}{n + 1})$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} 5 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ Sum: $\sum_{n = 1}^{\infty} \frac{1}{(n + 5) (n + 6)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} [\ln (n + 1) - \ln n]$ Sum: $\sum_{n = 1}^{\infty} \frac{1}{n (n + 3)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{n (n + 4)}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{4}{n (n + 2)}$ Sum: $\sum_{n = 1}^{\infty} 6 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{(3 n + 1) (3 n + 4)}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Telescoping Series

Date Period

Find what each question asks for: the sum, or the Nth partial sum and then the sum. Say so if the series diverges.

Sum:
$\sum_{n = 1}^{\infty} \ln ⁣ (1 + \frac{1}{n})$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} [\ln (n + 2) - \ln (n + 1)]$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{(n + 2) (n + 3)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{3}{(2 n - 1) (2 n + 1)}$
Sum:
$\sum_{n = 1}^{\infty} (e^{1 / n} - e^{1 / (n + 1)})$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} 2 ⁣ (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{n (n + 3)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{n (n + 2)}$
Sum:
$\sum_{n = 1}^{\infty} \frac{5}{n (n + 4)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{(3 n) (3 n + 3)}$
Sum:
$\sum_{n = 1}^{\infty} \ln ⁣ (\frac{n}{n + 1})$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} 5 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$
Sum:
$\sum_{n = 1}^{\infty} \frac{1}{(n + 5) (n + 6)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} [\ln (n + 1) - \ln n]$
Sum:
$\sum_{n = 1}^{\infty} \frac{1}{n (n + 3)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{n (n + 4)}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{4}{n (n + 2)}$
Sum:
$\sum_{n = 1}^{\infty} 6 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{(3 n + 1) (3 n + 4)}$



Sigma Prep · sigmaprep.io/worksheets Name Telescoping SeriesAnswer keyVersion 1 Date Period Find what each question asks for: the sum, or the Nth partial sum and then the sum. Say so if the series diverges. Sum: $\sum_{n = 1}^{\infty} \ln ⁣ (1 + \frac{1}{n})$ Diverges $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} [\ln (n + 2) - \ln (n + 1)]$ $\begin{array}{l} S_{N} = \ln (N + 2) - \ln 2 \\ Diverges \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{(n + 2) (n + 3)}$ $1$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{3}{(2 n - 1) (2 n + 1)}$ $\begin{array}{l} S_{N} = \frac{3}{2} - \frac{3}{2 (2 N + 1)} \\ Sum: \frac{3}{2} \end{array}$ Sum: $\sum_{n = 1}^{\infty} (e^{1 / n} - e^{1 / (n + 1)})$ $e - 1$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} 2 ⁣ (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ $\begin{array}{l} S_{N} = 2 - \frac{2}{\sqrt{N + 1}} \\ Sum: 2 \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{n (n + 3)}$ $\frac{11}{6}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{n (n + 2)}$ $\begin{array}{l} S_{N} = \frac{3}{2} - \frac{1}{N + 1} - \frac{1}{N + 2} \\ Sum: \frac{3}{2} \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{5}{n (n + 4)}$ $\frac{125}{48}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{(3 n) (3 n + 3)}$ $\begin{array}{l} S_{N} = \frac{2}{9} - \frac{2}{3 (3 N + 3)} \\ Sum: \frac{2}{9} \end{array}$ Sum: $\sum_{n = 1}^{\infty} \ln ⁣ (\frac{n}{n + 1})$ Diverges $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} 5 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ $\begin{array}{l} S_{N} = 5 e - 5 e^{1 / (N + 1)} \\ Sum: 5 e - 5 \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{1}{(n + 5) (n + 6)}$ $\frac{1}{6}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} [\ln (n + 1) - \ln n]$ $\begin{array}{l} S_{N} = \ln (N + 1) \\ Diverges \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{1}{n (n + 3)}$ $\frac{11}{18}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ $\begin{array}{l} S_{N} = 1 - \frac{1}{\sqrt{N + 1}} \\ Sum: 1 \end{array}$ Sum: $\sum_{n = 1}^{\infty} \frac{3}{n (n + 4)}$ $\frac{25}{16}$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{4}{n (n + 2)}$ $\begin{array}{l} S_{N} = 3 - \frac{2}{N + 1} - \frac{2}{N + 2} \\ Sum: 3 \end{array}$ Sum: $\sum_{n = 1}^{\infty} 6 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ $6 e - 6$ $S_{N}$ and the sum: $\sum_{n = 1}^{\infty} \frac{2}{(3 n + 1) (3 n + 4)}$ $\begin{array}{l} S_{N} = \frac{1}{6} - \frac{2}{3 (3 N + 4)} \\ Sum: \frac{1}{6} \end{array}$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

Telescoping SeriesAnswer keyVersion 1

Date Period

Find what each question asks for: the sum, or the Nth partial sum and then the sum. Say so if the series diverges.

Sum:
$\sum_{n = 1}^{\infty} \ln ⁣ (1 + \frac{1}{n})$ Diverges
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} [\ln (n + 2) - \ln (n + 1)]$ $\begin{array}{l} S_{N} = \ln (N + 2) - \ln 2 \\ Diverges \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{(n + 2) (n + 3)}$ $1$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{3}{(2 n - 1) (2 n + 1)}$ $\begin{array}{l} S_{N} = \frac{3}{2} - \frac{3}{2 (2 N + 1)} \\ Sum: \frac{3}{2} \end{array}$
Sum:
$\sum_{n = 1}^{\infty} (e^{1 / n} - e^{1 / (n + 1)})$ $e - 1$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} 2 ⁣ (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ $\begin{array}{l} S_{N} = 2 - \frac{2}{\sqrt{N + 1}} \\ Sum: 2 \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{n (n + 3)}$ $\frac{11}{6}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{n (n + 2)}$ $\begin{array}{l} S_{N} = \frac{3}{2} - \frac{1}{N + 1} - \frac{1}{N + 2} \\ Sum: \frac{3}{2} \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{5}{n (n + 4)}$ $\frac{125}{48}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{(3 n) (3 n + 3)}$ $\begin{array}{l} S_{N} = \frac{2}{9} - \frac{2}{3 (3 N + 3)} \\ Sum: \frac{2}{9} \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \ln ⁣ (\frac{n}{n + 1})$ Diverges
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} 5 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ $\begin{array}{l} S_{N} = 5 e - 5 e^{1 / (N + 1)} \\ Sum: 5 e - 5 \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{1}{(n + 5) (n + 6)}$ $\frac{1}{6}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} [\ln (n + 1) - \ln n]$ $\begin{array}{l} S_{N} = \ln (N + 1) \\ Diverges \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{1}{n (n + 3)}$ $\frac{11}{18}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ $\begin{array}{l} S_{N} = 1 - \frac{1}{\sqrt{N + 1}} \\ Sum: 1 \end{array}$
Sum:
$\sum_{n = 1}^{\infty} \frac{3}{n (n + 4)}$ $\frac{25}{16}$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{4}{n (n + 2)}$ $\begin{array}{l} S_{N} = 3 - \frac{2}{N + 1} - \frac{2}{N + 2} \\ Sum: 3 \end{array}$
Sum:
$\sum_{n = 1}^{\infty} 6 ⁣ (e^{1 / n} - e^{1 / (n + 1)})$ $6 e - 6$
$S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} \frac{2}{(3 n + 1) (3 n + 4)}$ $\begin{array}{l} S_{N} = \frac{1}{6} - \frac{2}{3 (3 N + 4)} \\ Sum: \frac{1}{6} \end{array}$

How to do these

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

Partial fractions: 2/(n(n + 1)) = 2/n - 2/(n + 1).
Adding the first N terms, every middle piece cancels, leaving 2 - 2/(N + 1).
As N grows, 2/(N + 1) goes to 0, so the sum is 2.

The answer is $S_{N} = 2 - \frac{2}{N + 1}$ ; sum: $2$ .

Where students go wrong

Cancelling too much when the shift is more than 1. With 1/n - 1/(n + 2), two pieces survive at each end, not one.

Also called telescoping sum, collapsing series or partial sums of a telescoping series.

What you can put on this worksheet

The sum, or diverges: Sum:
$\sum_{n = 1}^{\infty} 4 ⁣ (\frac{1}{n + 4} - \frac{1}{n + 5})$ → $\frac{4}{5}$
The partial sum S_N, then the sum: $S_{N}$ and the sum:
$\sum_{n = 1}^{\infty} 4 ⁣ (\frac{1}{n + 4} - \frac{1}{n + 5})$ → $\begin{array}{l} S_{N} = \frac{4}{5} - \frac{4}{N + 5} \\ Sum: \frac{4}{5} \end{array}$

Questions about these worksheets

Yes. Every telescoping series 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 telescoping series 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: the sum, or diverges and the partial sum s_n, then the sum. Tick as many as you want and set how many of each, or let it spread them evenly.

Cancelling too much when the shift is more than 1. With 1/n - 1/(n + 2), two pieces survive at each end, not one.

Telescoping series worksheet

Select your difficulty

Telescoping Series

Telescoping SeriesAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

Are these telescoping series worksheets free?+

Do the telescoping series worksheets come with an answer key?+

Are these telescoping series worksheets printable?+

What grade are these telescoping series worksheets for?+

Can I get a different version of the same telescoping series worksheet?+

Is this a free alternative to Kuta Software for telescoping series?+

Can I choose how hard the telescoping series questions are?+

What kinds of question are on a telescoping series worksheet?+

What do students get wrong with telescoping series?+