p-series worksheet

A series of 1 over n to a power converges exactly when the power is more than 1. Most of the work is getting a series into that form: a root becomes a fractional power, and a quotient of powers becomes one power.

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 p-Series Date Period Find p and decide whether each series converges, or answer the question asked. $\sum_{n = 1}^{\infty} \frac{6}{n^{9 / 4}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k + 3}}$ converge? $\sum_{n = 1}^{\infty} \frac{6}{n^{2}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{7}}$ converge? $\sum_{n = 1}^{\infty} \frac{7}{\sqrt[5]{n^{12}}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{3 k - 1}}$ converge? $\sum_{n = 1}^{\infty} \frac{2}{\sqrt{n^{5}}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4 k + 2}}}$ converge? $\sum_{n = 1}^{\infty} \frac{9}{n \sqrt[4]{n}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 5}}}$ converge? $\sum_{n = 1}^{\infty} \frac{2}{n \sqrt{n}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k - 5}}$ converge? $\sum_{n = 1}^{\infty} \frac{6}{n}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{2 k + 2}}$ converge? $\sum_{n = 1}^{\infty} \frac{8}{\sqrt[3]{n}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{3 k}}}$ converge? $\sum_{n = 1}^{\infty} \frac{6}{\sqrt{n}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{2}}$ converge? $\sum_{n = 1}^{\infty} \frac{3}{n^{2} \sqrt[3]{n}}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 2}}}$ converge? Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

p-Series

Date Period

Find p and decide whether each series converges, or answer the question asked.

$\sum_{n = 1}^{\infty} \frac{6}{n^{9 / 4}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k + 3}}$ converge?
$\sum_{n = 1}^{\infty} \frac{6}{n^{2}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{7}}$ converge?
$\sum_{n = 1}^{\infty} \frac{7}{\sqrt[5]{n^{12}}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{3 k - 1}}$ converge?
$\sum_{n = 1}^{\infty} \frac{2}{\sqrt{n^{5}}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4 k + 2}}}$ converge?
$\sum_{n = 1}^{\infty} \frac{9}{n \sqrt[4]{n}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 5}}}$ converge?
$\sum_{n = 1}^{\infty} \frac{2}{n \sqrt{n}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k - 5}}$ converge?
$\sum_{n = 1}^{\infty} \frac{6}{n}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{2 k + 2}}$ converge?
$\sum_{n = 1}^{\infty} \frac{8}{\sqrt[3]{n}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{3 k}}}$ converge?
$\sum_{n = 1}^{\infty} \frac{6}{\sqrt{n}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{2}}$ converge?
$\sum_{n = 1}^{\infty} \frac{3}{n^{2} \sqrt[3]{n}}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 2}}}$ converge?



Sigma Prep · sigmaprep.io/worksheets Name p-SeriesAnswer keyVersion 1 Date Period Find p and decide whether each series converges, or answer the question asked. $\sum_{n = 1}^{\infty} \frac{6}{n^{9 / 4}}$ $\begin{array}{l} p = \frac{9}{4} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k + 3}}$ converge? $k > - 2$ $\sum_{n = 1}^{\infty} \frac{6}{n^{2}}$ $\begin{array}{l} p = 2 \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{7}}$ converge? $k < 6$ $\sum_{n = 1}^{\infty} \frac{7}{\sqrt[5]{n^{12}}}$ $\begin{array}{l} p = \frac{12}{5} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{3 k - 1}}$ converge? $k > \frac{2}{3}$ $\sum_{n = 1}^{\infty} \frac{2}{\sqrt{n^{5}}}$ $\begin{array}{l} p = \frac{5}{2} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4 k + 2}}}$ converge? $k > \frac{1}{4}$ $\sum_{n = 1}^{\infty} \frac{9}{n \sqrt[4]{n}}$ $\begin{array}{l} p = \frac{5}{4} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 5}}}$ converge? $k > - 3$ $\sum_{n = 1}^{\infty} \frac{2}{n \sqrt{n}}$ $\begin{array}{l} p = \frac{3}{2} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k - 5}}$ converge? $k > 6$ $\sum_{n = 1}^{\infty} \frac{6}{n}$ $\begin{array}{l} p = 1 \\ Diverges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{2 k + 2}}$ converge? $k > - \frac{1}{2}$ $\sum_{n = 1}^{\infty} \frac{8}{\sqrt[3]{n}}$ $\begin{array}{l} p = \frac{1}{3} \\ Diverges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{3 k}}}$ converge? $k > 1$ $\sum_{n = 1}^{\infty} \frac{6}{\sqrt{n}}$ $\begin{array}{l} p = \frac{1}{2} \\ Diverges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{2}}$ converge? $k < 1$ $\sum_{n = 1}^{\infty} \frac{3}{n^{2} \sqrt[3]{n}}$ $\begin{array}{l} p = \frac{7}{3} \\ Converges \end{array}$ For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 2}}}$ converge? $k > 0$ Free math worksheets at sigmaprep.io/worksheets© 2026 Sigma Prep

Sigma Prep · sigmaprep.io/worksheets

Name

p-SeriesAnswer keyVersion 1

Date Period

Find p and decide whether each series converges, or answer the question asked.

$\sum_{n = 1}^{\infty} \frac{6}{n^{9 / 4}}$ $\begin{array}{l} p = \frac{9}{4} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k + 3}}$ converge? $k > - 2$
$\sum_{n = 1}^{\infty} \frac{6}{n^{2}}$ $\begin{array}{l} p = 2 \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{7}}$ converge? $k < 6$
$\sum_{n = 1}^{\infty} \frac{7}{\sqrt[5]{n^{12}}}$ $\begin{array}{l} p = \frac{12}{5} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{3 k - 1}}$ converge? $k > \frac{2}{3}$
$\sum_{n = 1}^{\infty} \frac{2}{\sqrt{n^{5}}}$ $\begin{array}{l} p = \frac{5}{2} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4 k + 2}}}$ converge? $k > \frac{1}{4}$
$\sum_{n = 1}^{\infty} \frac{9}{n \sqrt[4]{n}}$ $\begin{array}{l} p = \frac{5}{4} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 5}}}$ converge? $k > - 3$
$\sum_{n = 1}^{\infty} \frac{2}{n \sqrt{n}}$ $\begin{array}{l} p = \frac{3}{2} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{k - 5}}$ converge? $k > 6$
$\sum_{n = 1}^{\infty} \frac{6}{n}$ $\begin{array}{l} p = 1 \\ Diverges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{2 k + 2}}$ converge? $k > - \frac{1}{2}$
$\sum_{n = 1}^{\infty} \frac{8}{\sqrt[3]{n}}$ $\begin{array}{l} p = \frac{1}{3} \\ Diverges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{3 k}}}$ converge? $k > 1$
$\sum_{n = 1}^{\infty} \frac{6}{\sqrt{n}}$ $\begin{array}{l} p = \frac{1}{2} \\ Diverges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{n^{k}}{n^{2}}$ converge? $k < 1$
$\sum_{n = 1}^{\infty} \frac{3}{n^{2} \sqrt[3]{n}}$ $\begin{array}{l} p = \frac{7}{3} \\ Converges \end{array}$
For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{k + 2}}}$ converge? $k > 0$

How to do these

$\sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^{2}}$

Write the root as a power: the square root of n is n^(1/2).
Subtract the powers: n^(1/2) over n^2 is 1 over n^(3/2).
p = 3/2 is more than 1, so the series converges.

The answer is $p = \frac{3}{2}$ ; converges.

Where students go wrong

Calling p = 1 a convergent case. The harmonic series, p = 1, diverges; the test needs p strictly greater than 1.

Also called p series test, the p-test, p-series test or hyperharmonic series.

What you can put on this worksheet

Find p and decide (rewrite roots and quotients first): $\sum_{n = 1}^{\infty} \frac{3}{n^{1 / 4}}$ → $\begin{array}{l} p = \frac{1}{4} \\ Diverges \end{array}$
The values of k that make it converge: For what values of $k$ does $\sum_{n = 1}^{\infty} \frac{1}{n^{4 k + 3}}$ converge? → $k > - \frac{1}{2}$

Questions about these worksheets

Yes. Every p-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 p-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: find p and decide (rewrite roots and quotients first) and the values of k that make it converge. Tick as many as you want and set how many of each, or let it spread them evenly.

Calling p = 1 a convergent case. The harmonic series, p = 1, diverges; the test needs p strictly greater than 1.

p-series worksheet

Select your difficulty

p-Series

p-SeriesAnswer keyVersion 1

How to do these

Where students go wrong

What you can put on this worksheet

Questions about these worksheets

Are these p-series worksheets free?+

Do the p-series worksheets come with an answer key?+

Are these p-series worksheets printable?+

What grade are these p-series worksheets for?+

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

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

Can I choose how hard the p-series questions are?+

What kinds of question are on a p-series worksheet?+

What do students get wrong with p-series?+