Math Core

The integral test turns one family of series into a one-line decision: the p-series. Together with geometric series, p-series are the benchmarks you will compare almost everything else to, so it pays to know them cold.

The p-series test

A p-series has the form

\sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots

where $p$ is a constant. The harmonic series is the p-series with $p = 1$ .

To decide convergence, apply the integral test to $f(x) = \dfrac{1}{x^p}$ . For $p > 0$ this function is positive, continuous and decreasing on $[1, \infty)$ . You already know from improper integrals that

\int_1^{\infty} \frac{dx}{x^p} = \begin{cases} \dfrac{1}{p - 1} & \text{if } p > 1, \\[2mm] \infty & \text{if } p \le 1. \end{cases}

(If $p \le 0$ , the terms don't even go to $0$ , so the nth term test settles those cases.)

The p-series test

\sum_{n=1}^{\infty} \frac{1}{n^p} \;\text{ converges if } p > 1 \;\text{ and diverges if } p \le 1.

That's it: find the exponent, compare with $1$ . Some examples:

Series	$p$	Result
$\sum \frac{1}{n^2}$	$2$	converges
$\sum \frac{1}{\sqrt{n}}$	$\frac{1}{2}$	diverges
$\sum \frac{1}{n}$	$1$	diverges (harmonic)
$\sum \frac{1}{n^{1.01}}$	$1.01$	converges
$\sum \frac{1}{n\sqrt{n}}$	$\frac{3}{2}$	converges

Why $p = 1$ is the dividing line

Compare the partial sums of $\sum \frac{1}{n}$ and $\sum \frac{1}{n^2}$ . Both have terms that go to zero. But the harmonic terms shrink too slowly: the partial sums keep creeping upward forever. The terms of $\frac{1}{n^2}$ shrink fast enough that the partial sums level off near $1.645$ .

Upper dots: partial sums of Σ 1/n keep climbing. Lower dots: partial sums of Σ 1/n² level off below π²/6 ≈ 1.645 (dashed).Open in grapher →

Even $p = 1.001$ converges, while $p = 1$ does not. The borderline is sharp.

For $p > 1$ the integral also gives a quick sense of size. The rectangle picture from the integral test shows that the sum is between $\dfrac{1}{p-1}$ and $1 + \dfrac{1}{p-1}$ . For $p = 2$ that means the sum is between $1$ and $2$ , consistent with $\frac{\pi^2}{6} \approx 1.645$ . But notice what happens when $p$ is just barely above $1$ : for $p = 1.01$ the sum is more than $100$ . Convergent, yes, but only just.

Rewriting into p-series form

Often a series is a p-series in disguise. Simplify the terms first:

$\displaystyle\sum \frac{\sqrt{n}}{n^2} = \sum \frac{1}{n^{3/2}}$ , so $p = \frac{3}{2}$ : converges.
$\displaystyle\sum \frac{5}{\sqrt[3]{n^2}} = 5\sum \frac{1}{n^{2/3}}$ , so $p = \frac{2}{3}$ : diverges. A constant multiple never changes convergence.
$\displaystyle\sum n^{-e}$ has $p = e \approx 2.718$ : converges.

Common mistake

Don't confuse a p-series with a geometric series. In $\dfrac{1}{n^p}$ the variable is in the base and the exponent is fixed. In $\left(\dfrac{1}{2}\right)^n$ the variable is in the exponent. So $\sum \frac{1}{n^2}$ is a p-series ( $p = 2$ ), while $\sum \frac{1}{2^n}$ is geometric ( $r = \frac12$ ). Both converge, but for different reasons, and only the geometric one has an easy exact sum.

Worked examples

Worked example: Classifying p-series

Determine whether each series converges or diverges.

(a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{0.9}}$ (b) $\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{n^{4.2}}$ (c) $\displaystyle\sum_{n=1}^{\infty} \frac{4}{n\sqrt[4]{n}}$

Solution.

(a) $p = 0.9 \le 1$ : diverges.

(b) $\dfrac{n^3}{n^{4.2}} = \dfrac{1}{n^{1.2}}$ , so $p = 1.2 > 1$ : converges.

Worked example: Using a known sum

It is a famous fact that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ . Use it to find $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$ , the sum over even squares only.

Solution. Factor out the constant:

\sum_{n=1}^{\infty} \frac{1}{4n^2} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{24}.

Worked example: A parameter in the exponent

For which values of $k$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{3 - k}}$ converge?

Solution. This is a p-series with $p = 3 - k$ . It converges when $3 - k > 1$ , that is, when $k < 2$ .

Tip

When you see a sum of a rational or root expression in $n$ , ask "what power does this behave like for large $n$ ?" If the answer is $\frac{1}{n^p}$ , you already have a strong guess about convergence. The next lesson shows how to turn that guess into a proof.

Practice

Practice 1

Which of the following series converges?

Practice 2

Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{n^{3}\sqrt{n}}$ converge or diverge?

Practice 3

For which values of $p$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p - 1}}$ converge? Write an inequality in $p$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 4

Given $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ , find $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ . Give an exact answer.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Given $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ , find $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$ . Give an exact answer.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Using the previous two results, find the sum of the reciprocals of the odd squares, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = 1 + \frac{1}{9} + \frac{1}{25} + \cdots$ . Give an exact answer.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Consider the series I. $\displaystyle\sum \frac{1}{n^{\pi/3}}$ , II. $\displaystyle\sum \frac{1}{\sqrt[5]{n^4}}$ , III. $\displaystyle\sum \frac{\sqrt{n}}{n}$ . Which of them converge?

The p-series test

Why p=1p = 1p=1 is the dividing line

Rewriting into p-series form

Worked examples

Practice

Why $p = 1$ is the dividing line