Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm sure this has been asked a million times, but it's hard to google for a particular series without knowing its name.

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

I know this converges absolutely to $\frac{\pi^2}{6}$ and I know that it is absolutely convergent so that the terms can be rearranged.

So the sum is equal to $-1 + \sum_{n=2}^\infty \frac{1}{n^2} - \frac{1}{(n+1)^2} = -1 + \sum_{n=2}^\infty \frac{2n + 1}{n^2(n+1)^2}$. Which got me nowhere. Is it a clever rearrangment we're looking for here, or is there another tool to be used?

share|cite|improve this question
@JavaMan The OP says "converges absolutely to $\pi^2/6$", which is correct. – Alex Becker Sep 20 '12 at 18:51
WolframAlpha says this sum is equal to $-\frac12 \sum_{i=1}^\infty \frac{1}{n^2}$. Not sure if there's an easy rearrangement to show that though. – Alex Becker Sep 20 '12 at 18:52
The sum is not what you write when you separate the even and odd terms, but $\sum\limits_{n\geqslant1}\frac1{4n^2}-\frac1{(2n+1)^2}$. Which can be rearranged to $-\frac12\sum\limits_{n\geqslant1}\frac1{n^2}=\frac{\pi^2}{12}$. – Did Sep 20 '12 at 18:53
Javaman's statement is also correct :) – Julian Wergieluk Sep 20 '12 at 18:56
@AlexBecker: Thanks for the correction! My comment isn't adding to the discussion anyways, so I'll delete it. – JavaMan Sep 20 '12 at 21:17
up vote 8 down vote accepted

We can break the sum up into positive and negative terms, so $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=\sum_{n=1}^\infty \frac{1}{(2n)^2}-\sum_{n=1}^\infty\frac{1}{(2n+1)^2}=2\sum_{n=1}^\infty \frac{1}{(2n)^2}-\sum_{n=1}^\infty\frac{1}{n^2}=\frac{-1}{2}\sum_{n=1}^\infty\frac{1}{n^2}=-\frac{\pi^2}{12}$$

share|cite|improve this answer
Very nice trick to know! – StuartHa Sep 20 '12 at 18:59
Result must be divided by $2$? – M. Strochyk Sep 20 '12 at 19:02
@M.Strochyk What do you mean? – Alex Becker Sep 20 '12 at 19:03
@Alex Becker Now is ok – M. Strochyk Sep 20 '12 at 19:05


If you know $$ \sum_{n=1}^\infty \frac{1}{n^2} $$ Next you should find $$ \sum_\text{even} \frac{1}{n^2} $$ where you use only the even numbers.

Then some combination of these two will be the sum you want.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.