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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?

Michael Hardy
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countunique
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  • @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

2 Answers2

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

Deven Ware
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Alex Becker
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HINT.

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.

GEdgar
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