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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?

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    @AlexBecker: Thank$s$ for the correction! My comment isn't adding to the discussion anyways, so I'll delete it.2012-09-20

2 Answers 2

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

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    @Alex Becke$r$ Now is ok2012-09-20
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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.