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Find sum of series $$\sum_{n=1}^{\infty} \frac {(-1)^{n+1}} {n^2}$$.

I know the series converge absolutely so it is clearly convergent and in the absolute case the sum is $\pi^2/6$. However, I can't seem to find the sum in this case ?

Also the series is alternating.

Can someone help me out ?

2 Answers2

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It is well known that $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ and so $\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{\pi^2}{24}$ Thus $$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}-2\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{\pi^2}{6}-2\frac{\pi^2}{24}= \frac{\pi^2}{12}$$

Rene Schipperus
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    This is similar in spirit to the fact that $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{6} - \frac{\pi^{2}}{24} = \frac{\pi^{2}}{8}.$ – Geoff Robinson May 21 '14 at 11:30
  • Why does the first well known series imply the other ? Ohh, I see we get a factor of $1/4$ in the partial series.. –  May 21 '14 at 11:34
  • How did you get to $\frac1{n^2}-2$? – Shahar May 21 '14 at 11:48
  • He subtract one sum from two times the other sum (both convergent). –  May 21 '14 at 12:07
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Even if it is a little backwards engineering, I think one natural way to compute it is to compute the Fourier-Series of $$ f(x):=-\frac{1}{4}x^2+\frac{\pi^2}{12}\qquad -\pi<x<\pi $$ The convergence of the series to the function (pointwise is already enough) gives the desired result when evaluating at $x=0$.

EDIT: Rene's answer is much nicer! (Altough it uses the fact that $\sum_n n^{-2}=\frac{\pi^2}{6}$ which is usually also proved with the help of Fourier-Series).

frog
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  • there is a visual proof for $\sum_n n^{-2}=\frac{\pi^2}{6}$ :http://math.stackexchange.com/questions/1044022/are-there-any-visual-proofs-for-sum-n-1-infty-frac1n2-frac-pi2/1281530#1281530 – VividD May 14 '15 at 09:29