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

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 ?

• First find $\sum_{n=1}^\infty 1/(2n)^2$. – David Mitra May 21 '14 at 11:05
• This is a special case of the Dirichlet Eta function or alternating Zeta function, and you find $\eta(2)=\frac{\pi^2}{12}$ in the particular value section. – gammatester May 21 '14 at 11:11
• So it is not a series you just compute "easily" ? – user141901 May 21 '14 at 11:12

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 $$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}-2\sum\limits_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{\pi^2}{6}-2\frac{\pi^2}{24}= \frac{\pi^2}{12}$$
• 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.. – user141901 May 21 '14 at 11:34
• How did you get to $\frac1{n^2}-2$? – Shahar May 21 '14 at 11:48
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).