TITLE: Unusual evaluation of $\sum \frac{1}{n^2}$
QUESTION [11 upvotes]: Assume the formula $$\sum_{n=-\infty}^\infty\frac{1}{(n+u)^2}=\frac{\pi^2}{(\sin \pi u)^2},$$ where $u\notin\Bbb Z$. I have been trying to prove that $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}.$$Setting $u=1/2$, I was able to derive $$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8},$$but I haven't made any progress toward proving the value of $\zeta(2)$. Any suggestions?

REPLY [6 votes]: A more direct way of seeing this is to do the following. Break up the sum as
$$\sum_{n=-\infty}^{-1}\frac{1}{(n+u)^2}+\sum_{n=1}^\infty\frac{1}{(n+u)^2}=\frac{\pi^2}{\sin^2(\pi u)}-\frac{1}{u^2}$$
and take $u\to 0$. This is easiest to do by Taylor expanding
$$\frac{\pi^2}{\sin^2(\pi u)}=\frac{1}{u^2}+\frac{\pi^2}{3}+\frac{\pi^4 u^2}{15}+\cdots$$
so we see that the singularity cancels exactly and we have
$$2\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}.$$