Unusual evaluation of $\sum \frac{1}{n^2}$ 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?

Splitting the sum into odd and even parts:
$$\sum_{n=1}^{\infty}\frac 1{n^2}=\sum_{n=1}^{\infty}\frac 1{(2n)^2}+\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
$$\sum_{n=1}^{\infty}\frac 1{n^2}=\frac 14\sum_{n=1}^{\infty}\frac 1{n^2}+\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
$$\frac 34 \sum_{n=1}^{\infty}\frac 1{n^2}=\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
and it's trivial from there.
