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

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.

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 $$2\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}.$$