Proving that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$ I want to prove that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}, $$
using the Fourier Series for the $2\pi$-periodic function $f(\theta)=\theta^{2},\quad (-\pi<0<\pi)$, that is, 
$$\theta^{2}=\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}} $$
So, 
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}}=\frac{\theta^{2}}{4}-\frac{\pi^{2}}{12}.$$
I'm looking for a $\theta$ such that $(-1)^{n}\cos(n\theta)=(-1)^{n+1}\Rightarrow \cos(n\theta)=-1$ for all $n\in\mathbb{N}$. How can I choose that $\theta$?
 A: $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}}=\frac{\theta^{2}}{4}-\frac{\pi^{2}}{12}$$
Take $\theta=0$ so $\cos n \theta = 1$
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}=-\frac{\pi^{2}}{12}\\
(-1)\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}=(-1)\cdot\left(-\frac{\pi^{2}}{12}\right)\\
\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}\\
$$
A: Split it!
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\sum_{n ~\text{odd}}\frac{1}{n^{2}} - \sum_{n ~\text{even}}\frac{1}{n^{2}}.$$
Add and subtract the "even" part:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\left(\sum_{n ~\text{odd}}\frac{1}{n^{2}} + \sum_{n ~\text{even}}\frac{1}{n^{2}}\right) - \sum_{n ~\text{even}}\frac{1}{n^{2}} - \sum_{n ~\text{even}}\frac{1}{n^{2}} = \\
=\sum_{n=1}^{\infty}\frac{1}{n^2}-2\sum_{n ~\text{even}}\frac{1}{n^{2}} = \frac{\pi^2}{6} - 2\sum_{n ~\text{even}}\frac{1}{n^{2}}.$$
Now, notice that:
$$\sum_{n ~\text{even}}\frac{1}{n^{2}}=\sum_{i =1}^{\infty}\frac{1}{(2i)^{2}} = \frac{1}{4}\sum_{i =1}^{\infty}\frac{1}{i^{2}} = \frac{1}{4}\frac{\pi^2}{6}.$$
Therefore:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}} = \frac{\pi^2}{6} - 2\frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{12}.$$
