Explanation why $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ = $\frac{-\pi^2}{12}$ with Euler's $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ Can somebody explain why the sum is $\frac{-\pi^2}{12}$. Can you somehow use the zeta function? If so, how?
 A: Let's add them together:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{(-1)^n+1}{n^2}$$
and for odd $n$, $(-1)^n+1=0$ and for even $n$, $(-1)^n+1=2$, so this rewrites to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{2}{(2n)^2}$$
which simplifies to
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}+\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac12\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{12}$$
Thus,
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=\frac{\pi^2}{12}-\frac{\pi^2}{6}=-\frac{\pi^2}{12}$$
A: $$\sum_{i=1}^\infty \dfrac{(-1)^n}{n^2}$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\sum_{i=1}^\infty \dfrac{1}{(2n-1)^2}$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\big(\sum_{i=1}^\infty \dfrac{1}{n^2}-\frac14\sum_{i=1}^\infty \dfrac{1}{n^2}\big)$$
$$=\sum_{i=1}^\infty \dfrac{1}{(2n)^2}-\frac34\sum_{i=1}^\infty \dfrac{1}{n^2}$$
$$=\frac14\sum_{i=1}^\infty \dfrac{1}{n^2}-\frac34\sum_{i=1}^\infty \dfrac{1}{n^2}$$
$$=-\frac12\sum_{i=1}^\infty \dfrac{1}{n^2}$$
