What is the value of the series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n^2}$? I am given the following series:
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$$ 
I have used the alternating series test to show that the series converges. 
However, how do I go about showing what it converges to? 

Consider the Fourier series of $g(x)=x^2$ for $-\pi<x\le\pi$:
$$g(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)$$
note $b_n=0$ for an even function $g(t)=g(-t)$ and that:
$$a_n=\frac {1}{\pi} \int _{-\pi }^{\pi }\!{x}^{2}\cos \left( nx \right) {dx}
=4\,{\frac { \left( -1 \right) ^{n}}{{n}^{2}}},$$
$$\frac{a_0}{2}=\frac {1}{2\pi} \int _{-\pi }^{\pi }\!{x}^{2} {dx}
=\frac{1}{3}\pi^2,$$
$$x^2=\frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n
}\cos \left( nx \right) }{{n}^{2}}},$$
$$x=0\rightarrow \frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n
}}{{n}^{2}}}=0,$$
$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n}
}{{n}^{2}}}=-\frac{1}{12}\,{\pi }^{2}.$$
