Calculate the series: $\sum^\infty_{n=1}\frac{(-1)^n}{n^2}$ using dirichlet's theorem This question was in my exam:
Calculate the series: $$\sum^\infty_{n=1}\frac{(-1)^n}{n^2}$$.
I answered wrong and the teacher noted:  "You should use dirichlet's theorem".
I know my question is a bit general,
but can you please explain me how should I have solved this sum?
Thanks in advance.

By absolute convergence you can simply write:
$$\sum_{n=1}^{n}\frac{(-1)^n}{n^2}=\sum_{n \text{ even}}\frac{1}{n^2}-\sum_{n\text{ odd}}\frac{1}{n^2}=2\cdot\sum_{n \text{ even}}\frac{1}{n^2}-\sum_{n\geq 1}\frac{1}{n^2}=\frac{2}{4}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{n\geq 1}\frac{1}{n^2}$$ $$=-\frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}=-\frac{\zeta(2)}{2}=-\frac{\pi^2}{12}.$$
Have a look at this hot question, too.
