How can one calculate $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$?

Question

I am working through a textbook for my Fourier Series class and I came across the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$. I know from the context of the question that the answer has to be $-\frac{\pi^2}{12}$, but I am not sure how to arrive at this answer. I tried partial fraction decomposition to see if there was any telescoping behavior, but I had an issue with the $(-1)^n$ part.

Answer 1

Assuming that you know $S:=\sum_{n\geq 1} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$, we get (and I'll replace $(-1)^{n}$ with $(-1)^{n-1}$ which seems that you actually intended to be positive)

$$ \frac{1}{1^2} -\frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \cdots \\ = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \cdots \\ - 2 \left(\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \cdots\right) \\ = S - 2\cdot \frac{1}{2^{2}}S = \frac{1}{2}S = \frac{\pi^{2}}{12} $$