How to find sum of $\sum_{n=1}^{\infty}(-1)^n \frac{1}{n^2}$

Question

Probably using uniform convergence - I'm looking for $f(0)$, where $$f(x) = \sum_{n=1}^{\infty}(-1)^n \frac{\cos(\frac{x}{n})} {n^2}$$

Answer 1

It's well known that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$ We have that: $$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\sum_{n=1}^\infty\frac{1}{(2n)^2}-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$

and that $$\sum_{n=1}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(2n)^2}+\sum_{n=0}^\infty\frac{1}{(2n+1)^2}$$

See if you can use these to obtain your result