Finding the sum of an alternating series I want to find the sum of $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$$ I know that this is equal to $\frac{\pi^2}{12}$ thus I was thinking this must just be a taylor series of some trigonometric function but after looking it up, I cannot seem to find one that satisfies this. Any suggestions are greatly appreciated.
From below the user states that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \sum_{n=0}^{\infty}\left(\frac{1}{2n+1} \right)^2 -\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \sum_{n=1}^{\infty}\frac{1}{n^2} - 2\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \ldots = \frac{\pi^2}{12}$$
I want to know the details of the $\ldots$ part.
 A: The series can be written as
$$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2-\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2=\sum _1 ^{\infty} \frac {1}{n^2}-2\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2$$
By adding and subtracting:
$\sum _1 ^{\infty} \left(\frac {1}{2n}\right)^2$
Edited part :-
$$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2+\sum_1 ^{\infty}\left(\frac {1}{2n}\right)^2-2\sum_1 ^{\infty}\left(\frac{1}{2n}\right)^2$$
Now first two sums can be written as
$$\sum _1 ^{\infty} \frac {1}{n^2}$$ So the next two can be written as:
$$\frac {1}{2}\sum _1 ^{\infty} \frac {1}{n^2}$$
Now we know all the summations. Thus the answer is:
$$\frac {\pi ^2}{6}-\frac {\pi^2}{2.6}=\frac {\pi^2}{12} $$
A: Hint. One may see this as a special value of the dilogarithm function
$$
\text{Li}_2(z)=\sum_{n=1}^\infty\frac{z^n}{n^2},\qquad |z|\le1,
$$
recalling that
$$
\text{Li}_2(1)=\frac{\pi^2}6
$$ one may observe that
$$
2\left(\text{Li}_2(z)+\text{Li}_2(-z)\right)=\text{Li}_2(z^2)
$$ giving, by putting $z=1$,
$$
\text{Li}_2(-1)=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}.
$$
