Prove $\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$ I am trying to prove that the following sum equals to zero:
$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$$
This actually surprised me a-lot. Usually when these kinds of infinite sums equal to zero, it's because of some sort of symmetry, or something like that. But I can't see it here. The $n=0$ term isn't zero, and we don't have $n\mapsto -n$ symmetry. The only thing I thought of is that maybe the Taylor series of $\arctan(x)$ might help, but the sum there starts from $n=1$ (or $n=0$ if you look at $\arctan(x)/x^2$), and needs to be integrated. It was kind of complicated (maybe it's not and I missed something).
P.S. - The original thing I need to compute is the following integral $I$, which I proved that is connected to the mentioned sum through the Residue Theorem:
$$I=\int_{0}^{2\pi}e^{i\theta}\sec{(e^{-i\theta})}\text{d}\theta=\frac 8\pi \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{(2n+1)^2}=0$$
I added this because maybe the integral might be useful somehow (maybe easier to compute). Again I searched for symmetry (the integrand is clearly $2\pi$ symmetric, so I changed the integration interval to be $[-\pi,\pi]$, and then hoped that the integrand would be odd or something. But it's not, if I'm not mistaken).
Thanks
 A: Let us check some terms:

*

*For $n=-1$ and $n=0$ we get the terms of the series:
$$
\frac{(-1)^{-1}}{(-2+1)^2}=-\frac 1{1^2}
\text{ and }
\frac{(-1)^{0}}{(0+1)^2}=+\frac 1{1^2}\ .
$$

*For $n=-2$ and $n=1$ we get the terms of the series:
$$
\frac{(-1)^{-2}}{(-4+1)^2}=+\frac 1{3^2}
\text{ and }
\frac{(-1)^{1}}{(2+1)^2}=-\frac 1{3^2}\ .
$$

*For $n=-3$ and $n=2$ we get the terms of the series:
$$
\frac{(-1)^{-3}}{(-6+1)^2}=-\frac 1{5^2}
\text{ and }
\frac{(-1)^{2}}{(4+1)^2}=+\frac 1{5^2}\ .
$$

*For $n=-4$ and $n=3$ we get the terms of the series:
$$
\frac{(-1)^{-4}}{(-8+1)^2}=+\frac 1{7^2}
\text{ and }
\frac{(-1)^{3}}{(6+1)^2}=-\frac 1{7^2}\ .
$$
The cancelling scheme should be clear now.

A: Alternatively, $I=-i\oint_{|z|=1}z^{-2}\sec zdz=0$ because $\sec z$ has no poles in the contour, and the $z^{-2}$ factor doesn't contribute either. (The $-$ sign comes from the contour being clockwise.)
A: You can plot the summands:

So evidence suggests there are opposing terms, it's just that they're symmetric about $n=-1/2$ not $n=0$. Symbolically,
$$\sum_{n=-\infty}^{-1}\frac{(-1)^n}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$
Now just change the index of the first sum,
$$-\sum_{n=-\infty}^0\frac{(-1)^n}{(2n-1)^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}.$$
Hopefully you can see why this equals zero now.
A: Note that
$$\sum_{n=-\infty}^\infty{(-1)^n\over(2n+1)^2}=\sum_{m\equiv1\text{ mod }4}{1\over m^2}-\sum_{m\equiv3\text{ mod }4}{1\over m^2}$$
and $m\equiv1$ mod $4$ if and only if $-m\equiv3$ mod $4$, so, since $m^2=(-m)^2$, the two sums on the right hand side are equal, hence cancel each other out.
