Prove $\displaystyle\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^2}=0$

Question

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

Answer 1

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.)