The series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\mbox{sech}\left(\frac{(2n-1)\pi}{2}\right)$ Does anyone have a proof that
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\mbox{sech}\left(\frac{(2n-1)\pi}{2}\right)=\frac{\pi}{8}$$
 A: $$f(z) = \frac{1}{2z\cosh(z\pi)}$$
so that the sum in question is half of 
$$\sum_{n=-\infty}^\infty (-1)^n f\left(\frac{2n+1}{2}\right) = \sum \operatorname*{Res} (\pi \sec(\pi z)f(z))\text{ at poles of }f$$
$f$ has poles at $0$ and at $b_n = \frac{i(2n+1)}{2}$ for integer $n$.  Then
$$\operatorname*{Res}_{z=0} (\pi \sec(\pi z)f(z)) = \frac{\pi}{2}$$
$$\operatorname*{Res}_{z=b_n} (\pi \sec(\pi z)f(z)) = -\frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right)$$
Then
$$\sum_{n=-\infty}^\infty \frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right)=
-\sum_{n=-\infty}^\infty \frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right) + \frac{\pi}{2} \implies\\
\sum_{n=-\infty}^\infty \frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right) = \frac{\pi}{4}
$$
So finally, solving for the sum and dividing by 2:
$$\sum_{n=0}^\infty \frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right) = 
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} \operatorname{sech}\left(\pi \frac{2n-1}{2}\right)
=\frac{\pi}{8}$$
