The series $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\mbox{sech}\left(\frac{(2n-1)\pi}{2}\right)$

Question

Does anyone have a proof that $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}\mbox{sech}\left(\frac{(2n-1)\pi}{2}\right)=\frac{\pi}{8}$$

Answer 1

$$f(z) = \frac{1}{2z\cosh(z\pi)}$$

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:

$$\sum_{n=0}^\infty \frac{(-1)^n}{1+2n} \operatorname{sech}\left(\pi \frac{2n+1}{2}\right) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1} \operatorname{sech}\left(\pi \frac{2n-1}{2}\right) =\frac{\pi}{8}$$