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