Alternating reciprocal of cosh infinite sum

Today we will prove the following result seen in this post



\begin{align*} \sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh \pi n} &=\frac{1}{\Gamma^2\left(\frac{3}{4}\right)} \sqrt{\frac{\pi}{2}} \end{align*}


Click here to see the proof

Comments

Post a Comment