Show $\sum_{n=1}^{\infty}\frac{\sinh\pi}{\cosh(2n\pi)-\cosh\pi}=\frac1{\text{e}^{\pi}-1}$ and another Show that :
$$\sum_{n=1}^{\infty}\frac{\cosh(2nx)}{\cosh(4nx)-\cosh(2x)}=\frac1{4\sinh^2(x)}$$
$$\sum_{n=1}^{\infty}\frac{\sinh\pi}{\cosh(2n\pi)-\cosh\pi}=\frac1{\text{e}^{\pi}-1}$$

Alternatively, one can prove the following inductively:
$$\sum_1^n\frac{\sinh \pi}{\cosh 2k\pi-\cosh \pi}=\frac{\sinh n\pi}{\cosh (n+1)\pi-\cosh n\pi}$$
Then:
$$\begin{align*}\frac{\sinh n\pi}{\cosh (n+1)\pi-\cosh n\pi}&=\frac{e^{n\pi}-e^{-n\pi}}{e^{(n+1)\pi}+e^{-(n+1)\pi}-e^{n\pi}-e^{-n\pi}}\\[7pt]&=\frac{1-e^{-2n\pi}}{e^\pi-1+e^{-(2n+1)\pi}-e^{-2n\pi}}\end{align*}$$
All the terms that include $n$ tend to $0$ so we have:
$$\sum_1^{\infty}\frac{\sinh \pi}{\cosh 2n\pi-\cosh \pi}=\frac{1}{e^{\pi}-1}$$
