Show $\sum_{n=1}^{\infty}\frac{\sinh\pi}{\cosh(2n\pi)-\cosh\pi}=\frac1{\text{e}^{\pi}-1}$ and another Show that :
$$\sum_{n=1}^{\infty}\frac{\cosh(2nx)}{\cosh(4nx)-\cosh(2x)}=\frac1{4\sinh^2(x)}$$
$$\sum_{n=1}^{\infty}\frac{\sinh\pi}{\cosh(2n\pi)-\cosh\pi}=\frac1{\text{e}^{\pi}-1}$$
 A: OK, I have figured out the second sum using a completely different method.  I begin with the following result (+):
$$\sum_{k=1}^{\infty} e^{-k t} \sin{k x} = \frac{1}{2} \frac{\sin{x}}{\cosh{t}-\cos{x}}$$
I will prove this result below; it is a simple geometrical sum.  In any case, let $x=i \pi$ and $t=2 n \pi$; then
$$\begin{align}\frac{\sinh{\pi}}{\cosh{2 n \pi}-\cosh{\pi}} &= 2 \sum_{k=1}^{\infty} e^{-2 n \pi k} \sinh{k \pi}\end{align}$$
Now we can sum:
$$\begin{align}\sum_{n=1}^{\infty} \frac{\sinh{\pi}}{\cosh{2 n \pi}-\cosh{\pi}} &= 2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} e^{-2 n \pi k} \sinh{k \pi}\\ &= 2 \sum_{k=1}^{\infty} \sinh{k \pi} \sum_{n=1}^{\infty}e^{-2 n \pi k}\\ &= 2 \sum_{k=1}^{\infty} \frac{\sinh{k \pi}}{e^{2 \pi k}-1} \\ &= \sum_{k=1}^{\infty} \frac{e^{\pi k} - e^{-\pi k}}{e^{2 \pi k}-1} \\ &= \sum_{k=1}^{\infty} e^{-\pi k} \\ \therefore \sum_{n=1}^{\infty} \frac{\sinh{\pi}}{\cosh{2 n \pi}-\cosh{\pi}}  &= \frac{1}{e^{\pi}-1} \end{align}$$
To prove (+), write as the imaginary part of a geometrical sum.
$$\begin{align} \sum_{k=1}^{\infty} e^{-k t} \sin{k x} &= \Im{\sum_{k=1}^{\infty} e^{-k (t-i x)}} \\ &= \Im{\left [ \frac{1}{1-e^{-(t-i x)}} \right ]} \\ &= \Im{\left [ \frac{1}{1-e^{-t} \cos{x} - i e^{-t} \sin{x}} \right ]}\\ &= \frac{e^{-t} \sin{x}}{(1-e^{-t} \cos{x})^2 + e^{-2 t} \sin^2{x}}\\ &= \frac{\sin{x}}{e^{t}-2 \cos{x} + e^{-t}} \\ \therefore \sum_{k=1}^{\infty} e^{-k t} \sin{k x} &= \frac{1}{2} \frac{\sin{x}}{\cosh{t}-\cos{x}}\end{align}$$
QED
A: For the first sum, it turns out that the residue theorem works, although I am still not sure why it does not work for the second sum.  In any case, consider
$$\sum_{n=-\infty}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} - \cosh{2 x}} = 2 \sum_{n=1}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} - \cosh{2 x}} - \frac{1}{\cosh{2 x} - 1}$$
By the residue theorem, the sum on the LHS is zero.  The reason is that 
$$\mathrm{Res}_{z=1/2+i k} \frac{\pi \cot{\pi z} \cosh{2 x z}}{\cosh{4 x z}-\cosh{2 x}} = 0 \quad \forall k \in \mathbb{Z}$$
$$\therefore \sum_{n=1}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} - \cosh{2 x}} = \frac{1}{2}\frac{1}{\cosh{2 x} - 1} = \frac{1}{4 \sinh^2{x}} $$
A: 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:
$$\sum_1^{\infty}\frac{\sinh \pi}{\cosh 2n\pi-\cosh \pi}=\frac{1}{e^{\pi}-1}$$
