Another sum took the form

$$\sum_{n=1}^{\infty}\frac{\sinh\pi}{\cosh(2n\pi)-\cosh\pi}$$

For some reason, I couldn’t get the residue theorem to help me here. I took a different apporach instead.

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.

$$\begin{align} \sum_{k=1}^{\infty} e^{-k t} \sin{k x} &= \Im{\sum_{k=1}^{\infty} e^{-k (t-i x)}} \\ &= \Im{\left [ \frac{1}{1-e^{-(t-i x)}} \right ]} \\ &= \Im{\left [ \frac{1}{1-e^{-t} \cos{x} – i e^{-t} \sin{x}} \right ]}\\ &= \frac{e^{-t} \sin{x}}{(1-e^{-t} \cos{x})^2 + e^{-2 t} \sin^2{x}}\\ &= \frac{\sin{x}}{e^{t}-2 \cos{x} + e^{-t}} \\ \therefore \sum_{k=1}^{\infty} e^{-k t} \sin{k x} &= \frac{1}{2} \frac{\sin{x}}{\cosh{t}-\cos{x}}\end{align}$$

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