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

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    $\begingroup$ Dude, please more info, show work etc. PS: I didn't downvote, just letting you know what's wrong with this post. Nice identities thouh! +1 $\endgroup$ – Arjang Feb 21 '13 at 5:08
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    $\begingroup$ have you tried anything? $\endgroup$ – Dominic Michaelis Feb 21 '13 at 5:08
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    $\begingroup$ To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Also, many would consider your post rude because it is a command ("Show that..."), not a request for help, so please consider rewriting it. $\endgroup$ – Zev Chonoles Feb 21 '13 at 5:09
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    $\begingroup$ Also, one question per post please. $\endgroup$ – Arjang Feb 21 '13 at 5:10

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.

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



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

$$\mathrm{Res}_{z=1/2+i k} \frac{\pi \cot{\pi z} \cosh{2 x z}}{\cosh{4 x z}-\cosh{2 x}} = 0 \quad \forall k \in \mathbb{Z}$$

$$\therefore \sum_{n=1}^{\infty} \frac{\cosh{2 n x}}{\cosh{4 n x} - \cosh{2 x}} = \frac{1}{2}\frac{1}{\cosh{2 x} - 1} = \frac{1}{4 \sinh^2{x}} $$


Alternatively, one can prove the following inductively:

$$\sum_1^n\frac{\sinh \pi}{\cosh 2k\pi-\cosh \pi}=\frac{\sinh n\pi}{\cosh (n+1)\pi-\cosh n\pi}$$


$$\begin{align*}\frac{\sinh n\pi}{\cosh (n+1)\pi-\cosh n\pi}&=\frac{e^{n\pi}-e^{-n\pi}}{e^{(n+1)\pi}+e^{-(n+1)\pi}-e^{n\pi}-e^{-n\pi}}\\[7pt]&=\frac{1-e^{-2n\pi}}{e^\pi-1+e^{-(2n+1)\pi}-e^{-2n\pi}}\end{align*}$$

All the terms that include $n$ tend to $0$ so we have:

$$\sum_1^{\infty}\frac{\sinh \pi}{\cosh 2n\pi-\cosh \pi}=\frac{1}{e^{\pi}-1}$$


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