How to prove this identity involving the series $\sum_{m=1}^\infty \frac{\sin m\theta}{\sinh m u}$? In Methods of Mathematical Physics by Jeffreys and Jeffreys I have found this absolutely fascinating exercise (which is from the Cambridge Mathematical Tripos for 1938):
By expressing $\sin m\theta$ and $\sinh mu$ in terms of exponentials, prove the identity
$$
\sum_{m=1}^\infty \frac{\sin m\theta}{\sinh m u}=\sum_{n=1}^\infty \frac{\sin\theta}{\cosh(2n-1)u-\cos\theta} \quad(u>0,\theta\text{ real}).
$$
(p. 55 of the 3rd edition)
As I have never encountered such series before, this one has given me quite some difficulty. I half suspect that there is some clever trick involving the geometric series that I've missed. But even after much fiddling around I can't seem to crack it. Any help would be much appreciated. 
 A: Let $S$ be the sum, then
$$S = \Im  \left(\sum _{m = 1}^{\infty }\frac{2}{{e}^{m u}} \cdot \frac{{e}^{i m {\theta}}}{1-{e}^{{-2} m u}}\right) = \Im  \left(\sum _{m = 1}^{\infty } \frac{2}{{e}^{m u}} {e}^{i m {\theta}} \sum _{n = 0}^{\infty } {e}^{{-2} n m u}\right) = 2 \Im  \left(\sum _{n = 0}^{\infty } \sum _{m = 1}^{\infty } {e}^{i m {\theta}-\left(2 n+1\right) m u}\right)$$
Hence 
$$S = 2 \Im  \left(\sum _{n = 0}^{\infty } \frac{{e}^{i {\theta}-\left(2 n+1\right) u}}{1-{e}^{i {\theta}-\left(2 n+1\right) u}}\right) = 2 \Im  \left(\sum _{n = 0}^{\infty } \frac{{e}^{i {\theta}-\left(2 n+1\right) u} \left(1-{e}^{{-i} {\theta}-\left(2 n+1\right) u}\right)}{1-2 \cos  \left({\theta}\right) {e}^{{-\left(2 n+1\right)} u}+{e}^{{-2} \left(2 n+1\right) u}}\right)$$
Hence
$$S = 2 \sum _{n = 0}^{\infty } \frac{\sin  \left({\theta}\right) {e}^{{-\left(2 n+1\right)} u}}{1-2 \cos  \left({\theta}\right) {e}^{{-\left(2 n+1\right)} u}+{e}^{{-2} \left(2 n+1\right) u}} = \sum _{n = 0}^{\infty } \frac{\sin  \left({\theta}\right)}{\cosh  \left(2 n+1\right) u-\cos  \left({\theta}\right)}$$
