How can we show the following equation


$$\sum\_{n\text{ odd}}\frac1{n\sinh(n\pi)}=\frac{\mathrm{ln}2}8\;?$$


I found it in a physics book(David J. Griffiths,'Introduction to electrodynamics',in Chapter 3,problem 3.48), which provided no proof.



Evaluating an infinite sum related to $\sinh$
Here is a proof, surely not the simplest. Using 
$$\frac1{\sinh(n\pi)}=\frac{2e^{-\pi n}}{1-e^{-2\pi n}}=2\sum\_{m\text{ odd}}e^{-mn\pi}$$ 
and 
$$2\sum\_{n\text{ odd}}\frac{x^n}{n}=-2\ln(1-x)+\ln(1-x^2)=\ln\frac{1+x}{1-x},\qquad |x|<1,$$ 
we get that
$$\sum\_{n\text{ odd}}\frac1{n\sinh(n\pi)}=\sum\_{m\text{ odd}}\ln\frac{1+e^{-m\pi}}{1-e^{-m\pi}}=\ln\prod\_{m\text{ odd}}\frac{1+e^{-m\pi}}{1-e^{-m\pi}}.$$ This is the same as $\ln\frac{G\_1}{g\_1}$, where $G\_N$ and $g\_N$ are the class invariants as in Chapter 34 of Berndt: Ramanujan's notebooks, Part V. It is known (see same chapter) that $G\_1=1$ and $(g\_1G\_1)^8(G\_1^8-g\_1^8)=\frac{1}{4}$. Hence $g\_1=2^{-1/8}$, and the original sum equals 
$$\sum\_{n\text{ odd}}\frac1{n\sinh(n\pi)}=\ln\frac{G\_1}{g\_1}=\frac{\ln 2}{8}.$$


