Showing that $\sum\limits_{n \text{ odd}}\frac{1}{n\sinh\pi n}=\frac{\ln 2}{8}$

$$-\frac{8\varepsilon_0V_0}{\pi}\sum_{n \text{ odd}}\frac{1}{n\sinh(n\pi)}=\boxed{\displaystyle-\frac{\varepsilon_0V_0}{\pi}\ln 2.}$$
I have not found a way to sum this series analytically. Mathematica gives the numerical value $0.0866434$, which agrees precisely with $\ln2/8.$

Can someone do this series please?
it is from the book on electrodynamics by Griffiths.
Its in the solution manual 3.48.

Solutions Collecting From Web of "Showing that $\sum\limits_{n \text{ odd}}\frac{1}{n\sinh\pi n}=\frac{\ln 2}{8}$"

Lemma. For $x>0$, we have
$$\sum_{n\text{ odd}}\frac{1}{n\sinh nx}=-\frac12\,\ln\frac{\vartheta_4\left(0|e^{-x}\right)}{\vartheta_3\left(0|e^{-x}\right)}.\tag{1}$$
Proof. Let us write
$$\frac{1}{\sinh nx}=\frac{2}{e^{nx}}\sum_{k=0}^{\infty}e^{-2knx}=
2\sum_{k=0}^{\infty}e^{-(2k+1)nx}.$$
Substituting this into the left side of (1) and using that $\displaystyle\sum_{n\text{ odd}}\frac{q^n}{n}=-\frac12\ln\frac{1-q}{1+q}$, we get
$$\sum_{n\text{ odd}}\frac{1}{n\sinh nx}=-\ln\prod_{k=0}^{\infty}\frac{1-e^{-(2k+1)x}}{1+e^{-(2k+1)x}}=-\frac12\,\ln\frac{\vartheta_4\left(0|e^{-x}\right)}{\vartheta_3\left(0|e^{-x}\right)},$$
where at the last step we have used product representations of the Jacobi theta functions (see e.g. (92), (93) here). $\blacksquare$

The formula
$$\sum_{n\text{ odd}}\frac{1}{n\sinh \pi n}=\frac{\ln2}{8}$$
then immediately follows from the special values
\begin{align*}
\vartheta_3\left(0|e^{-\pi}\right)=\frac{\pi^{\frac14}}{\Gamma\left(\frac34\right)},\qquad
\vartheta_4\left(0|e^{-\pi}\right)=\frac{\pi^{\frac14}}{2^{\frac14}\Gamma\left(\frac34\right)}.
\end{align*}