The value of the sum $\sum_{n=1}^\infty \frac{1}{n\sinh(\pi n /4)}$ Can we compute the exact value of the sum
$$S = \sum_{n=1}^\infty \dfrac{1}{n\sinh(\pi n /4)}.$$
WolframAlpha spits out $S = 1.4667$. But I have no clue how it obtains this and I suspect this may be an approximation? It may be useful to know that
$$
\dfrac{1}{n\sinh(\pi n /4)} = \dfrac{|\Gamma(in/4)|^2}{4\pi},
$$
where $\Gamma$ is the gamma function.

It is a special value of modular forms.
For $\Re(x) > 0$
$$f(x)=\sum_{n\ge 1} \frac1{n (e^{ nx}-e^{-nx})} = \sum_{n\ge 1}\frac1n\sum_{k\ge 0} e^{-(2k+1)nx} = -\sum_{k\ge 0} \log(1-e^{-(2k+1)x})$$
For $\Im(z) > 0$ we consider the modular discriminant
$$\Delta(z) = (2\pi)^{12} e^{2i\pi z} \prod_{k\ge 1} (1-e^{2i\pi kz})^{24}$$
$$\log \frac{\Delta(z)}{\Delta(2z)} = -2i\pi z + 24\sum_{k\ge 0}\log(1-e^{2i\pi (2k+1)z})= -2i\pi z- 24 f(-2i\pi z)$$
So that $$S=2f(\pi/4) = \frac{-1}{12}\left(\log \frac{\Delta(i/8)}{\Delta(i/4)}-\pi/4\right)$$
Using $\Delta(z)= z^{-12}\Delta(-1/z)$ (weight 12 cusp form) we get $$\begin{align}S&=
\frac{-1}{12}\left(\log \frac{2^{12}\Delta(8i)}{\Delta(4i)}-\pi/4\right)
\\&=\frac{-1}{12}\left(24\log \frac{\eta(8i)}{\eta(4i)}+12 \log 2-\pi/4\right)\end{align}$$
$\frac{\eta(8i)}{\eta(4i)}$ (the Dedekind eta function) has a $\color{red}{\text{radical closed-form}}$ that you can find here What is the exact value of $\eta(6i)$?
