Series with trilogarithm

Let ${\rm Li}_3$ denote the trilogarithm function. Prove that:

$$\sum_{n=1}^{\infty} {\rm Li}_3 \left(e^{-2n \pi} \right)= \frac{7 \pi^3}{360} - \frac{\zeta(3)}{2}$$

(Seraphim Tsipelis)

Solution [by r9m]

We begin by recalling that the Fourier series of $\coth \pi z$ is:

$$\pi\coth (\pi z) = \frac{1}{z} + 2z\sum\limits_{j=1}^{\infty} \frac{1}{z^2+j^2}$$

Thus:

\begin{align*}
\sum\limits_{n=1}^{\infty} \text{Li}_3\left(e^{-2n\pi}\right) &= \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \frac{e^{-2nk\pi}}{k^3}\\
&= \sum\limits_{k=1}^{\infty} \frac{1}{k^3\left(e^{2k\pi} - 1\right)}\\&= \frac{1}{2}\sum\limits_{k=1}^{\infty} \frac{1}{k^3}\left(\coth (k\pi) - 1\right)\\
&= \frac{1}{2\pi}\sum\limits_{k=1}^{\infty} \frac{1}{k^3}\left(\frac{1}{k}+2k\sum\limits_{j=1}^{\infty} \frac{1}{k^2+j^2}\right) - \frac{1}{2}\zeta(3)\\
&= \frac{1}{2\pi}\left(\zeta(4) + 2\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty} \frac{1}{k^2(k^2+j^2)}\right) - \frac{1}{2}\zeta(3)\\
&= \frac{1}{2\pi}\zeta(4) + \frac{1}{2\pi}\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty}\left(\frac{1}{k^2(k^2+j^2)}+\frac{1}{j^2(k^2+j^2)}\right) - \frac{1}{2}\zeta(3)\\&= \frac{1}{2\pi}\zeta(4) + \frac{1}{2\pi}\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty}\frac{1}{k^2j^2} - \frac{1}{2}\zeta(3) \\
&= \frac{1}{2\pi}\zeta(4) + \frac{1}{2\pi}\zeta(2)^2 - \frac{1}{2}\zeta(3) \\
&= \frac{7\pi^3}{360} - \frac{1}{2}\zeta(3)
\end{align*}

The exercise can also be found in mathematica.gr