Who's up to the challenge? 54

we know $\displaystyle \sum_{r=1}^{\infty} x^{r-1} = \frac{1}{1-x}$ , for $-1<x<1$

Using the above and the definition of Polylogarithm Function we have the following series:-

$\displaystyle -2\sum_{r=1}^{\infty}\sum_{k=1}^{\infty}\int_{0}^{1}\frac{x^{k+r-1}\ln^{3}(x)}{k^{4}}dx$

Now substituting $x=e^{-t}$ we have :-

$\displaystyle -2\sum_{r=1}^{\infty}\sum_{k=1}^{\infty}\int_{\infty}^{0}\frac{e^{-t(k+r-1)}(-t)^{3}(-e^{-t})}{k^{4}}dt$

$\displaystyle =2\sum_{r=1}^{\infty}\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{e^{-t(k+r)}t^{3}}{k^{4}}dt$

Now substituting $t(k+r) = z$ we have:-

$\displaystyle 2\sum_{r=1}^{\infty}\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{e^{-z}z^{3}}{(r+k)^{4}k^{4}}dz$

$\displaystyle = 12\sum_{r=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k^{4}(r+k)^{4}}$

$\displaystyle 12\sum_{k=1}^{\infty}\frac{\left(\zeta(4)-H_{k}^{(4)}\right)}{k^{4}} = 12\left(\zeta(4)^{2} - \sum_{k=1}^{\infty}\frac{H_{k}^{(4)}}{k^{4}}\right)$

We know $\displaystyle \sum_{k=1}^{\infty}\frac{H_{k}^{(s)}}{k^{s}} = \frac{1}{2}\left(\zeta(s)^{2} +\zeta(2s)\right)$ where $s\geq 2$ is a positive integer.

Hence using the above we arrive at:-

$\displaystyle 12\cdot \frac{1}{2}\left(\zeta(4)^{2}-\zeta(8)\right)$

Now $\displaystyle \zeta(4) =\frac{\pi^{4}}{90}$ and $\displaystyle \zeta(8) = \frac{\pi^{8}}{9450}$ .

Using them we have our answer as $\displaystyle 6\cdot\frac{\pi^{8}}{56700} = \frac{\pi^{8}}{9450} = \zeta(8)$

Here $H_{n}^{(r)}$ denotes the nth Hyperharmonic number of order r .