Expressing $\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k\,2^k}$ as a finite sum involving $\zeta(\cdot)$, $Li_k(\cdot)$, $\pi$, and $\ln 2$

Question

While working on the integral posted here, through a large amount of skulduggery, I managed to arrive at the following intriguing sum

$$\begin{align}\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k 2^k} &= 2\operatorname{Li}_5 \left (\frac{1}{2} \right ) + \ln 2 \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{7}{16} \zeta (3) \ln^2 2 \\ &- \frac{\pi^2}{36} \ln^3 2 + \frac{\pi^4}{90} \ln 2 - \frac{7}{96} \pi^2 \zeta (3) - \frac{27}{32} \zeta (5) + \frac{1}{30} \ln^5 2 \end{align}$$

Of course I do not know how to arrive at this result in a direct manner coming from it at the sum on the left.

So my question is, does anyone have any suggestions on how one would approach the evaluation of the sum on the left?

Answer 1

We have $$S=\sum_{n\geq1}\frac{1}{n^{4}}\sum_{k=1}^{n}\frac{1}{k2^{k}}=\sum_{n\geq1}\frac{1}{n^{4}}\int_{0}^{1}\frac{1-x^{n}2^{-n}}{2-x}dx$$ $$=\int_{0}^{1}\frac{1}{2-x}\left(\zeta\left(4\right)-\mathrm{Li}_{4}\left(\frac{x}{2}\right)\right)dx.$$

Now, since we can split the integral, we obtain $$ S=\zeta\left(4\right)\log\left(2\right)-\int_{0}^{1}\frac{\mathrm{Li}_{4}\left(\frac{x}{2}\right)}{2-x}dx$$ $$=\zeta\left(4\right)\log\left(2\right)-\sum_{k\geq0}\int_{0}^{1/2}x^{k}\mathrm{Li}_{4}\left(x\right)dx$$ $$=\zeta\left(4\right)\log\left(2\right)-\frac{1}{2}\sum_{k\geq0}\frac{1}{2^{k}}\sum_{m\geq1}\frac{1}{m^{4}\left(k+m+1\right)2^{m}}\tag{1}$$ and now we can expand the inner series in $(1)$ via partial fractions decomposition and evaluate every single term.