Exact value of a sum involving harmonic numbers

Question

Could somebody tell me the exact value of this series? $$ \sum_{k=1}^{\infty} (-1)^k\frac{H_k^{(5)}}{k} $$ where $$ H_k^{(n)}=\sum_{i=1}^{k}\frac{1}{i^n} $$

Thanks!

Answer 1

Hint. You may write $$ \sum_{k=1}^{\infty} (-1)^k\frac{H_k^{(5)}}{k}=\sum_{k=1}^{\infty} (-1)^k\frac{H_{k-1}^{(5)}}{k}+\sum_{k=1}^{\infty} \frac{(-1)^k}{k^6}=\zeta(\bar{1},5)-\frac{31 \pi ^6}{30240}. $$ I am not sure the Multi Zeta Values $\zeta(\bar{1},5)$ has a closed form in terms of known constants.

Answer 2

Denote ${\bf H}^{(q)}_n(t) := \sum\limits_{m=1}^\infty H_m^{(q)}/m^n \cdot t^m$. By using the formula on the bottom of the answer to Calculating alternating Euler sums of odd powers we have: \begin{equation} {\bf H}^{(5)}_1(t) = Li_6(t) + Li_1(t) Li_5(t) - Li_2(t) Li_4(t) + \frac{1}{2} [Li_3(t)]^2 \end{equation} Since $Li_n(-1) = (1-2^{n-1})/2^{n-1} \zeta(n)$ the result is expressed through zeta functions only.