Evaluation of two Euler type sums

Question

We know that the harmonic number sum (also called Euler type sum) enter link description here $$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^2}{2^n}}}} = {\rm{L}}{{\rm{i}}_4}\left( {\frac{1}{2}} \right)\; + \frac{1}{{16}}\zeta (4) + \frac{1}{4}\zeta (3)\log 2 - \frac{1}{4}\zeta \left( 2 \right){\log ^2}2 + \frac{1}{{24}}{\log ^4}2,$$ How to calculate the closed form of the following Euler type Sums $$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^3}{2^n}}}} ,\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^4}{2^n}}}}.$$ Here the harmonic numbers are defined by $$H^{(k)}_n:=\sum\limits_{j=1}^n\frac {1}{j^k}\quad {\rm and}\quad H^{(k)}_0:=0.$$

Answer 1

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$

divide both sides by $x$ then integrate from $x=0$ to $1/2$ and use the fact that $\int_0^{1/2}x^{n-1}=\frac1{n2^n}$ we have

$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx=\sum_{n=1}^\infty \frac{1}{n2^n}\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$

rearrange to get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=3\operatorname{Li}_5\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\frac12\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx$$

Substitute the first sum

\begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align}

along with the result from Song's solution

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\ dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$

we get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52$$