# Evaluation of two Euler type sums

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 $$H^{(k)}_n:=\sum\limits_{j=1}^n\frac {1}{j^k}\quad {\rm and}\quad H^{(k)}_0:=0.$$

• In the following post you can find a closed form for $\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{3}2^{n}}$. In the same post you may find the closed form of $\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{3}}x^{n}$ so maybe integrating the formula it is possible to find a closed form for $\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n^{4}2^{n}}.$ This is the link math.stackexchange.com/questions/909228/… – Marco Cantarini Apr 5 '17 at 12:16
• $$\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 2 \right)}}}{{{n^3}{2^n}}}}+\sum\limits_{n = 1}^\infty {\frac{{H_n^{\left( 3 \right)}}}{{{n^2}{2^n}}}}=\frac{9}{16}\zeta_2\zeta_3 -\frac{9}{16}\zeta_3\ln^2 2 +\frac{1}{6}\zeta _2\ln^3 2 -\frac{29}{32}\zeta_5+\frac{1}{4}\zeta_4\ln2-\frac{1}{60}\ln^5 2+2\operatorname{Li}_5\left(\frac{1}{2}\right)$$ Relationship obtained after long calculations – user178256 Apr 5 '17 at 18:48