# 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
• @user178256 that relation that relates the two sums is amazing. I have another relation that related these two sums which means we can solve for them by elimination. my relation can be obtained very easily by the way. did you reply on many sums/ integrals to find this relation? – Ali Shather Jul 24 at 10:18

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

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