# How do I evaluate this sum :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n}$?

I have tried evaluating this series

$$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n}$$

using some methods but it's seems to me that it is very hard. However, I noticed that the series converges faster than the Riemann series.

My question here is:

Is there some mathematical technique for evaluating the above series?

Note1: Here, $H_n$ denotes the harmonic numbers.

Edit : I have a wrong type I meant in the denomenator $2^n$

Thank you for any help.

• it evaluates to 0.38360530295885199... --- which is not recognized as having a closed form representation by the Inverse Symbolic Calculator – Carlo Beenakker Oct 13 '16 at 19:58
• sorry , I have a wrong typo I meant :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n}$ not $3^n$ in the denominator and i edited the question – Youssra El Yossra Youssra Oct 13 '16 at 20:08
• with the $2^n$ in the exponent it's 0.96630012055722... do you really need more accuracy? – Carlo Beenakker Oct 13 '16 at 20:22
• What is "the Riemann series"? – Gerry Myerson Oct 13 '16 at 21:46
• i meant sum(1/n^k) , for k greater then 1 – Youssra El Yossra Youssra Oct 13 '16 at 21:47

By multiplying out the factor $H_n^3$, it is not too hard to see that your sum can be written as a rational linear combinations of special values of weight $4$ multiple polylogarithms. The Maple package HyperInt (by Erik Panzer) can perform simplifications with multiple polylogarithms. According to this software, $$\sum_{n=1}^\infty \frac{H_n^3}{(n+1)2^n}=\frac{\log(2)^4}{12}-\frac{\pi^4}{144}+\frac{\log(2)^2\pi^2}{6}+\log(2)\zeta(3).$$ The algorithm the software uses is described in Panzer's paper "Algorithms for the symbolic integration of hyperlogarithms with applications to Feynman integrals" (there is also a version of the paper on arXiv).
Let $g(a)=\frac{a}{1-a}, f(a)=\log(1-a)$ and $h(a)=\log^2(1-a)$. Let $\square^m$ denote a $m$-dimensional unit hypercube.
The following is an application or reformulation of the above infinite series evaluation. $$\int_0^1\int_0^1\frac{h(\frac12xz)-xh(\frac12z)+xzh(\frac12)-zh(\frac12x)}{x(1-x)z(1-z)}dxdz= \frac{\log^42}{12}-\frac{\pi^4}{144}+\zeta(2)\log^22+\zeta(3)\log2.$$ Proof outline: Since $\frac1k=\int_0^1x^{k-1}dx$, we may write \begin{align} \sum_{n\geq1}\frac{t^nH_n^3}{n+1}&=\sum_{n\geq1}t^n\int_0^1w^ndw\sum_{i,j,k=1}^n\int_{\square^3}x^{i-1}y^{j-1}z^{k-1}dxdydz \\ &=\int_{\square^4}dxdydzdw\sum_{n\geq1}(tw)^n\sum_{i,j,k=1}^nx^{i-1}y^{j-1}z^{k-1}\\ &=\int_{\square^4}\frac{dxdydzdw}{(1-x)(1-y)(1-z)}\sum_{n\geq1}(tw)^n(1-x^n)(1-y^n)(1-z^n)\\ &=\int_{\square^4}dxdydzdq\frac{g(tw)-g(twy)-g(twx)+g(twxy)-g(twz)+g(twyz)+g(twxz)-g(twxyz)}{(1-x)(1-y)(1-z)}. \end{align} Next, make a repeated use of the integral evaluations $$\int_0^1\frac{g(tw)-g(twy)}{1-y}dy=-\frac{f(tw)}{1-tw}$$ and $$\int_0^1\left(\frac{f(twx)}{1-twx}-\frac{f(tw)}{1-tw}\right)\frac{dw}{1-w}= \frac{xh(t)-h(tx)}{2tx}$$ to arrival at the double integral upon replacing $t=\frac12$. The rest follows from the series evaluation algorithm that Julian Rosen alluded to. $\square$
Remark. This approach actually could prove (directly) the easier-looking $$\sum_{n\geq1}\frac{H_n^2}{(n+1)2^n}=\frac13\log^32+\zeta(2)\log2-\frac12\zeta(3).$$