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. 



How do I evaluate this sum :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n} $?
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](https://bitbucket.org/PanzerErik/hyperint/wiki/Home) (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](http://www.sciencedirect.com/science/article/pii/S0010465514003658)" (there is also [a version of the paper on arXiv](https://arxiv.org/abs/1403.3385)).


