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
$$H^{(k)}_n:=\sum\limits_{j=1}^n\frac {1}{j^k}\quad {\rm and}\quad H^{(k)}_0:=0.$$
Monday, 19 August 2013
sequences and series - Evaluation of two Euler type sums
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