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Any idea how to solve the following Euler sum

$$\sum_{n=1}^\infty \left( \frac{H_n}{n+1}\right)^3 = -\frac{33}{16}\zeta(6)+2\zeta(3)^2$$

I think It can be solved it using contour integration but I am interested in solutions using real methods.

  • 1
    Summation by parts driven by $$\begin{eqnarray*}H_{n+1}^{3}-H_n^3 &=& (H_{n+1}-H_n)\left[(H_{n+1}-H_n)^2+3 H_{n} H_{n+1}\right]\\&=&\frac{1}{(n+1)^3}+\frac{3 H_n^2}{(n+1)}+\frac{3 H_n}{(n+1)^2}\end{eqnarray*} $$ maybe? – 2017-01-24
  • 0
    Are there similar such, maybe easier, known identities ? – 2017-01-24
  • 0
    @ReneSchipperus, see http://algo.inria.fr/flajolet/Publications/FlSa98.pdf – 2017-01-24
  • 4
    Here you go: https://www.researchgate.net/publication/311703645_Evaluation_of_a_cubic_Euler_sum – 2017-01-25
  • 0
    @tired, thanks very nice. – 2017-01-25

0 Answers 0