# A cubic nonlinear Euler sum

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.

• 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? – Jack D'Aurizio Jan 24 '17 at 21:53
• Are there similar such, maybe easier, known identities ? – Rene Schipperus Jan 24 '17 at 22:14
• @ReneSchipperus, see algo.inria.fr/flajolet/Publications/FlSa98.pdf – Zaid Alyafeai Jan 24 '17 at 22:20
• Here you go: researchgate.net/publication/… – tired Jan 25 '17 at 1:21
• @tired, thanks very nice. – Zaid Alyafeai Jan 25 '17 at 1:39

\begin{align} S&=\sum_{n=1}^\infty\left(\frac{H_n}{n+1}\right)^3=\sum_{n=1}^\infty\left(\frac{H_{n-1}}{n}\right)^3\\ &=\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_n^2}{n^4}+3\sum_{n=1}^\infty \frac{H_n}{n^5}-\sum_{n=1}^\infty\frac{1}{n^6} \end{align}
Substituting the following results: $$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$$ $$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
We get $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$