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.
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Sign up to join this communityAny 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.
\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 $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$
Note that the first and second sum are proved here and here respectively. As for the third sum, can be obtained using the Euler identity.