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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asked Jan 24 '17 at 21:40
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1$\begingroup$ 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? $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 21:53
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$\begingroup$ Are there similar such, maybe easier, known identities ? $\endgroup$ – Rene Schipperus Jan 24 '17 at 22:14
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$\begingroup$ @ReneSchipperus, see algo.inria.fr/flajolet/Publications/FlSa98.pdf $\endgroup$ – Zaid Alyafeai Jan 24 '17 at 22:20
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4$\begingroup$ Here you go: researchgate.net/publication/… $\endgroup$ – tired Jan 25 '17 at 1:21
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$\begingroup$ @tired, thanks very nice. $\endgroup$ – Zaid Alyafeai Jan 25 '17 at 1:39
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\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.
answered Jul 21 '19 at 22:11
