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
    $\begingroup$ 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
  • $\begingroup$ Are there similar such, maybe easier, known identities ? $\endgroup$ – Rene Schipperus Jan 24 '17 at 22:14
  • $\begingroup$ @ReneSchipperus, see algo.inria.fr/flajolet/Publications/FlSa98.pdf $\endgroup$ – Zaid Alyafeai Jan 24 '17 at 22:20
  • 4
    $\begingroup$ Here you go: researchgate.net/publication/… $\endgroup$ – tired Jan 25 '17 at 1:21
  • $\begingroup$ @tired, thanks very nice. $\endgroup$ – 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 $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.