3

I want to find the closed forms of the following Euler type sums $$\sum\limits_{n=1}^\infty \frac{H_n}{n^p\binom{2n}{n}}4^n$$ and $$\sum\limits_{n=1}^\infty \frac{H_{2n}}{n^p\binom{2n}{n}}4^n.$$ A similar result please see A Challenging Euler Sum $\sum\limits_{n=1}^\infty \frac{H_n}{\tbinom{2n}{n}}$. By Mathematica,
the following equations seems to be true $$\sum\limits_{n=1}^\infty \frac{H_n}{n^2\binom{2n}{n}}4^n=\pi^2\log(2)+\frac{7}{2}\zeta(3)$$ and $$\sum\limits_{n=1}^\infty \frac{H_{2n}}{n^2\binom{2n}{n}}4^n=\frac{1}{2}\pi^2\log(2)+\frac{35}{4}\zeta(3).$$ How to find their closed forms for general $p=0,1,2,3,\ldots$?

Ali Shadhar
  • 25,498
xuce1234
  • 537
  • 2
  • 8
  • I corrected a typo. Both involve $\zeta(3)$ – Tito Piezas III Jun 10 '19 at 16:54
  • 1
    According to Mathematica, we have

    $$\sum _{n=1}^{\infty } \frac{H_n }{n^2 \binom{2 n}{n}}z^n=-\text{Li}_3\left(\frac{2}{-z+\sqrt{z-4} \sqrt{z}+2}\right)-\text{Li}_3\left(-\frac{2}{z+\sqrt{z-4} \sqrt{z}-2}\right)+2 i \pi \csc ^{-1}\left(\frac{2}{\sqrt{z}}\right)^2,$$

    so with $z=4$ you get your first identity. Mathematica can't seem to find anything for $n^p$ in the denominator with $p=3, 4, 5, \ldots$. But overall you may need to focus your attention on the Polylogarithm. Perhaps try to derive the above result without Mathematica and see if you can generalise it for $n^p$...

     – pshmath0 Aug 11 '19 at 20:21
  • Hi Pixel, do you know how to prove the above formula? More general, can you find the generating function $$\sum_{n=1}^\infty \frac{H_n}{n^3\binom{2n}{n}}z^n?$$I want to find the closed form $$\sum_{n=1}^\infty \frac{H_{n-1}}{n^3\binom{2n}{n}}(-1)^n.$$ – xuce1234 Mar 24 '21 at 23:23

0 Answers0