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Generating functions of the sequences $\binom{2n}{n}^2H_n$ and $\binom{2n}{n}^2H_{2n}$, where $H_n$ is $n$-th harmonic number, are known in terms of elliptic integrals $$ \sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1} $$ $$ \sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_{2n}}{16^n}k^{2n}=\frac12K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k}{4(1-k^2)}.\tag{2} $$ Clausen's formula written in the form $$ \left[{}_2F_1\left({2a,2b\atop a+b+\tfrac12};x\right)\right]^2={}_3F_2\left({2a,~2b,~a+b\atop a+b+\tfrac12,2a+2b};4x(1-x)\right)\tag{3} $$ allows one to find the generating function of the sequence $\binom{2n}{n}^3(H_{2n}-H_n)$ by differentiating $(3)$ with respect to $a$ at $a=b=\tfrac14$: $$ \sum_{n=1}^\infty \binom{2n}{n}^3\frac{H_{2n}-H_n}{64^n}(4x(1-x))^n=\frac{2}{3\pi}K(\sqrt{x})\sum_{n=1}^\infty \binom{2n}{n}^2\frac{4H_{2n}-3H_n}{16^n}x^n.\tag{4} $$ Thus according to $(1)$ and $(2)$ this gf is known in terms of elliptic integrals.

Q: What can be said about the generating functions $\displaystyle f(x)=\sum_{n=1}^\infty \binom{2n}{n}^3\frac{H_n}{64^n}x^n$ and $\displaystyle g(x)=\sum_{n=1}^\infty \binom{2n}{n}^3\frac{H_{2n}}{64^n}x^n$ of the sequences $\binom{2n}{n}^3H_n$ and $\binom{2n}{n}^3H_{2n}$ separately?

In view of $(4)$ one can find closed form of $f(1)-g(1)$ and Whipple's sum http://dlmf.nist.gov/16.4.E7 (after differentiating it wrt $c$ at $a=c=\tfrac12,~d=1$) gives closed form of $2g(1)-3f(1)$. This means that at least the closed forms of $f(1)$ and $g(1)$ can be found.

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  • They both depend on integrals involving $K^2$ (the squared complete elliptic integral of the first kind) - see example $4.15$ at page $32$ here. – Jack D'Aurizio Jan 20 '18 at 17:53
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    Nice question. I found some trouble in trying to write down the Fourier-Legendre series expansion of $K^2$ in a explicit way. I fear not to be able to say something deeper than maybe, at the moment. – Jack D'Aurizio Jan 20 '18 at 18:08
  • How about using a recursive form for $\binom{2n}{n}^3H_n$? Like $$a_n=\binom{2n}{n}^3H_n=\binom{2n}{n}^3H_{n-1}+\frac{1}{n}\binom{2n}{n}^3=\ 8\left(\frac{2n-1}{n}\right)^3\binom{2n-2}{n-1}^3H_{n-1}+\frac{(n+1)^3}{n}C_n^3=\ 8\left(\frac{2n-1}{n}\right)^3a_{n-1}+\frac{(n+1)^3}{n}C_n^3$$ where $C_n$ are Catalan numbers – rtybase Jan 21 '18 at 10:58
  • Nemo, since the question is "What can be said about the generating functions", should I make a multiple integral derivation such as I did for your other question? – Yuriy S Feb 07 '18 at 11:38
  • @ Nemo, I don't know, I haven't obtained the integral yet. – Yuriy S Feb 07 '18 at 15:28
  • @Nemo If you are still interested; I have been practicing working with Riordan's transform. Are you interested in the (...)^3*H closed forms from (1),(2) or are you interested in working with (4),(5)? – rrogers Sep 28 '22 at 14:12
  • Do you know of any references in the literature to Eqs (1) and (2)? If not, how best can one prove these results? – omegadot Nov 15 '23 at 13:42
  • @Nemo Thanks for the link, but are you sure that is the correct paper? I cannot find any Theorem 2 in this paper and the paper seems to have nothing to do with harmonic numbers. – omegadot Nov 20 '23 at 10:36

1 Answers1

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This is close but not totally the same. It has a strong relation. It can be proven that: $$\frac{2}{\pi}K(k)=4\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{n^3k^{2n-2}H_{n}}{2^{4n}(2n-1)^2}-\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{(n+1)k^{2n}H_{n}}{2^{4n}}\tag{1}.$$ And from $(1)$ making the change $k=k\sin{t}$ and integrating in $(0,\pi/2)$: $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{H_{n}}{64^n}\frac{k^{2n-2}(k^2(n+1)(2n-1)^{3}-8n^4)}{(1-2n)^3}\\=\frac{4K^{2}(\sqrt{1/2(1-\sqrt{1-k^2})})}{\pi^2}.\tag{2}$$

Some effor shows the Ramanujan like series involving harmonic numbers. $$\sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{q(n)H_{n}}{2^{12n}(2n-1)^3}=\frac{16}{\pi}\tag{3},$$ where $q(n)=21168n^5-18816n^4+272n^3-180n^2+17n + 5$.

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