TITLE: Infinite series with harmonic numbers related to elliptic integrals
QUESTION [10 upvotes]: It is known that the following functions are elliptic integrals
$$
\, _2F_1\left({a,1-a\atop 1};x\right),\quad a=\tfrac12,\tfrac13,\tfrac14,\tfrac16,\tag{1}
$$
$$
\, _2F_1\left({\tfrac{1}{3},\tfrac{2}{3}\atop \tfrac{4}{3}};x\right),\, _2F_1\left({\tfrac{1}{2},b\atop b+1};x\right),\quad b=\tfrac14,\tfrac16.\tag{2}
$$
It was proved in the arxiv preprint by Martin Nicholson that the generating function
$$
\sum_{n=1}^\infty \frac{(a)_n(1-a)_n}{(n!)^2}H_nx^n,\quad a=\tfrac12,\tfrac13,\tfrac14,\tfrac16\tag{3}
$$
has closed form in terms of elliptic integrals $(1)$ [here $H_n$ is $n$-th Harmonic number]. The formulas for $a=\tfrac12$ can be found in this question. However no similar series with harmonic numbers are known for $(2)$. Now consider the series of Kummer's and Pfaff's transformations of hypergeometric function
$$
\, _2F_1\left({\tfrac{1}{2},b\atop b+1};4 x (1-x)\right)=\, _2F_1\left({1,2b\atop b+1};x\right)=\frac{1}{1-x}\, _2F_1\left({1,1-b\atop b+1};\frac{x}{1-x}\right).
$$
We see that the series
$$
\, _2F_1\left({1,\tfrac34\atop \tfrac54};x\right)=\sum_{n=0}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}x^n
$$
is related to elliptic integral $(2)$ with $b=\tfrac14$. It follows from Theorem $4$ of the preprint mentioned above that there is the following strange evaluation of a sum with harmonic numbers
\begin{equation}
\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{\tfrac14-(-1)^n}{2n+1}H_n=\frac{\Gamma^4(\tfrac14)}{64\pi}\ln 2.\tag{4}
\end{equation}
We see that two series with different 'arguments' $x=1$ and $x=-1$ conspire to have a closed from. However attempts to find closed form for each sum
$$
\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1},~\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{(-1)^nH_n}{2n+1}\tag{5}
$$
separately, including the method outlined in the preprint, was not successful so far.

Q1: Do the series $(5)$ have closed form?


Q2: Does the series
$$
\sum_{n=1}^\infty\frac{(\tfrac34)_n}{(\tfrac54)_n}\frac{H_n}{2n+1}\left(2(-x)^n-\frac{(4x)^n}{(1+x)^{2n+1}}\right)\tag{6}
$$
have closed form for all $~|x|\le 1$?

Of course $(6)$ is not a generating function, but at least the sum with $x=1$ in $(6)$ can be calculated according to $(4)$.

REPLY [3 votes]: The Clausen's product formula is (see [1] pg.188)
$$
\left({}_2F_1\left(a,b;a+b+1/2;z\right)\right)^2={}_3F_2(2a,a+b,2b;a+b+1/2,2a+2b;z)\tag 1
$$
Set in $(1)$ $a=1/2$, $b=1/4$, then
$$
{}_3F_2\left(1,\frac{1}{2},\frac{3}{4};\frac{5}{4},\frac{3}{2};z\right)=\left({}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};z\right)\right)^2\tag 2
$$
Now from [2] pg.87 Entry 30, we have the next theorem of Ramanujan:
Theorem (Ramanujan).
Let $\alpha+\beta+1=\gamma+\delta$, $c=\Gamma(\alpha)\Gamma(\delta)/\{\Gamma(\gamma)\Gamma(\delta)\}$, and
$$
y=c\frac{{}_2F_1\left(\alpha,\beta;\delta;1-x\right)}{{}_2F_1\left(\alpha,\beta;\gamma;x\right)}.
$$
Then
$$
y'=-\frac{x^{-\gamma}(1-x)^{-\delta}}{{}_2F_1^2(\alpha,\beta;\gamma;x)}.
$$
Using the theorem of Ramanujan with $\alpha=1/4$, $\beta=1/2$, $\delta=5/4$, $\gamma=1/2$, we get
$$
c=\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}
$$
and because ${}_2F_1\left(\alpha,\beta;\gamma;x\right)=\frac{1}{\sqrt[4]{1-x}}$, we get
$$
y'=-\frac{1}{(1-x)^{3/4}\sqrt{x}}.
$$
Also
$$
y=\frac{(1-x)^{1/4}\Gamma\left(\frac{1}{4}\right){}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};1-x\right)}{\Gamma\left(\frac{5}{4}\right)}=-\int^{x}_{c_0}\frac{1}{(1-t)^{3/4}\sqrt{t}}dt
$$
Hence
$$
{}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};1-x\right)=-\frac{\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}(1-x)^{-1/4}\int^{x}_{c_0}\frac{1}{(1-t)^{3/4}\sqrt{t}}dt.
$$
Hence
$$
{}_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};x\right)=\frac{1}{4x^{1/4}}\int^{x}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\tag 3
$$
Hence
$$
{}_3F_2\left(1,\frac{1}{2},\frac{3}{4};\frac{5}{4},\frac{3}{2};x\right)=\left(\frac{1}{4x^{1/4}}\int^{x}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2\tag 4
$$
Setting this into $S(p)$ in the answer of Yuriy S, we get
$$
S(p)=
$$
$$
\int^{1}_{0}\left(\frac{1}{1-x}\left(\frac{1}{4p^{1/4}}\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2-\frac{1}{1-x}\left(\frac{1}{4(xp)^{1/4}}\int^{xp}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2\right)dx=
$$
$$
=\frac{1}{16\sqrt{p}}\int^{1}_{0}\frac{1}{1-x}\left[\left(\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-t}}\right)^2-\left(\int^{p}_{0}\frac{dt}{t^{3/4}\sqrt{1-tx}}\right)^2\right]dx\tag 5
$$
Hence
$$
\frac{d}{dp}(\sqrt{p}S(p))=\int^{1}_{0}\frac{1}{1-x}\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p\right) \sqrt{1-p x}-\sqrt{1-p}\cdot {}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p x\right)}{2 \sqrt{(1-p) p} \sqrt{1-p x}}dx=
$$
$$
=\frac{1}{2\sqrt{p}}\int^{1}_{0}\frac{1}{1-x}\left(\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p\right)}{ \sqrt{1-p}}-\frac{{}_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};p x\right)}{\sqrt{1-p x}}\right)dx.\tag 6
$$
References
[1]: J.M. Borwein and P.B. Borwein. ''Pi and the AGM''. 1987 ed., New York: John Wiley and Sons Inc., (1987).
[2]: B.C. Berndt. ''Ramanujan`s Notebooks Part II''. Springer-Verlag, New York. (1989).