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This question originates from this recent question of Paramanand Singh about a series computed by Ramanujan, probably related to elliptic integrals and Legendre functions.

Is there a closed form for $$ {}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1,1;1\right)=\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4=\frac{8}{\pi^3}\int_{0}^{\frac{1}{2}}\frac{K(m)^2}{\sqrt{m(1-m)}}\,dm $$ ?

Many proofs of $\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^3=\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}$ are well-known, for instance through Clausen formula or Fourier-Legendre series expansions (pages 27-28 here). Such methods do not seem to apply smoothly for computing a closed form for the RHS, neither Parseval's identity applied to $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 e^{ni\theta} = \frac{2}{\pi}\,K(e^{i\theta})$$ where $e^{i\theta}$ is regarded as the elliptic modulus. Suggestions are welcome.

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  • $\begingroup$ Of course, the given ${}_4 F_3(1)$ can be written in terms of $$\iint_{(0,\pi/2)^2}\iint_{(0,\pi/2)^2}\frac{dx\,dy\,dz\,dw}{1-\sin^2(x)\sin^2(y)\sin^2(z)\sin^2(w)}.$$ $\endgroup$ – Jack D'Aurizio Nov 5 '17 at 18:49
  • $\begingroup$ Slight variation that may or may not be helpful $$ _4F_3(\cdot)=\frac{2}{\pi^3}\sum_{a=0}^\infty\sum_{b=0}^\infty \frac{\Gamma(\frac{1}{2}+a)^2\Gamma(\frac{1}{2}+b)^2}{a!^2b!^2}B\left(\frac{1}{2},\frac{1}{2}+a+b,\frac{1}{2}\right) $$ $\endgroup$ – Benedict W. J. Irwin Nov 7 '17 at 10:18
  • $\begingroup$ The residues seem to converge pretty quickly, this sum has terms with different signs $$ _4F_3(\cdot)= - \sum_{k=0}^\infty \mathrm{Res}\left(\frac{\psi(-z)}{4^{4z}}\binom{2z}{z}^4,\frac{2k+1}{2}\right) $$ this one seems to have positive terms $$ _4F_3(\cdot)= - \sum_{k=0}^\infty \mathrm{Res}\left(\frac{\gamma+\psi(-z)}{4^{4z}}\binom{2z}{z}^4,\frac{2k+1}{2}\right) $$ but evaluating these gives a fairly complicated mixture. $\endgroup$ – Benedict W. J. Irwin Nov 14 '17 at 12:06
  • $\begingroup$ In the first case the first term gives $$ _4F_3(\cdot) \approx \frac{48 \zeta (3) (\gamma +\log (512))-\pi ^4+512 \log ^3(2) (\gamma +\log (4))+32 \pi ^2 \log (2) (\log (2)-\gamma )}{6 \pi ^4} \approx 1.12244 $$ in the second case which seems more natural $$ _4F_3(\cdot) \approx \frac{432 \zeta (3) \log (2)-\pi ^4+1024 \log ^4(2)+32 \pi ^2 \log ^2(2)}{6 \pi ^4} \approx 1.11326 $$ $\endgroup$ – Benedict W. J. Irwin Nov 14 '17 at 12:08
  • $\begingroup$ In case you have not noticed my question got answered by folks at mathoverflow.net. You may want to have a look at mathoverflow.net/q/285746/15540 $\endgroup$ – Paramanand Singh Feb 16 '18 at 13:26
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This is by no means a full answer to this question but I believe that this approach is fruitful and can be salvaged once I manage to learn more about elliptic functions. We start from the first integral representation given by Jack D'Aurizio. \begin{eqnarray} _4 F_3(.;1) &=& \frac{8}{\pi^3} \int\limits_0^{\frac{1}{2}} \frac{[K(m)]^2}{\sqrt{m(1-m)}} dm\\ &=& \frac{8}{\pi^3} \int\limits_0^{\frac{1}{2}} \frac{1}{(1-m)}\frac{[K(\frac{m}{m-1})]^2}{\sqrt{m(1-m)}} dm \\ &=& -\imath \frac{8}{\pi^3} \int\limits_0^{-1} \frac{[K(u)]^2}{\sqrt{u}} du \\ &=& -\imath \frac{16}{\pi^3} \int\limits_0^{\imath} [K(u^2)]^2 du \end{eqnarray} In the second line from above we used the first functional identity from http://functions.wolfram.com/EllipticIntegrals/EllipticK/17/01/ , in the third line we substituted $u:=m/(m-1)$ and in the last line we substituted for $\sqrt{u}$. Now the expression on the very bottom lends itself to integration by parts. Using Mathematica I found out that the following identity holds: \begin{equation} \frac{d^n}{d u^n}[ K(u^2)]^2 = \frac{P_1^{(n)}(u^2)\cdot [K(u^2)]^2+P_2^{(n)}(u^2) \cdot K(u^2) E(u^2) + P_3^{(n)}(u^2) \cdot [E(u^2)]^2 1_{n\ge 2}}{u^n \cdot (1-u^2)^n} \end{equation} for $n=1,2,3,\cdots$. Here $P^{(n)}_j(u)$ (for $j=1,2,3$) are polynomials of order $n-j+1$ in $u$. Those polynomials satisfy the following recurrence relations: \begin{eqnarray} P_1^{(n+1)}(u) &=& (n(3u-1)+2(u-1)) P_1^{(n)}(u) &+ (u-1) P_2^{(n)}(u)+&0& - 2 u (u-1) \frac{d}{d u} P_1^{(n)}(u)\\ P_2^{(n+1)}(u) &=& 2 P_1^{(n)}(u) +& n(3u-1) P_2^{(n)}(u)+& 2(u-1) P_3^{(n)}(u)+& - 2 u (u-1) \frac{d}{d u} P_2^{(n)}(u)\\ P_3^{(n+1)}(u) &=& 0+& P_2^{(n)}(u) +& (n(3u-1)-2(u-1))P_3^{(n)}(u)+& - 2 u (u-1) \frac{d}{d u} P_3^{(n)}(u)\\ \end{eqnarray} subject to $P_1^{(1)}(u)= 2 u-2$, $P_2^{(1)}(u)=2$ and $P_3^{(1)}(u)=0$.

Now, we have: \begin{eqnarray} &&_4 F_3(.;1) = -\imath \frac{16}{\pi^3}\left(\sum\limits_{n=1}^\infty (-1)^{n-1} \left. \left(\frac{u^n}{n!} \cdot \frac{d^{n-1}}{d u^{n-1}}[ K(u^2)]^2 \right)\right|_{u=0}^{u=\imath}\right) \\ &&= \frac{16}{\pi^3} \left( \right.\\ \left. K(-1)^2 \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_1^{(n-1)}(-1)+\right.\\ \left. K(-1) E(-1) \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_2^{(n-1)}(-1)+\right.\\ \left. E(-1)^2 \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_3^{(n-1)}(-1) \right) \end{eqnarray}

Here the first line comes from integrating by parts and the second line comes form the expression for the $n$th derivative above. Now, we have: \begin{eqnarray} K(-1)^2 &=& \frac{8}{\pi} \Gamma(\frac{5}{4})^4\\ K(-1)E(-1) &=& \frac{\pi}{4} + \frac{8}{\pi} \Gamma(\frac{5}{4})^4\\ E(-1)^2 &=& \frac{\pi}{2} + \frac{2 \pi^3}{\Gamma(\frac{1}{4})^4} + \frac{8}{\pi} \Gamma(\frac{5}{4})^4 \end{eqnarray}

Now, the only thing that remains is to find the sums in the expressions above. i believe that they can be evaluated in closed form using the recurrence relations for the polynomials. I will attempt to finish this work asap. Meanwhile I only show the first fifty cumulative sums of the (integration-by-parts) series in question. We have:

enter image description here

whereas the quantity in question computed numeraically to forty digits of precision from the integral representations reads $1.118636387164187068349619257525640916795$.

Update: I have checked numerically that the infinite series in question do converge. Indeed we have : \begin{equation} \frac{(-1)^{n-1}}{n!2^{n-1}} \left( P_j^{(n-1)}(-1)\right)_{j=1}^3 \simeq \frac{1}{n^{1.79}} \cdot\left( 6.5, -8, 2.5\right) \end{equation} as $n\rightarrow \infty$. However, for the time being, I don't have any theoretical explanation for that behavior nor do I have a way to computing those series in close form.

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  • $\begingroup$ (+1) Very promising approach, eager to see further developments. I will try something on my own, too, thanks for your time. $\endgroup$ – Jack D'Aurizio Feb 14 '18 at 18:41

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