Sorry if this comes out of the blue. Looking at old notes of mine, I found the identity $$\dfrac{\Gamma(1/4)^4}{\pi^3}=4+\sum_{n\ge0}\binom{2n+1}{n}^3\dfrac{1}{2^{6n+1}}\;.$$ I cannot remember how I proved it, but I am pretty sure it was the specialization of a more general series evaluating to products and quotients of $\Gamma(z)$ for certain parameters. Does somebody have a clue for a proof ? More generally, are there special value evaluations of series such as $\sum_{n\ge0}\binom{2n+1}{n}^3z^n$ or $\sum_{n\ge0}\binom{2n}{n}^3z^n$ ?
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3$\begingroup$ I believe it follow from $$\sum\limits_{n=0}^{\infty} \frac{\binom{2 n+1}{n}^3}{2^{6 n+1}} x^n=-\frac{4 \left(\pi ^2-4\, K\left(\frac{1}{2} \left(1-\sqrt{1-x}\right)\right)^2\right)}{\pi ^2 x}$$ where $K$ is the elliptic K function. $\endgroup$– Steven ClarkCommented Sep 27 at 22:19
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3$\begingroup$ One also has $$\sum\limits_{n=0}^{\infty} \binom{2 n+1}{n}^3 x^n=-\frac{\pi^2-4\, K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2}{8 \pi^2 x}$$ and $$\sum\limits_{n=0}^{\infty} \binom{2 n}{n}^3 x^n=\frac{4\, K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2}{\pi^2}.$$ $\endgroup$– Steven ClarkCommented Sep 27 at 22:25
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7$\begingroup$ The right side is equal to $4\,{}_3F_2(1/2,1/2,1/2;1,1\mid 1)$, which can be evaluated by Dixon's theorem. $\endgroup$– Ira GesselCommented Sep 28 at 0:36
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1$\begingroup$ @Steven Clark: thanks: can you put this as an answer with a reference so that I can accept it ? $\endgroup$– Henri CohenCommented Sep 28 at 7:01
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The formula: $$ \sum_{n \geq 0} \frac{\binom{2n}{n}^3}{2^{6n}} = \frac{\pi}{\Gamma^4(\frac34)} $$ was given by Ramanujan (see Ramanujan Notebook II, Example 18, p.24). Henri Cohen's formula is related to it in the way shown by Steven Clark.
We can give simpler proofs of Ramanujan's and Henri Cohen's formulas noticing that they are Gosperable. Thus obtaining the partial sums from 0 to k (use for example "Wolfram Alpha" or "Maple"), and then the limit as k to infinity we obtain the values of the sums. Indeed, for Cohen's formula we have the following proof: $$ \sum_{n=0}^k \frac{\binom{2n+1}{n}^3}{2^{6n+1}}=-2^{-6k-7}\binom{2k+3}{k+1}^3 {_4}F_3(2,k+\frac52,k+\frac52,k+\frac52;k+3,k+3,k+3;1) \\ -4+ \frac{4\pi}{\Gamma(\frac34)^4}. $$ Taking the limit as $k \to \infty$, we obtain $$ \sum_{n=0}^{\infty} \frac{\binom{2n+1}{n}^3}{2^{6n+1}}=-4+\frac{4 \pi}{\Gamma(\frac34)^4}=-4+\frac{\Gamma(\frac14)^4}{\pi^3}, $$ and a similar proof for Ramanujan's formula.
See my question at mathoverflow with title "Gosperable fórmulas".
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$\begingroup$ Steven Clark, Ira Gessel, and Jesus Guillera's answers are excellent, how can I accept all of them? Thanks a lot all of you. $\endgroup$ Commented Sep 29 at 20:26