Proving $\sum_{n=0}^{\infty}\frac{1}{2^{n}}\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^2=\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2$

Question

Infinite Sum :

$$S=\sum_{n=0}^{\infty}\frac{1}{2^{n}}\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^2$$

This Sum could be said to be related to Moments of Elliptic Integrals as we have : $$\int_{0}^{1}k^{2r}K'(k)dk=\frac{\pi^{2}}{4}\binom{2r}{r}^2\frac{1}{2^{4r}}$$ Where, $K(k)$ is the Complete Elliptical Integral of the First Kind and $K'(k)=K(k')$ with $k^2+k'^2=1$.
Using this one can arrive at the following Integral : $$S=\frac{16}{\pi^{2}}\int_{0}^{1}\frac{K'(k)}{2-k^{2}}dk$$

or the Equivalent using Quadratic Transformation: $$S=\frac{32}{\pi^{2}}\int_{0}^{1}K(k)\left(\frac{1+k}{k^{2}+6k+1}\right)dk$$

But I have been unable to evaluate any of these Integrals.

I suspect it might have a closed form as does its similar cousin :

$$\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^{2}=\frac{7}{2}\frac{\zeta(3)}{\pi}$$

EDIT: After Inputting the Value into Wolfram we get this :

$$S=\frac{1}{\pi\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2$$

Answer 1

Evaluate $S$ as a Cauchy product, using the generating function for the squared central binomial coefficient:

$$\begin{align*} S &= \sum_{n=0}^{\infty} \frac1{2^n} \sum_{r=0}^{n} \frac{1}{2^{4r}} \binom{2r}{r}^2 \\ &= \sum_{n=0}^{\infty} \sum_{r=0}^{n} \frac{1}{2^{n-r}} \cdot \frac1{2^{5r}} \binom{2r}r^2 \\ &= \sum_{i=0}^\infty \frac1{2^{5i}} \binom{2i}i^2 \cdot \sum_{j=0}^\infty \frac1{2^j} \\ &= \frac{16-8\sqrt2}{\pi} K\left(\frac{\sqrt2-1}{\sqrt2+1}\right) \\ &= \frac{16-8\sqrt2}{\pi} \cdot \frac{(\sqrt2+1)\Gamma\left(\frac14\right)^2}{8\sqrt{2\pi}} \\ &= \frac{\Gamma\left(\frac14\right)^2}{\pi^{3/2}} \end{align*}$$

where $K(k)=\int_0^{\pi/2} \frac{dt}{\sqrt{1-k^2\sin^2t}}$. The conversion from $K$ to $\Gamma$ in the penultimate step relies on a formula mentioned in the Borweins' Pi and the AGM (see Evaluation of $K$ in terms of $\Gamma$ at the bottom of p.296) but the exact details elude me.

Answer 2

Too long for comments.

$$A_n=\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^2=\frac {a_n}{b_n}$$

Numerators and denominators form sequences $A277233$ and $A056982$ in $OEIS$

$$B_m=\sum_{n=0}^{m}\frac{1}{2^{n}}\,\sum_{r=0}^{n}\frac{1}{2^{4r}}\binom{2r}{r}^2=\frac {c_m}{d_m}$$

Denominators form sequence $A130034$ in $OEIS$ but, for numerators $$\{1,13,505,4421,591497,4838381,156584033,1259879669,646944895433\} $$ $OEIS$ does not find anything.

Using inverse symbolic calculators, nothing found for $$2.3606811980321924520906758811169771744674332697629\cdots$$