Evaluate $\sum_{n\geq 0} \frac{1}{64^n(2n+1)}\left[\frac{{4n\choose 2n}}{{2n\choose n}}\right]$ Since we have encountered several classes of infinite series containing  central binomial coefficients which inspired to me introduce following class of series.
$$\sum_{n=0}^{\infty}\frac{1}{64^n(2n+1)}\left[\frac{{4n\choose 2n}}{2n\choose n}\right]$$
My final closed form for this series is
$$\frac{8}{\sqrt{3}}F\left(\frac{\pi}{4},i\sqrt{\frac{2}{3}}\right)+2\sqrt{5}E\left(\frac{\pi}{4},\sqrt{\frac{2}{5}}\right)-2\sqrt{3}E\left(\frac{\pi}{4},i\sqrt{\frac{2}{3}}\right)-\frac{8}{\sqrt{5}}F\left(\frac{\pi}{4},\sqrt{\frac{2}{5}}\right)$$
where $F(\alpha, u)$, $E(\alpha^{\prime},u^{\prime})$ and $i=\sqrt{-1}$ denotes incomplete  elliptical integrals of the first kind , second kind and imaginary number respectively.
By Wolfram alpha check here which is further equal to $${}_3F_2\left(\frac{1}{4},\frac{3}{4},1;\frac{1}{2},\frac{3}{2};\frac{1}{16}\right)$$
The closed form in terms of incomplete Elliptical integrals and hypergeometric expression are found to be equal by Wolfram alpha  check here and here respectively.
I have no idea if the hypergeometric expression can be deduced to some closed form with elementary  functions and the closed form I obtained.

Not an answer but a lead: Using DLMF 16.5.2 we may write
$$
{_3F}_2\left({1,\frac{1}{4},\frac{3}{4}\atop\frac{3}{2},\frac{1}{2}};\frac{1}{16}\right)=\frac{1}{2}\int_0^1\frac{1}{\sqrt{1-t}}{_2F}_1\left({\frac{1}{4},\frac{3}{4}\atop\frac{1}{2}};\frac{t}{16}\right)\,\mathrm dt.
$$
The reduction formula here then allows us to deduce the integral representation
$$
{_3F}_2\left({1,\frac{1}{4},\frac{3}{4}\atop\frac{3}{2},\frac{1}{2}};\frac{1}{16}\right)=\frac{1}{4}\int_0^1\frac{1}{\sqrt{1-t}}\left(\frac{1}{\sqrt{1+\sqrt{t/16}}}+\frac{1}{\sqrt{1-\sqrt{t/16}}}\right)\,\mathrm dt,
$$
which according to mathematica can be evaluated in terms of elliptic integrals.
