How to prove that this series sums to an elliptic integral? According to Mathematica,
$$\frac{\pi}{2}\sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}x^n = \frac{1}{\sqrt{1+8\sqrt{x}}}K\left(\frac{16\sqrt{x}}{1+8\sqrt{x}}\right)$$
and similarly,
$$\frac{\pi}{2} \sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}(-1)^n x^n = \frac{1}{\left(1+64x\right)^{1/4}}K\left(\frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\right)$$
where $K$ is the complete elliptic integral of the first kind. How can one prove these two expressions?
A relevant formula seems to be the "Landen transformation" listed in the NIST website. However it's unclear to me how this can be used to obtain the above.
 A: Define the function
$$ f(x):=\sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}x^n =
{}_2F_1\Big(\frac14,\frac34;1;64x\Big). \tag{1}$$
This is the generating function of OEIS sequence A000897. We want to prove that
$$ \frac\pi2 f(x^2) = \frac1{\sqrt{1+8x}}K\Big(\frac{16x}{1+8x}\Big). \tag{2} $$
Note that we can use the identity
$$ K(x) = \frac\pi2{}_2F_1\Big(\frac12,\frac12;1;x\Big), \tag{3} $$
equation $(1)$, and omitting factors of $\,\frac\pi2\,$
to rewrite equation $(2)$ as
$$ \,{}_2F_1\Big(\frac14,\frac34;1;64x^2\Big) =\frac1{\sqrt{1+8x}}
{}_2F_1\Big(\frac12,\frac12;1;\frac{16x}{1+8x}\Big). \tag{4} $$
From Berndt, Bhargava, and Garvan "Ramanujan's Theories of Elliptic Functions to Alternative Bases" (PDF link), pertaining to Ramanujan's
theory of signature 4 there is Theorem 9.1,
$${}_2F_1\Big(\frac12,\frac12;1;\frac{2x}{1+x}\Big)=\sqrt{1+x}\,{}_2F_1\Big(\frac14,\frac34;1;x^2\Big). \tag{5}$$
This theorem with $\,x\,$ replaced by $\,8x\,$ is easily shown to
be equivalent to equation $(4)$, and this proves equation $(2)$.
We also want to prove
$$ \frac\pi2 f(-x) = \frac{1}{\sqrt[4]{1+64x}}K\Big(\frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\Big). \tag{6}$$
Using equation $(1)$,
equation $(3)$, and omitting factors of $\,\frac\pi2\,$
rewrite equation $(6)$ as
$$ {}_2F_1\Big(\frac14,\frac34;1;-64x\Big) =
\frac{1}{\sqrt[4]{1+64x}}
{}_2F_1\Big(\frac12,\frac12;1;
\frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\Big).
\tag{7}$$
Replace $\,64x\,$ with $\,x\,$ to get
$$ {}_2F_1\Big(\frac14,\frac34;1;-x\Big) =
\frac{1}{\sqrt[4]{1+x}}
{}_2F_1\Big(\frac12,\frac12;1;
\frac{-1+\sqrt{1+x}}{2\sqrt{1+x}}\Big).
\tag{8}$$
I don't know yet how to prove this.
