A Ramanujan infinite series

$$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$

I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.

But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$

This is not an answer but just a result obtained using a CAS.

Let $$f_k=\sum_{n=0}^\infty \binom{2 n}{n}^kx^n$$ The following expressions have been obtained $$f_1=\frac{1}{\sqrt{1-4 x}}$$ $$f_2=\frac{2 }{\pi }K(16 x)$$ $$f_3=\frac{4 }{\pi ^2}K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2$$ $$f_4=\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1;256 x\right)$$ $$f_5=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;1 024 x\right)$$ where appear, for $k=2,3$, the complete elliptic integrals of the first kind and, for $k>3$, the generalized hypergeometric functions

This is one of the simplest and famous series given by Ramanujan and it's value is $2/\pi$. Unfortunately Ramanujan's technique requires a reasonable amount of effort to understand. I have presented the proof for this series and it's friend $$\frac{4}{\pi}= 1+\frac{7}{4}\left(\frac{1}{2}\right)^{3}+\frac{13}{4^{2}}\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}+\dots$$ in this post.