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How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$

The formula $$\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}$$ (in older notation) appears as eq. 38 in Ramanujan's paper Modular equations ...
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Determine whether $\frac{K\left(\sqrt{1-x^2}\right)^2}{K(x)^2}$ is positive rational (given $x$)

Denote the complete elliptic integral of the first kind by $$K(x)=\int_0^{\pi /2}\frac{d\varphi}{\sqrt{1-x^2\sin^2\varphi}}$$ and $$f(x)=\frac{K\left(\sqrt{1-x^2}\right)^2}{K(x)^2}$$ Question: Given a ...
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