Can anyone prove this equation? $$\sum_{n=0}^{\infty} (-1)^n \left( \frac{(2n-1)!!}{(2n)!!} \right)^3 = \left( \frac{\Gamma\left(\frac{9}{8}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{7}{8}\right)} \right)^2$$ This is a closed-form sum by Ramanujan.
https://mathworld.wolfram.com/DoubleFactorial.html
According to the above website, G.H. Hardy gives a generalization of this sum.
If anyone can help proving this, I would really be grateful.
Let $$a_n=\frac{(2 n-1)\text{!!}}{(2 n)\text{!!}}=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi } \,\,\Gamma (n+1)}$$ which explains the appearance of hypergeometric functions.
$$S_k(x)=\sum_{n=0}^\infty (-1)^n\, a_n^k\,x^n$$ produces
$$S_1(x)=\frac{1}{\sqrt{x+1}}$$ $$S_2(x)=\frac{2 K(-x)}{\pi }$$ $$S_3(x)=\frac{4 K\left(\frac{1}{2} \left(1-\sqrt{x+1}\right)\right)^2}{\pi ^2}$$ $$S_4(x)=\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1;-x\right)$$ $$S_5(x)=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1; -x\right)$$ I am sure that you see the pattern.
Using $x=1$ $$S_3(1)=\left(\frac{2 K\left(\frac{1}{2} \left(1-\sqrt{2}\right)\right)}{\pi } \right)^2=\left( \frac{\Gamma\left(\frac{9}{8}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{7}{8}\right)} \right)^2$$
