Getting the RHS into summation form Some time ago, I wrote down this identity$$\frac 4\pi=1+\left(\frac 12\right)^2\frac 1{1!\times2}+\left(\frac 12\times\frac 32\right)^2\frac 1{2!\times2\times3}+\ldots$$And being the idiot I was, I didn't write down the RHS into a compact sum.

Question: How do you write the RHS with a summation?$$1+\left(\frac 12\right)^2\frac 1{1!\times2}+\left(\frac 12\times\frac 32\right)^2\frac 1{2!\times2\times3}+\ldots=\sum\limits_{k=0}^{\infty}\text{something}$$

Obviously, there is a $k!$ in there, but that's as much as I know. The sum also includes pochhammer symbols$$(a)_n=a(a+1)\cdots(a+n-1)$$ because the RHS is a hypergeometric function.

If the general term has the form
$$ a_k=\left(\frac{(2k-1)!!}{2^k}\right)^2\frac{1}{k!(k+1)!} = \left(\frac{(2k)!}{4^k k!}\right)^2\frac{1}{k!(k+1)!}$$
then
$$ \sum_{k\geq 0}a_k = \sum_{k=0}\frac{1}{4^k}\binom{2k}{k}\frac{1}{4^k(k+1)}\binom{2k}{k}=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{2 d\theta}{\sqrt{1-e^{i\theta}}\left(1+\sqrt{1-e^{-i\theta}}\right)} $$
equals $\frac{8}{2\pi}=\color{red}{\frac{4}{\pi}}$ by the residue theorem or by the explicit computation of a primitive. This proves
$$ \phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};2;1\right)=\frac{4}{\pi}$$
that is related with a complete elliptic integral of the first kind by
$$ \phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};2;1\right)=\frac{2}{\pi}\int_{0}^{1}K(\sqrt{k})\,dk $$
previously solved here.
