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.

A straight forward use of hypergeometric functions is as follows:
Since it has been established that
$$a_{n} = \left(\frac{(2n)!}{2^{2n} \, n!}\right)^2\frac{1}{n!(n+1)!}$$
then
\begin{align}
a_{n} &= \left(\frac{(2n)!}{2^{2n} \, n!}\right)^2\frac{1}{n!(n+1)!} \\
&= \frac{1}{\Gamma(2) \, n! \, (2)_{n}} \, \left( \frac{\Gamma(1+2n)}{2^{2n} \, \Gamma(n+1)}\right)^{2} = \frac{1}{\Gamma(2) \, n! \, (2)_{n}} \, \left( \frac{(1)_{2n}}{2^{2n} \, (1)_{n}}\right)^{2} \\
&= \frac{1}{ n! \, (2)_{n}} \, \left( \frac{2^{2n} \, (1)_{n} \, \left(\frac{1}{2}\right)_{n}}{2^{2n} \, (1)_{n}}\right)^{2} = \frac{\left(\frac{1}{2}\right)_{n} \left(\frac{1}{2}\right)_{n}}{n! \, (2)_{n}}.
\end{align}
Using 
$${}_{2}F_{1}(a, b; c; 1) = \frac{\Gamma(c) \, \Gamma(c-a-b)}{\Gamma(c-a) \, \Gamma(c-b)}$$
then
\begin{align}
\sum_{n=0}^{\infty} a_{n} &= \sum_{n=0}^{\infty} \left(\frac{(2n)!}{2^{2n} \, n!}\right)^2\frac{1}{n!(n+1)!} = {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 2; 1\right) = \frac{\Gamma(2) \, \Gamma(1)}{\Gamma^{2}\left(\frac{3}{2}\right)} = \frac{4}{\pi}.
\end{align}
