Getting the RHS into summation form

Question

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.

Answer 1

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.

Answer 2

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}

Answer 3

The formula for the $k$th term of this sum looks like it should be $$\bigg(\frac{(2k-3)!!}{2^{k-1}}\bigg)^2 \frac{1}{k!(k-1)!}$$ However, the first term, $1$, does not seem to follow this pattern, so you can write the RHS as $$1+\sum_{k=1}^\infty \bigg(\frac{(2k-3)!!}{2^{k-1}}\bigg)^2 \frac{1}{k!(k-1)!}$$ Does this look right to you?

I'm not very familiar with the pochhammer symbol, but by your definition of it, it seems unnecessary, since $$(a)_n=a(a+1)...(a+n-1)=\frac{(a-n+1)!}{(a-1)!}$$ and so everything in the sum can just be expressed using factorials.