Generalized hypergeometric functions II

Series involving ${}_{3}F_{2}$

Today we show the proof of this nice series posted by @infseriesbot \[ \sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} = \frac{\pi}{\Gamma\left(\frac{5}{8}\right)^2\Gamma\left(\frac{7}{8}\right)^2}\] The proof will rely on some properties of the Pochhammer polynomials (rising factorials) and the generalized hypergeometric functions

Proof

Consider the Pochhammer polynomial: \[(x)_{n} = x(x+1)(x+2)\cdots (x+n-1) = \prod_{j=0}^{n-1}(x+j) \quad n=1,2,3,...\] also called rising factorial.

Pochhammer polynomials can be expressed as the quotient of two gamma functions: \[ (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}\] Therefore \[ (4n)! = \Gamma(4n+1) = (1)_{4n} \] Pochhammer polynomials also obey the following duplication formula: \[ (x)_{2n} = 4^n\left(\frac{x}{2}\right)_{n}\left(\frac{1+x}{2}\right)_{n}\] Therefore \begin{align*} (4n)! = (1)_{4n} =& 4^{2n}\left(\frac{1}{2}\right)_{2n}\left(1\right)_{2n}\\ =& 4^{2n}\left[4^{n}\left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\right]\left[4^{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n}\right]\\ =& 4^{4n} \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n} \end{align*} From the definition of the Pochhammer polynomials is easy to show that $\displaystyle (1)_{n} = n!$
Therefore
\begin{align*} S = \sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} =&\sum_{n=0}^{\infty} \frac{4^{4n} \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(1\right)_{n}}{2^{8n}(1)_{n}(1)_{n}(1)_{n}}\frac{1}{n!} \\ =& \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}}{(1)_{n}(1)_{n}}\frac{1}{n!}\\ =& {}_3F_{2}\left({\frac{1}{4},\frac{3}{4},\frac{1}{2}\atop 1,1};1\right) \end{align*} where $\displaystyle {}_3F_{2}$ is the Generalized Hypergeometric function

$\displaystyle {}_3F_{2}$ satisfy the following identity: \[{{}_{3}F_{2}}\left({a,b,c\atop d,e};1\right)=\frac{\Gamma\left(e\right)\Gamma% \left(d+e-a-b-c\right)}{\Gamma\left(e-a\right)\Gamma\left(d+e-b-c\right)}{{}_{% 3}F_{2}}\left({a,d-b,d-c\atop d,d+e-b-c};1\right)\] whenever $\operatorname{Re}(d+e-a-b-c)>0$ and $\operatorname{Re}(e-a)>0$

Therefore \[S = {}_3F_{2}\left({\frac{1}{4},\frac{3}{4},\frac{1}{2}\atop 1,1};1\right) = \frac{\sqrt{\pi}}{\Gamma\left(\frac{3}{4}\right)^2}{{}_{% 3}F_{2}}\left({\frac{1}{4},\frac{1}{4},\frac{1}{2}\atop 1,\frac{3}{4}};1\right)\] ${}_{3}F_{2}$ also satisfy the Dixon’s Well-Poised Sum: \[{{}_{3}F_{2}}\left({a,b,c\atop a-b+1,a-c+1};1\right)=\frac{\Gamma\left(\frac{1% }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left(% \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1% \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)}\] Hence \[ {}_{ 3}F_{2}\left({\frac{1}{4},\frac{1}{4},\frac{1}{2}\atop 1,\frac{3}{4}};1\right) = \frac{\Gamma\left(\frac{9}{8}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{8}\right)}{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{7}{8}\right)\Gamma\left(\frac{5}{8}\right)\sqrt{\pi} } = \frac{\Gamma\left(\frac{3}{8}\right)^2}{\sqrt{2\pi}\Gamma\left(\frac{5}{8}\right)^{2}}\] Therefore \[ S = \frac{\Gamma\left(\frac{3}{8}\right)^2}{\sqrt{2}\Gamma\left(\frac{3}{4}\right)^2\Gamma\left(\frac{5}{8}\right)^{2}} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}\] Hence, we can conclude \[ \boxed{ \sum_{n=0}^{\infty} \frac{(4n)!}{2^{8n}n!^4} = \frac{\pi}{\Gamma\left(\frac{5}{8}\right)^2\Gamma\left(\frac{7}{8}\right)^2}}\]