Generalized hypergeometric functions IV

Another nice involving ${}_{3}F_{2}$

We show the proof of this nice series posted by @PDemichel : \[ \sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}\]

It turned out that we proved it earlier using different formulas to calculate ${}_{3}F_{2}$.

Proof: In the past we have used the duplication or triplication formula for Pochhamer polyonomials or rising factorials. It turned out that this formulas are particular cases for the general formula (Proof in the Appendix): \[ (x)_{nm} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n} \] With this formula, we can expand $(4n)!$ in the following way: \[ (4n)! = \frac{\Gamma(4n+1)}{\Gamma(1)} = (1)_{4n} = 4^{4n}\left(\frac{1}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(1\right)_{n}\] Additionally, we know that $(1)_{n} = n!$. Hence, \[ \sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \sum_{n=0}^{\infty} \frac{4^{4n}\left(\frac{1}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(1\right)_{n}}{256^n (1)_{n}(1)_{n}(1)_{n}n!} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}\left(\frac{1}{2}\right)_{n}}{ (1)_{n}(1)_{n}n!}={}_{3}F_{2} \left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;1\right)\] Now recall the Watson’s Sum formula for ${}_{3}F_{2}$: \[{{}_{3}F_{2}}\left({a,b,c\atop\frac{1}{2}(a+b+1),2c};1\right)=\frac{\Gamma% \left(\frac{1}{2}\right)\Gamma\left(c+\frac{1}{2}\right)\Gamma\left(\frac{1}{2% }(a+b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a-b)\right)}{\Gamma\left(\frac{1}{2% }(a+1)\right)\Gamma\left(\frac{1}{2}(b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a)% \right)\Gamma\left(c+\frac{1}{2}(1-b)\right)},\] Therefore \[ \sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} ={}_{3}F_{2} \left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;1\right) = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(1)\Gamma(1)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\left(\frac{1}{4}+\frac{4}{4}\right)\right)\Gamma\left(\frac{1}{2}\left(\frac{3}{4}+\frac{4}{4}\right)\right)\Gamma\left(\frac{1}{2}+\frac{1}{2}\left(\frac{3}{4}\right)\right)\Gamma\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{4}\right)\right)} \] Hence \[\sum_{n=0}^{\infty} \frac{(4n)!}{256^n n!^4} = \frac{\pi}{\Gamma^2\left(\frac{5}{8}\right)\Gamma^2\left(\frac{7}{8}\right)}\]

Appendix

We show the proof of this result for Pochhammer polynomials:
\[ (x)_{nm} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n} \] Consider: \[(x)_{nm} = x(x+1)(x+2)\ldots (x+nm-1)\] Factorize $(x)_{nm}$ in the following way: \[(x)_{nm} = A_{1}\cdot A_{2} \cdot A_{3}\ldots A_{n} \] where \[\begin{matrix} A_{1} =& x& (x+1)&(x+2)&\ldots &(x+m-1)\\ A_{2} =& (x+m)& (x+m+1) &(x+m+2) & \ldots & (x+2m-1)\\ A_{3} =& (x+2m)&(x+2m+1)&(x+2m+2)& \ldots &(x+3m-1)\\ \vdots & \vdots & \vdots &\vdots & \ldots & \vdots\\ A_{n} =&(x+(n-1)m) &(x+(n-1)m+1)& (x+(n-1)m+2) & \ldots & (x+nm-1) \end{matrix}\]
Now factor m out of each factor in each $A_{j}$
\[ \begin{matrix} A_{1} =& m^m & \frac{x}{m}& \left(\frac{x+1}{m}\right)& \left(\frac{x+2}{m}\right)&\ldots &\left(\frac{x+m-1}{m}\right)\\ A_{2} =& m^m & \left(\frac{x}{m}+1\right)&\left(\frac{x+1}{m}+1\right)&\left(\frac{x+2}{m}+1\right)& \ldots& \left(\frac{x+m-2}{m}\right)\\ A_{3} =& m^m & \left(\frac{x}{m}+2\right)&\left(\frac{x+1}{m}+2\right)&\left(\frac{x+2}{m}+2\right)&\ldots &\left(\frac{x+3m-1}{m}\right)\\ \vdots &\vdots & \vdots& \vdots &\vdots &\ldots &\vdots \\ A_{n} =& \underbrace{m^m}_{B_{0}} & \underbrace{\left(\frac{x}{m}+n-1\right)}_{B_{1}} & \underbrace{\left(\frac{x+1}{m}+n-1\right)}_{B_{2}} &\underbrace{\left(\frac{x+2}{m}+n-1\right)}_{B_{3}} &\ldots & \underbrace{\left(\frac{x+mn-1}{m}\right)}_{B_{m}} \end{matrix}\] Now it touned out that column $B_{0}=m^{nm}$ and \[B_{j} = \left(\frac{x+j-1}{m} \right)_{n} \] Therefore \[ \boxed{(x)_{nm} = B_{0}\cdot B_{1} \cdot B_{2} \ldots B_{m} = m^{nm} \prod_{j=1}^{m}\left(\frac{x+j-1}{m} \right)_{n} }\]