Series involving ${}_{2}F_{1}$
Today we show the proof of this nice series posted by @infseriesbot: \[ \sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \frac{\sqrt{ \pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}\]
Proof
First recall that the Pochhammer polyonomial or rising factorial can be expressed as a quiotinent of two gamma functions: \[ (x)_{n} = x(x+1)(x+2)\cdots(x+n-1) = \frac{\Gamma(n+x)}{\Gamma(x)}\] Hence \[(3n)! = \frac{\Gamma(3n+1)}{\Gamma(1)} = (1)_{3n}\] Now we will use the triplication formula for Pochhamer polynomials: \[ (x)_{3n} = 3^{3n}\left(\frac{x}{3}\right)_{n}\left(\frac{x+1}{3}\right)_{n}\left(\frac{x+2}{3}\right)_{n}\] Is easy to show the proof of this: By definition \[(x)_{3n} = x(x+1)(x+2)\cdots(x+3n-1) \] If we factorize in this form: \[ (x)_{3n} = \underbrace{\left[x(x+3)(x+6)...(x+3n-3)\right]}_{n-\textrm{terms}}\underbrace{\left[(x+1)(x+4)(x+7)...(x+3n-2)\right]}_{n-\textrm{terms}}\underbrace{\left[(x+2)(x+5)(x+8)...(x+3n-1)\right]}_{n-\textrm{terms}}\] Consider \[ A = x(x+3)(x+6)\cdots(x+3n-3) = 3^n\left(\frac{x}{3}\right)\left(\frac{x}{3}+1\right)\dots \left(\frac{x}{3}+n-1\right) = 3^n\left(\frac{x}{3}\right)_{n}\] \[ B = (x+1)(x+4)(x+7)\cdots(x+3n-2) = 3^n\left(\frac{x+1}{3}\right)\left(\frac{x+1}{3}+1\right)\dots \left(\frac{x+1}{3}+n-1\right) = 3^n\left(\frac{x+1}{3}\right)_{n}\] \[ C = (x+2)(x+5)(x+8)\cdots(x+3n-1)= 3^n\left(\frac{x+2}{3}\right)\left(\frac{x+2}{3}+1\right)\dots \left(\frac{x+2}{3}+n-1\right) = 3^n\left(\frac{x+2}{3}\right)_{n}\] Hence \[ (x)_{3n} = ABC = 3^{3n}\left(\frac{x}{3}\right)_{n}\left(\frac{x+1}{3}\right)_{n}\left(\frac{x+2}{3}\right)_{n}\] Therefore \[(3n)! = (1)_{3n} = 3^{3n}\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}\left(1\right)_{n}\] Now, using the fact that $(1)_{n}=n!$ \[ \sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \sum_{n=0}^{\infty} \frac{3^{3n}\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}\left(1\right)_{n}}{54^n (1)_{n} (1)_{n} n!}= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{3}\right)_{n}\left(\frac{2}{3}\right)_{n}}{ (1)_{n}}\left(\frac{1}{2^n}\right)\frac{1}{n!} = {}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};1;\frac{1}{2}\right)\] Now using the following identity for ${}_{2}F_{1}$: \[{}_{2}F_{1} \left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2};\tfrac{1}{2}\right)=\sqrt{% \pi}\frac{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2}\right)}{\Gamma% \left(\tfrac{1}{2}a+\tfrac{1}{2}\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{1}{2}% \right)}.\] we have \[\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} ={}_{2}F_{1}\left(\frac{1}{3},\frac{2}{3};1;\frac{1}{2}\right) = \frac{\sqrt{\pi}{\Gamma(1)}}{\Gamma\left(\frac{1}{6}+\frac{1}{2}\right)\left(\frac{2}{6}+\frac{1}{2}\right)} = \frac{\sqrt{\pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}\] \[ \boxed{\sum_{n=0}^{\infty} \frac{(3n)!}{54^n n!^3} = \frac{\sqrt{\pi}}{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{5}{6}\right)}}\]
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