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How would one simplify $$\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n)}$$ into a form not containing gamma expressions (for example, into something with $n$, $\sqrt{\pi}$, and so on, but not $\Gamma$)? My intuition is that the desired simplification involves a clever application of the recursion property $\Gamma(n) = (n-1)\Gamma(n-1)$ but I am not seeing how to do this.

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$$\frac{\Gamma\left(n+\tfrac{1}{2}\right)}{\Gamma(n)} = \frac{\left(n-\tfrac{1}{2}\right)\left(n-\tfrac{3}{2}\right)\cdots\tfrac{1}{2}\sqrt{\pi}}{(n-1)!}=\frac{(2n-1)!!\sqrt{\pi}}{2^n (n-1)!}=n\sqrt{\pi}\,\frac{(2n-1)!!}{(2n)!!}$$ leads to $$\frac{\Gamma\left(n+\tfrac{1}{2}\right)}{\Gamma(n)} =\color{red}{\binom{2n}{n}\frac{n\sqrt{\pi}}{4^n}}.$$

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$$\Gamma(z)\Gamma(z+\frac{1}{2}) = 2^{1-2z}\sqrt{\pi}\Gamma(2z) $$

This equation will help you get rid of $\Gamma({\frac{1}{2}+z})$ symbol. This equation is called duplication formula.

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  • $\begingroup$ So? How do you get rid of the dependency on $\Gamma$? $\endgroup$ – Jack D'Aurizio Nov 4 '17 at 18:18
  • $\begingroup$ You can get $$\Gamma(\frac{1}{2}+n) = \frac{2^{1-2n}\sqrt{\pi}\Gamma(2n)}{\Gamma(n)}$$, then $$\frac{\Gamma(\frac{1}{2}+n)}{\Gamma(n)}$$ can be resolved. Sorry, I expressed wrong meaning. I will edit it. $\endgroup$ – hello_god Nov 5 '17 at 1:20

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