Show that $\Gamma(n+\frac{1}{3})\cdot \Gamma(n+\frac{2}{3}) = a\left(\frac{(3n)!}{3^{3n}\cdot n!}\right)$. Furthermore, find an expression for $a$.

I just can’t seem to equate these, I've tried using the fact that $\Gamma(n) = (n - 1)!$ but it’s just not working.

  • 3
    $\begingroup$ Based on this, I assume you mean $\Gamma(n+\frac13)\Gamma(n+\frac23)=a\frac{(3n)!}{3^{3n}n!},\,a:=\frac{2\pi}{\sqrt{3}}$. $\endgroup$ – J.G. Apr 26 at 12:23

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