# Show that $\Gamma(n+\frac{1}{3})\cdot \Gamma(n+\frac{2}{3}) = a\left(\frac{(3n)!}{3^{3n}\cdot n!}\right)$

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.

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