$\sum_{n=0}^m \binom{m}{n}\Gamma\Big(n-\frac{1}{2}\Big)\Gamma\Big(m-n-\frac{1}{2}\Big)=0$

Question

How to prove that for all $m\in\mathbb{N}$, $m\geq2$:

$$\sum_{n=0}^m \binom{m}{n}\Gamma\Big(n-\frac{1}{2}\Big)\Gamma\Big(m-n-\frac{1}{2}\Big)=0$$

Any hint?

Answer 1

The following identity is pretty simple to prove (for instance, through Euler's Beta function):

$$ \sum_{h\geq 0}\frac{\Gamma\left(h-\tfrac{1}{2}\right)}{\Gamma(h+1)} z^h =-2\sqrt{\pi}\sqrt{1-z}$$ and for any $m\in\mathbb{N}$ your (convolution) sum is just $m!$ times the coefficient of $z^m$ in the Taylor series at the origin of $\left(-2\sqrt{\pi}\sqrt{1-z}\right)^2$. If $m\geq 2$, that is zero.