How to prove this result?
$$ \frac{\Gamma(1-n-a) \Gamma(1-\frac{n}{2})}{\Gamma(1-n) \Gamma(1 - \frac{n}{2} -a)} = 0\quad \mbox{if}\ n\ \mbox{is}\ odd $$.
My thinking is following:
We know that the poles for gamma function are $0$ and negative integers. $(1 - \frac{n}{2})$ and $(1 - \frac{n}{2} -a)$ cannot be negetive integers, so $\Gamma(1- \frac{n}{2})$ and $\Gamma(1 - \frac{n}{2} -a)$ are analytic. So we dont have to worry about them. They dont have poles.
Now for rest of terms, both $(1-n-a)$ and $(1-n)$ are either zero or negative integers, so they are poles of gamma.
So how can I prove that this quatity is zero?