How to evaluate $\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$ How to evaluate $$\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$$, where n is integer > 0?
I know the gamma function formula will give
$$ \frac{(\frac{n-2}{2})!}{(\frac{n-3}{2})!}$$ How to simplify it?

This is an elaboration of the hint by Raymond Manzoni.
One of $n/2$ and $(n-1)/2$ is an integer, and the other is a half integer, and while there is a nice expression of the gamma function on integers as a factorial, evaluating the gamma function on half integers is more complicated.  However, they can be evaluated using the following facts:


*

*$\Gamma(1/2)=\sqrt{\pi}$

*$\Gamma(x)=(x-1)\Gamma(x-1)$


We then have $\Gamma(n+1/2)=(n-1/2)\Gamma(n-1/2)=(n-1/2)(n-3/2)\Gamma(n-3/2)=\cdots$
Continuing the pattern, we can write
$$\Gamma(n+1/2)=\Gamma(1/2)\left(\frac{1}{2}\frac{3}{2}\cdots \frac{2n-1}{2} \right)=\frac{\sqrt{\pi}(2n-1)!!}{2^n}.$$
Here, we use the double factorial notation $n!!=n(n-2)(n-4)\cdots (1\text{ or } 2)$  However, we can simplify $(2n-1)!!$ and write it in terms of normal factorials and powers of $2$. Namely, we have 
$$(2n-1)!!=1 \cdot 3 \cdots (2n-1)=\frac{(2n-1)!}{2\cdot 4 \cdots 2(n-1)}=\frac{(2n-1)!}{2^{n-1}(n-1)!}.$$
