Evaluate $\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$ $n,h \in \mathbb{N}$ Suppose $n,h \in \mathbb{N}$. How can I evaluate $$\frac{\Gamma(n/2)}{\Gamma(n/2+h)},$$
where $\Gamma$ is the Euler gamma function.
All I know is that $$\Gamma(n+1) = n!$$

If you consider the case where $n$ is large (and much larger than $h$), take logarithms and use Stirling approximation. This would give
$$\log \left(\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n}{2}+h\right)}\right)=\log
   \left(2^h\right)-h \log ({n})+\frac{h-h^2}{n}+\frac{2 h^3-3 h^2+h}{3
   n^2}+O\left(\frac{1}{n^{3}}\right)$$
For example, using $n=100$ and $h=10$, the exact value would be $\approx 4.38613\times 10^{-18}$ while the truncated expression would lead to  $\approx 4.40747\times 10^{-18}$.
Edit
Continuing from the previous expansion and Taylor series, usind $A=e^{\log(A)}$, we can avoid exponentiation and get (as a slightly worse approximation)
$$\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma
   \left(\frac{n}{2}+h\right)}=\left(\frac{2}{n}\right)^h\left(1-\frac{(h-1) h}{n}+\frac{(h-1) h (h+1) (3 h-2)}{6
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$
