Ratio of $\Gamma(n)$ to $\Gamma(n/2)^2, n\to \infty?$ I'm trying to find the limit
$$\lim_{n\to \infty}\dfrac{\Gamma(n)}{\Gamma(\frac{n}{2})^2},$$
and also trying to find the precise decay rate for the above, i.e. the function $D(n)$ so that if $G(n):=\frac{\Gamma(n)}{\Gamma(\frac{n}{2})^2}$
$$\lim_{n\to \infty}\frac{G(n)}{D(n)}=1$$
How do we find them? My guess is that the limit will be zero, but not sure.

If you take logarithms and use Stirling approximation
$$\log \left(\frac{\Gamma (n)}{\Gamma \left(\frac{n}{2}\right)^2}\right)=n \log (2)+\frac{1}{2} \log \left(\frac{n}{8 \pi }\right)-\frac{1}{4 n}+\frac{1}{24
   n^3}+O\left(\frac{1}{n^5}\right)$$
$$\frac{\Gamma (n)}{\Gamma \left(\frac{n}{2}\right)^2}=\frac{2^{n-\frac{3}{2}} \sqrt{n}}{\sqrt{\pi }}\,\exp\Bigg[-\frac{1}{4 n}+\frac{1}{24
   n^3}+O\left(\frac{1}{n^5}\right) \Bigg]$$
So if, you use
$$D(n)=\frac{2^{n-\frac{3}{2}} \sqrt{n}}{\sqrt{\pi }} \quad \implies \quad \frac{G(n)}{D(n)}=1-\frac{1}{4 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$
Making it more general
$$\frac{\Gamma (n)}{\Gamma \left(\frac{n}{a}\right)^a}=a^n \sqrt{a^{-a} (2 \pi )^{1-a} n^{a-1}}\Bigg[1-\frac{a^2-1}{12 n} +\frac{\left(a^2-1\right)^2}{288 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
