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.

You do not need strictly need Stirling's approximation or the full power of the Legendre duplication formula, the usual approximations for central binomial coefficients are enough. If $n=2m+2$
$$ \color{blue}{\frac{\Gamma(n)}{\Gamma(n/2)^2}} = \frac{(2m+1)!}{m!^2} = (2m+1)\binom{2m}{m}\sim \frac{4^m}{\sqrt{\pi m}}\left(2m+\frac{3}{4}\right)\color{blue}{\sim\frac{2^{n}}{2\sqrt{2\pi n}}\left(n-\frac{1}{4}\right)}$$
and since
$f(z)=\log\Gamma(z)$ is a convex function on $\mathbb{R}^+$ (as a consequence of the Bohr-Mollerup theorem or just by the integral representation of the $\Gamma$ function) the previous asymptotic approximation holds also if $n$ is an odd number.
