Ratio of $\Gamma(n)$ to $\Gamma(n/2)^2, n\to \infty?$

Question

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.

Answer 1

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.

Answer 2

If like me you love Stirling's formula $\Gamma(z) \underset{z\to\infty}{\sim} (\tfrac{z}{e})^z\,(\tfrac{2\pi}{z})^{1/2}$, then you get $$ \frac{\Gamma(n)}{\Gamma(n/2)^2} \sim \frac{(\tfrac{n}{e})^n\,(\tfrac{2\pi}{n})^{1/2}}{(\tfrac{n}{2e})^{n}\,(\tfrac{4\pi}{n})} \sim 2^{n-1}\,\sqrt{\frac{n}{2\pi}} $$ and so your quantity diverges to $+\infty$.

Answer 3

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

$$\frac{\Gamma (n)}{\Gamma \left(\frac{n}{a}\right)^a}=a^n \sqrt{a^{-a} (2 \pi )^{1-a} n^{a-1}}\Bigg[1-\frac{a^2-1}{12 n} +\frac{\left(a^2-1\right)^2}{288 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$