TITLE: Evaluating a limit where the ratio test fails $\lim_{n\to\infty} \frac{1}{2^{2n}}\frac{(2n)!}{(n!)^2}\sqrt{n}$.
QUESTION [5 upvotes]: $\lim_{n\to\infty} \frac{1}{2^{2n}}\frac{(2n)!}{(n!)^2}\sqrt{n}$.
This is what happens when I try the ratio test,
$\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}|=
\lim_{n\to\infty} \frac{1}{4}\frac{(2n+2)(2n+1)}{(n+1)^2}\frac{\sqrt{n+1}}{\sqrt{n}}$
$\lim_{n\to\infty} \frac{4n^3+5n+2}{4n^2+8n+4}\frac{\sqrt{n+1}}{\sqrt{n}}=1$
There are so many factorials in this limit that I don't have the slightest guess of how to proceed with this unfortunate failure.

REPLY [3 votes]: Using the Stirling asymptotic formula
$$
n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n
$$
and substituting
$$
\lim_{n\rightarrow \infty}\frac{1}{2^{2n}}\frac{(2n)!}{(n!)^2}\sqrt{n}\approx \lim_{n\rightarrow \infty}\frac{1}{2^{2n}}\frac{\sqrt{2\pi 2n}}{(\sqrt{2\pi n})^2}\frac{(\frac{2n}{e})^{2n}}{(\frac{n}{e})^{2n}}\sqrt{n} = \frac{1}{\sqrt{\pi}}
$$