discrete mathematics – Show that $dfrac{(2n)!}{2^{2n}(n!)^2}$ is asymptotically equivalent to $dfrac{1}{sqrt{pi n}}$


Problem

Show that $dfrac{(2n)!}{2^{2n}(n!)^2}$ is asymptotically equivalent to $dfrac{1}{sqrt{pi n}}$.

Can someone please verify my solution attempt?

Solution

Given functions $f(n)$ and $g(n)$, $f(n)$ is asymptotically equivalent to $g(n)$ if and only if:

$$lim_{ntoinfty}dfrac{f(n)}{g(n)}=1.$$

Let $f(n)=dfrac{(2n)!}{2^{2n}(n!)^2}$ and $g(n)=dfrac{1}{sqrt{pi n}}$.

By Stirling’s formula, $n!$ can be expressed as:

$$n!=sqrt{2pi n}left(dfrac{n}{e}right)^n e^{epsilon(n)}$$

where

$$dfrac{1}{12n+1}leqepsilon(n)leqdfrac{1}{12n}.$$

So, $f(n)$ may be rewritten as:

$$begin{aligned}
f(n)&=dfrac{(2n)!}{2^{2n}(n!)^2}\
&=dfrac{sqrt{4pi n}left(dfrac{2n}{e}right)^{2n} e^{epsilon(2n)}}
{2^{2n} 2pi nleft(dfrac{n}{e}right)^{2n} e^{2epsilon(n)}}\
&=dfrac{2^{2n+1}left(dfrac{n}{e}right)^{2n} e^{epsilon(2n)}}
{2^{2n+1} sqrt{pi n}left(dfrac{n}{e}right)^{2n} e^{2epsilon(n)}}\
&=dfrac{e^{epsilon(2n)}}{ e^{2epsilon(n)}sqrt{pi n}}
end{aligned}$$

So, since both $epsilon(n)$ and $epsilon(2n)$ approach $0$ as $ntoinfty$, we have:

$$lim_{ntoinfty}dfrac{f(n)}{g(n)} = lim_{ntoinfty} left(dfrac{e^{epsilon(2n)}}{ e^{2epsilon(n)}sqrt{pi n}} cdot sqrt{pi n}right)=lim_{ntoinfty} dfrac{e^{epsilon(2n)}}{ e^{2epsilon(n)}}=dfrac{lim_{ntoinfty}e^{epsilon(2n)}}{lim_{ntoinfty}e^{2epsilon(n)}}=1,$$

as required.