# 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.