I am interested in taking the following limit:
\begin{equation}
\lim_{n \to \infty}\frac{{n/2 \choose m}}{n \choose m}.
\end{equation}
Provided that $m$ is fixed the solution is:
\begin{equation}
\lim_{n \to \infty}\frac{{n/2 \choose m}}{n \choose m}=2^{-m}.
\end{equation}
This can be seen by writing:
\begin{aligned}
\frac{{n/2 \choose m}}{n \choose m}&=\frac{\left(\frac{n}{2}\right)!(n-m)!}{\left(\frac{n}{2}-m\right)!n!}\\
&=\prod_{j=0}^{m-1}\frac{\frac{n}{2} -j}{n-j}\\
&=\frac{1}{2^m}\prod_{j=1}^{m-1}\frac{n-2j}{n-j}
\end{aligned}
and letting $n$ tend to infinity.
What however, can be said when $m$ depends on $n$? For example, if $m=\sqrt{n}$? Do we still have the quantity behaving asymptotically like $2^{-\sqrt{n}}$?
Best Answer
For fixed $m$, asymptotically, $$ \lim_{n\to\infty}\frac1{n^m}\binom{n}{m}=\frac1{m!} $$ this can be seen by writing $$ \begin{align} \frac1{n^m}\binom{n}{m} &=\frac1{m!}\frac{n(n-1)(n-2)\cdots(n-m+1)}{n^m}\\ &=\frac1{m!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{m-1}n\right) \end{align} $$ and taking the product of the $m-1$ limits.
If $m=\sqrt{n}$, then $1-\frac kn\sim e^{-k/n}$ for large $n$. Thus, asymptotically, $$ \begin{align} \frac1{n^m}\binom{n}{m} &\sim\frac1{m!}e^{-\frac{m(m-1)}{2n}}\\ &\sim\frac1{m!}e^{-1/2} \end{align} $$ and $$ \begin{align} \frac1{(n/2)^m}\binom{n/2}{m} &\sim\frac1{m!}e^{-\frac{m(m-1)}{n}}\\ &\sim\frac1{m!}e^{-1} \end{align} $$ Thus, the ratio is asymptotic to $$ \frac1{2^m}e^{-1/2} $$
Numerical Verification
Mathematica 8, gives $$ 2^{1000}\frac{\binom{500000}{1000}}{\binom{1000000}{1000}}\doteq0.60653076090233049286 $$ and $$ e^{-1/2}\doteq0.60653065971263342360 $$