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 $2^{-\sqrt{n}}$?
3 Answers
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 $$ e^{-1/2}\doteq0.60653065971263342360 $$
Rewrite $$ \lim_{n \to \infty}\frac{\dbinom{\frac{n}{2}}{m}}{\dbinom{n}{m}}=\lim_{n \to \infty}\frac{\left(\frac{n}{2}\right)!}{\left(\frac{n}{2}-m\right)!\ m!}\cdot\frac{(n-m)!\ m!}{n!}=\lim_{n \to \infty}\frac{\left(\frac{n}{2}\right)!\ (n-m)!}{n!\ \left(\frac{n}{2}-m\right)!}.\tag1 $$ Since $n\to\infty$, we can use Stirling's formula: $$ k!\sim\sqrt{2\pi k}\left(\frac{k}{e}\right)^k\quad\text{as}\quad k\to\infty. $$ Thus, we can use Stirling's formula to obtain the limit in $(1)$. I hope this help.
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
Let's consider the case where $m=np$ for $p<\frac 12$. We have: $$ {n/2 \choose np}\approx e^{\frac n2H(2p)} $$ where $H(q)$ is binary entropy function. Similarly: $$ {n \choose np}\approx e^{nH(p)} $$ Therefore: $$ \frac{{n/2 \choose m}}{n \choose m}\approx \exp\left({\frac n2\left(H(2p)-2H(p)\right)}\right). $$ We can check that $\left(H(2p)-2H(p)\right)$ is negative for $p\neq 0$ and therefore the limit is 0 for $m=np$ however the limit behaves as $e^{-cn}$.
