Evaluating a Binomial Coefficient Limit: $\lim_{n\to\infty} 2^{-2n} \binom{2n}{n}$.

Question

Can anyone quickly help me evaluate the limit:

$$\lim_{n\rightarrow\infty} 2^{-2n} \binom{2n}{n}$$

I know it follows from Stirling's approximation, but I cannot quite arrive at the answer. Thanks.

Answer 1

It is clear from Stirling's formula that

$$ \binom{2n}{n} \sim \frac{1}{\sqrt{\pi n}}. $$

But if one is not interested in the rate of convergence, then it is possible to produce an elementary solution. Indeed, fix $m \geq 0$. Then for any $n \geq m$,

$$ 2^{-2n}\binom{2n}{n} = \frac{\binom{2n}{n}}{\sum_{k=0}^{2n}\binom{2n}{k}} \leq \frac{\binom{2n}{n}}{\sum_{j=-m}^{m} \binom{2n}{n+j}} $$

For each fixed $j$, it is straightforward to check that $\binom{2n}{n+j}/\binom{2n}{n} \to 1$ as $n\to\infty$. So

$$ \limsup_{n\to\infty} 2^{-2n}\binom{2n}{n} \leq \frac{1}{2m+1} $$

But since $m$ is arbitrary, letting $m \to \infty$ gives the desired claim.

Answer 2

Consider$$a_n= 2^{-2n} \binom{2n}{n}=2^{-2n}\frac{(2n)!}{(n!)^2}$$ Take logarithms $$\log(a_n)=-2n\log(2)+\log((2n!))-2\log(n!)$$Now, use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ to get $$\log(a_n)=-\frac{1}{2} \left(\log \left({n}\right)+\log (\pi )\right)-\frac{1}{8 n}+O\left(\frac{1}{n^3}\right)=\log\left(\frac 1 {\sqrt{n \pi}} \right)+O\left(\frac{1}{n}\right)$$ making $$a_n \sim \frac 1 {\sqrt{n \pi}}$$

Answer 3

As shown in equation $(10)$ of this answer: $$ \frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}} $$ Therefore, $$ \frac1{\sqrt{\pi(n+\frac13)}}\le4^{-n}\binom{2n}{n}\le\frac1{\sqrt{\pi(n+\frac14)}} $$