Evaluate $\lim_{n \to \infty}\frac{(2n)!}{(n!)^2 \cdot 2^{2 n}}$ What will be the limit of lim$\frac{(2n)!}{(n!)^2 \cdot 2 ^{2 n}} $ ? 
Actually I  was told that if the probability of being vegetarian of every person in a city is $0.5$ . Then what is the probability of half of the population being vegetarian? Actually I have seen many times that   the converse of the above statement is supposed to be true i.e if half of the population are vegetarian then the probability of every person being vegetarian is $0.5$: $P(X=n) = \frac{(2n)!}{(n!)^2 \cdot 2 ^{2 n}} $. $P(X=n)$ will certainly not close to $0.5$ for some finite $n$. But I am quite curious to know what will happen when $n$ tends to infinity..
Can someone please help me out?
 A: With Stirling's Approximation 
$$\lim_{n \to \infty} \frac{(2n)!}{(n!)^24^n}= \lim_{n \to \infty} \frac{\displaystyle \sqrt{4\pi n} \left( \frac{2n}{e}\right)^{2n}}{(2\pi n)\left( \frac{n}{e}\right)^{2n} 4^n}  = \lim_{n \to \infty} \frac{1}{\sqrt{\pi n}} = 0$$
And intuitively, actually, this is fairly obvious. If you approximate the binomial distribution with a normal one by the demoivre-laplace theorem, the probability mass of exactly one point goes to zero for sure. So, while it's very likely to, say, be within 1% of the half mark as $n$ is large, it's not so likely to exactly hit it
A: $$0 \leq \lim_{n \to \infty}\frac{(2n)!}{(n!)^2 \cdot 2^{2 n}} = 
\lim_{n \to \infty}\frac{(2n)(2n-1)\ldots(n+1)}{n!}\left( \frac{1}{n!2^{2n}} \right) $$ 
$$ = \lim_{n \to \infty}\frac{[2(n)](2n-1)[2(n-1)](2n-3)[2(n-2)](2n-5)\ldots (2)(1)}{n!} \left( \frac{1}{n!2^{2n}} \right) $$
$$ = \lim_{n \to \infty} \frac{(2n-1)(2n-3)\ldots (3)(1)}{2^{n}} \left( \frac{1}{n!2^{n}} \right) $$
$$ = \lim_{n \to \infty} \left( \frac{1}{n} \right) \left( \frac{2n-1}{n-1} \right) \left( \frac{2n-3}{n-2} \right)  \ldots \left( \frac{1}1{} \right) \left( \frac{1}{2^{n}} \right) $$
$$ \leq \lim_{n \to \infty}\frac{2^{n}}{n2^{n}} = \lim_{n \to \infty} \frac{1}{n} = 0.$$
Note: $\frac{2n-k}{n-k} = 2 - \frac{k}{n-k} < 2.$
