What is value of limit $\lim_{n\to \infty}\frac{(2n)!}{4^n(n!)^2}$

Question

$\lim_{n\to \infty}\frac{(2n)!}{4^n(n!)^2}$ I tried a lot but I haven't desired result. I tried with sandwich theorem

Answer 1

First Approach

In this answer, it is shown that $$ \frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}} $$

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Another Approach

Squaring and cross multiplying show that $$ \begin{align} \frac{\,\frac{(2n)!}{4^{n}n!^2}\,}{\frac{(2n-2)!}{4^{n-1}(n-1)!^2}} &=\frac{2n-1}{2n}\\ &\le\sqrt{\frac{n}{n+1}} \end{align} $$ Therefore, $$ \begin{align} \frac{(2n)!}{4^{n}n!^2} &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &\le\prod_{k=1}^n\sqrt{\frac{k}{k+1}}\\ &=\sqrt{\frac1{n+1}} \end{align} $$

Answer 2

By De Moivre's formula $2\cos\theta=e^{i\theta}+e^{-i\theta}$ and the binomial theorem we have $$\frac{(2n)!}{4^n n!^2}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos \theta\right)^{2n}\,d\theta $$ hence the wanted limit is zero by the dominated/monotone convergence theorem.

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Alternative approach: by letting $a_n=\frac{1}{4^n}\binom{2n}{n}$ we have $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)^2}=\frac{n+\frac{1}{2}}{n+1}\leq\sqrt{\frac{n+\frac{1}{2}}{n+\frac{3}{2}}} $$ $$ \frac{a_{n+1}}{a_0}=\frac{a_{n+1}}{a_n}\cdot\frac{a_n}{a_{n-1}}\cdots\frac{a_1}{a_0}\leq\frac{1}{\sqrt{2n+3}} $$ and $a_n\leq\frac{1}{\sqrt{2n+1}}$.

Answer 3

The estimations

$\frac{4^{n}}{\sqrt{4n}} \leq {2n \choose n} \leq \frac{4^{n}}{\sqrt {3n+1}}$ for $n \ge 1$ can be found here:

https://en.wikipedia.org/wiki/Central_binomial_coefficient

Answer 4

Note that $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$ is the number of $2n$-digit binary numbers with exactly $n$ zeros and $n$ ones. Second, note that $2^{2n} = 4^n$ is the number of $2n$-digit binary numbers. So the limit is "clearly" $0$.

See the second answer to this:

Inequality $\binom{2n}{n}\leq 4^n$