What is the limit of $\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$? What is the limit of the following sequence?.
$$\frac{\left ( 2n \right )!}{(n!)^{2}4^{n}}$$
I guess the limit is zero but I don't know exactly how to prove it.
Any help please.

Because I think there are people interested in an elementary solution:
The ratio between terms $a_n=\binom{2n}{n}\frac{1}{4^n}$ is given by
$$\begin{array}{ll} \displaystyle \frac{a_{n+1}}{a_n} & =\frac{\displaystyle\frac{(2n+2)!}{(n+1)!(n+1)!}\frac{1}{4^{n+1}}}{\displaystyle\frac{(2n)!}{n!n!}\frac{1}{4^n}} \\[6pt] & \displaystyle = \frac{(2n+2)(2n+1)}{(n+1)(n+1)4} \\[3pt] & \displaystyle =\frac{2n+1}{2n+2} \\[2pt] & \displaystyle =1-\frac{1}{2n+2}. \end{array} $$
Therefore, the term $a_n$ is
$$\begin{array}{ll} a_n & \displaystyle =a_0\cdot\frac{a_1}{a_0}\cdot\frac{a_2}{a_1}\cdots\frac{a_n}{a_{n-1}} \\ & = \left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\cdots\left(1-\frac{1}{2n}\right) \\ & \le \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right) \end{array} $$
and therefore
$$ \begin{array}{ll} a_n^2 & \le  \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{2n}\right)\left(1-\frac{1}{2n+1}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\cdots\left(\frac{2n-1}{2n}\right)\left(\frac{2n}{2n+1}\right)=\frac{1}{2n+1}. \end{array} $$
From $a_n\le\frac{1}{\sqrt{2n+1}}$ we conclude the limit is $0$.
