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.
 A: 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 $0$.
A: Simpler than this cannot be possible. By Stirling:
$$n! \approx \sqrt{2\pi n}n^n e^{-n}$$
Which of course has to be arranged to get an approximation for $(2n)!$ and for $(n!)^2$, which is trivial algebra. Then you get:
$$\frac{\sqrt{\pi } 2^{2 n+1} e^{-2 n} n^{2 n+\frac{1}{2}}}{2 \pi  e^{-2 n} n^{2 n+1} 4^n}$$
Easy algebra will let you to rewrite this as a "goes like":
$$\color{red}{\sim \frac{1}{\sqrt{\pi } \sqrt{n}}}$$
Which as $n\to \text{whatever}\ $ you need, gives you the limit.
I think $n\to +\infty$ was understood, hence the limit is zero.
A: Alternative approach: since $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $\int_{0}^{2\pi}e^{nix}e^{-mix}\,dx=2\pi\delta(m,n)$, as a consequence of the binomial theorem we have
$$ a_n=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2\pi}\int_{0}^{2\pi}\cos(x)^{2n}\,dx. $$
By the dominated convergence theorem, the limit of the RHS as $n\to +\infty$ is zero. So it is the limit of the LHS. 
Actually the integral representation also tells us that $\{a_n\}_{n\geq 1}$ is a log-convex sequence: $a_n a_{n+2} \geq a_{n+1}^2$ follows from Cauchy-Schwarz.

Here it comes the horrific overkill. Since $\sum_{n\geq 0}a_n^3$ is a convergent series, equal to $\frac{1}{4\pi^3}\Gamma\left(\frac{1}{4}\right)^4$, we clearly have $\lim_{n\to +\infty}a_n=0$.
