Limit of $ (2^n (n!)^2)/(2n+1)!$ I want to show that
$$
\lim_{n \rightarrow \infty} \frac{2^n (n!)^2}{(2n+1)!} = 0,
$$
but it's been a long time since I took calculus, and I don't know how to do it. I've tried to squeeze it, but I didn't succeed... 
Thanks in advance!
 A: Let $a_n = \dfrac{2^n(n!)^2}{(2n+1)!}$.
Then we have
$$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}\bigl((n+1)!\bigr)^2(2n+1)!}{2^n(n!)^2(2n+3)!} = \frac{2(n+1)^2}{(2n+2)(2n+3)} = \frac{n+1}{2n+3} < \frac12.$$
That shows that the limit is $0$ by majorisation by a geometric sequence $\dfrac{1}{2^n}$.
For a more precise asymptotic, you can use Stirling's formula. Short-cutting, with $a_n = \frac{2^n}{(2n+1)\binom{2n}{n}}$, and using the asymptotic
$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}},$$
we have
$$a_n \sim \frac{\sqrt{\pi}}{2^{n+1}\sqrt{n}}.$$
A: Another quick-and-dirty approach: $2^n(n!)^2$ $=(1\cdot2\cdots n)\cdot(2\cdot 4\cdots 2n)$ $\leq \left(1\cdot3\cdots (2n-1)\right)\cdot(2\cdot4\cdots 2n)$ $=(2n)!$, so $\dfrac{2^n(n!)^2}{(2n+1)!}\leq \dfrac{(2n!)}{(2n+1)!}=\dfrac1{2n+1}$, which obviously goes to $0$ as $n\to\infty$.  (As other answers have shown, of course, it goes to $0$ much faster than this - exponentially faster, in fact.)
A: Using Sterling Approximation, we can intuitively see ($C$ is some positive constant):
$$
\lim_{n \rightarrow \infty} \frac{2^n (n!)^2}{(2n+1)!}= \lim_{n \rightarrow \infty}  \frac{C2^n (n^{n+\frac 12} e^{-n})^2}{(2n+1)^{2n+1+\frac 12} e^{-2n-1}}\\
=\lim_{n \rightarrow \infty}  C2^ne \frac{1}{(2n+1)^{\frac 12}} (\frac{n}{2n+1})^{2n+1}\\
=\lim_{n \rightarrow \infty}  C2^ne \frac{1}{(2n+1)^{\frac 12}}(\frac{1}{2})^{2n+1} (\frac{n}{n+\frac 12})^{2n+1}\\
=\lim_{n \rightarrow \infty}  Ce \frac{1}{(2n+1)^{\frac 12}}(\frac{1}{2})^{n+1} (\frac{1}{e})\\
=\lim_{n \rightarrow \infty}  \frac{C}{2^{n+1}(2n+1)^{\frac 12}}
$$
So the limit goes to zero as $n$ goes to infinity.
