Evaluating $\lim\limits_{n \to \infty}\frac{2^{2n}(n!)^2}{(2n+1)!}$ 
Can you help me find this limit:$$\lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)!}$$

I tried to use
$$ \lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)!} = \ln\left(\exp\left( \lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)!} \right)\right)$$ but it didn't work out so well.
This problem came up when I was trying to find the radius of convergence of the sum $$\sum_{n=0}^{\infty}(-1)^n\left[\dfrac{2^{n}(n!)^2}{(2n+1)!}\right]^p.x^n$$
Thanks!
(Anyway I omitted the exponential $p$ in the limit because it just a constant).
(*Edited. Is there any others way instead of using Stirling approximation(1) or the Center Binomial Coefficient(2)? I just stated learning Analysis and this limit is only a part of the finding the radius of convergence problems,though. It take way too long and hard to prove (1) and (2)). Sorry for not  ask clearly the first time.
 A: \begin{align*}
\lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)!}=\lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)(2n)!}=\lim_{n \rightarrow \infty} \dfrac{2^{2n}\sqrt{\pi n}}{(2n+1)4^n}=\lim_{n \rightarrow \infty} \dfrac{\sqrt{\pi n}}{(2n+1)}=0\\
\end{align*}
using this.
A: My first approach would be to begin with Stirling's approximation:
$$n! \sim \sqrt{2 \pi n} \frac{n^n}{e^n} \implies (2n+1)! = \sqrt{2\pi(2n+1)} \frac{(2n+1)^{2n+1}}{e^{2n+1}}$$
Using this, you get that your limit - call it $L$ - is equivalent to
$$L = \lim_{n \to \infty} 2^{2n} \cdot 2\pi n \cdot \frac{n^{2n}}{e^{2n}} \cdot \frac{1}{\sqrt{2 \pi(2n+1)}} \cdot \frac{e^{2n+1}}{(2n+1)^{2n+1}}$$
A bit messier, but note:

*

*The $e$ terms can cancel to give you $e$

*The $\pi$ and $\sqrt{\pi}$ cancel to give the latter

*You can group the powers of $2$ and $n$, and then the pair together themselves

*The constants not dependent on $n$ can be brought outside the limit

With these in mind and some minor algebraic manipulations, you get
$$L = e \sqrt{\frac \pi 2} \lim_{n \to \infty} \left( \frac{2n}{2n+1} \right)^{2n+1} \frac{1}{\sqrt{2n+1}}$$
Noting that $2n = 2n+1-1$, you can simplify the parenthetical even more:
$$L = e \sqrt{\frac \pi 2} \lim_{n \to \infty} \left(1 - \frac{1}{2n+1} \right)^{2n+1} \frac{1}{\sqrt{2n+1}}$$
I imagine it is sufficiently obvious that the limit is zero as a result. If not, let's continue from here. Show that the limit of each factor exists, and then you have
$$L = e \sqrt{\frac \pi 2} \lim_{n \to \infty} \left(1 - \frac{1}{2n+1} \right)^{2n+1} \lim_{n \to \infty} \frac{1}{\sqrt{2n+1}}$$
With a redefining of $m=2n+1$ in the left-hand limit, you find the limit expression to be $(1 - 1/m)^m$, which is well-known to approach $1/e$. The right-hand limit is more obviously $0$. Thus,
$$L = e \sqrt{\frac \pi 2} \lim_{n \to \infty} \left(1 - \frac{1}{n} \right)^{n} \lim_{n \to \infty} \frac{1}{\sqrt{2n+1}} = e \sqrt{\frac \pi 2} \cdot \frac 1 e \cdot 0 = 0$$

Alternatively, per Brian's comment, we get a slicker solution. Notice,
$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$$
which means, per a simplification he also notes and $4^n = 2^{2n}$,
$$\frac{(n!)^2}{(2n+1)!} = \frac{1}{(2n+1) \binom{2n}{n}} \sim \frac{\sqrt{\pi n}}{(2n+1) 2^{2n}}$$
This in turn gives us that
$$L = \lim_{n \to \infty} 2^{2n} \cdot \frac{\sqrt{\pi n}}{(2n+1) 2^{2n}} = \lim_{n \to \infty} \frac{\sqrt{\pi n}}{2n+1} = 0$$
A: If $x_n$ is the sequence in question then we have $$\frac {x_{n+1}}{x_n}=\frac {2n+2}{2n+3}=1-\frac{1}{2n+3}$$ and therefore $$\log\frac {x_{n+1}}{x_n}<-\frac{1}{2n+3}$$ On summing such inequalities for $n=1,2,\dots,n$ we get $$\log\frac{x_{n+1}}{x_1}<-\sum_{m=1}^{n}\frac {1}{2m+3}$$ The RHS diverges to $-\infty $ and therefore so does the LHS. It should now be clear that $x_n\to 0$.

Use of Stirling formula can be avoided in most cases and one should try to apply the standard tests for convergence.
A: You have:
$$
\begin{align}
\lim_{n \rightarrow \infty} \dfrac{2^{2n}(n!)^2}{(2n+1)!}
&=\lim_{n \rightarrow \infty}\frac{\left(2^nn!\right)^2}{(1)(2)(3)\cdots(2n)(2n+1)}\\
&=\lim_{n \rightarrow \infty}\frac{\left((2)(4)(6)\cdots(2n)\right)^2}{(1)(2)(3)\cdots(2n)(2n+1)}\\
&=\lim_{n \rightarrow \infty}\frac{(2)(4)(6)\cdots(2n)}{(1)(3)\cdots(2n+1)}\\
&=\lim_{n \rightarrow \infty}\frac23\frac45\cdots\frac{2n}{2n+1}\\
&=\lim_{n\to\infty}\left(\exp\left(\ln(1-1/3)+\ln(1-1/5)+\cdots+\ln(1-1/(2n+1))\right)\right)
\end{align}
$$
Note that
$$\ln(1-1/3)+\ln(1-1/5)+\cdots+\ln(1-1/(2n+1))<-\frac{1}{3}-\frac{1}{5}-\cdots-\frac{1}{2n+1}$$
because $\ln(1-x)<-x$. And the right side is negative harmonic, so its partial sums are arbitrarily large negative. Therefore so are the partial sums from the left side. Therefore the $\exp$ above will be given arbitrarily large negative numbers. And therefore the limit is $0$.
A: By induction easy to show that:
$$\binom{2n+1}{n}>\frac{2^n\left(1+\sqrt[n]{n+1}\right)^n}{n+1}.$$
Thus, $$ \dfrac{2^{2n}(n!)^2}{(2n+1)!}=\dfrac{2^{2n}n!(n+1)!}{(n+1)(2n+1)!}<\left(\frac{2}{1+\sqrt[n]{n+1}}\right)^n\rightarrow0$$
