Calculate $\lim \sqrt[n] \frac{(2n)!}{(n !)^2}.$ I want to calculate $$\lim_{n\to \infty} \sqrt[n] \frac{(2n)!}{(n !)^2}$$
According to Wolfram alpha https://www.wolframalpha.com/input?i=lim+%5B%282n%29%21%2F%7Bn%21%5E2%7D%5D%5E%7B1%2Fn%7D , this value is $4$, but I don't know why.
I have $\sqrt[n]{\dfrac{(2n)!}{(n !)^2}}=\sqrt[n]{\dfrac{2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)}{n!}}$ but I have no idea from here.
Another idea is taking $\log.$
$\log  \sqrt[n] \frac{(2n)!}{(n !)^2}=\dfrac{\log \frac{(2n)!}{(n !)^2}}{n}
=\dfrac{\log \dfrac{2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)}{n!}}{n}
=\dfrac{\log [2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)]-\log n!}{n}
$.
This doesn't seem to work.
Do you have any idea or hint ?
 A: Option $1$ : Use Stirling's Approximation to solve it easily
Option $2$:-
You already have
$\dfrac{\log [2n\cdot (2n-1)\cdot \cdots \cdot (n+2)\cdot (n+1)]-\log n!}{n}$
Well this is just :-
$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log(\frac{n+r}{r})=\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log(1+\frac{n}{r})=\int_{0}^{1}\log(1+\frac{1}{x})\,dx = \ln(4)$$ .
Option $3$. Use Cauchy's Limit Theorems which say that when you have a sequence $\{x_{n}\}$ whose limit exists finitely  , then the Arithmetic mean  of the first $n$ terms also converge to the same limit as $n\to\infty$. That is $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{x_{r}}{n}=\lim_{n\to\infty}x_{n}$ . See here and here for example.
Thus the answer is $e^{\ln(4)}=4$
My Advice: Use Stirling's Approximation if you're allowed to use it because it is much quicker.
A: By the Stolz theorem, one has
\begin{eqnarray}
\lim_{n\to \infty} \ln\bigg[\sqrt[n] \frac{(2n)!}{(n !)^2}\bigg]&=&\lim_{n\to \infty} \frac{\ln(2n)!-2\ln(n !)}{n}\\
&=&\lim_{n\to \infty} \frac{\left[\ln(2n+2)!-2\ln((n+1)!)\right]-\left[\ln(2n)!-2\ln(n !)\right]}{(n+1)-n}\\
&=&\lim_{n\to \infty} \ln[(2n+2)(2n+1)]-2\ln(n+1)\\
&=&\lim_{n\to \infty} \ln\bigg[\frac{(2n+2)(2n+1)}{(n+1)^2}\bigg]\\
&=&\ln 4
\end{eqnarray}
and hence
$$\lim_{n\to \infty} \sqrt[n] \frac{(2n)!}{(n !)^2}=4.$$
