Calculating the limit of $[(2n)!/(n!)^2]^{1/n}$ as $n$ tends to $\infty$ Analysis textbook by Shanti Narayan, is asking to prove the limit
$\lim {\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \to \frac{1}{4}$ as $n \to \infty$.
I tried but was unable to find the solution. Even Wolfram Alpha is telling the limit to be $4$.
Please help!
 A: As a previous exercise you can prove that
$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\ln \left(\frac{n}{k}\right)=-\int_0^1\ln t\, dt=1.\tag{1}\label{eq1}$$
Note that
$$\begin{align*}
\ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n} &=\frac{1}{n}\left[ \sum_{k=1}^{2n} \ln k-2\sum_{k=1}^n \ln k \right]\\
&= \frac{1}{n}\sum_{k=1}^n \ln(n+k)-\ln(k)\\
&= \frac{1}{n}\left[\sum_{k=1}^n\ln n+\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right)-\sum_{k=1}^n\ln k\right]\\
&= \frac{1}{n}\sum_{k=1}^n \ln\left(\frac{n}{k}\right)+\frac{1}{n}\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right).
\end{align*}$$
Note that in the last equality the second term is a Riemann's sum of $t\mapsto \ln t$ over $[1,2]$. Using \eqref{eq1} we have
$$\begin{align*}
\lim_{n\to\infty}\ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n} &= -\int_0^1\ln t\, dt+\int_1^2 \ln t\, dt\\
&=1+(-1+\ln 4),
\end{align*}$$
so $$\lim_{n\to\infty}\ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=\ln 4.$$ By the continuity of $t\mapsto e^t$ you get $$\lim_{n\to\infty} \left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=4.$$
A: A simple proof is based on the observation that $\dfrac{(2n)!}{(n!)^2}$
 is the central binomial coefficient $\displaystyle{ {2n} \choose n}$.
Look at row $2n$ in the Pascal triangle. The sum of all terms is $2^{2n}= 4^n$ and so ${{2n} \choose n} \le 4^n$. Moreover, the central binomial coefficient is the largest number in that row and so $4^n \le (2n+1){{2n} \choose n}$.
Hence
$$
\frac{4^n}{2n+1} \le {{2n} \choose n} \le 4^n
$$
Since $(2n+1)^{1/n} \to 1$, we conclude that ${{2n} \choose n} ^ {1/n} \to 4$.
A: If you use Stirling's approximation $$n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$ then you get $${\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \approx {\left({\dfrac{\sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n}}{{2 \pi n} \left(\frac{n}{e}\right)^{2n}}}\right)}^{1/n} = 4\left(\frac{1}{n\pi}\right)^\frac{1}{2n} \approx 4.$$
It is not difficult to translate this into the language of limits.
A: Note that
$$
\begin{align}
\frac{(2(n+1))!}{(n+1)!^2}
&=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\frac{(2n)!}{n!^2}\tag1\\
&=4\left(1-\frac{1}{2n+2}\right)\frac{(2n)!}{n!^2}\tag2
\end{align}
$$
Bernoulli's Inequality says
$$
\begin{align}
\left(\frac{(2n)!}{n!^2}\right)^{1/n}
&=4\left(\prod_{k=0}^{n-1}\left(1-\frac{1}{2k+2}\right)\right)^{1/n}\tag3\\
&\ge4\left(\frac12\prod_{k=1}^{n-1}\left(1-\frac{1}{k+1}\right)^{1/2}\right)^{1/n}\tag4\\
&=4\left(\frac1{2\sqrt{n}}\right)^{1/n}\tag5
\end{align}
$$
Equation $(3)$ and inequality $(5)$ give
$$
4\left(\frac1{2\sqrt{n}}\right)^{1/n}\le\left(\frac{(2n)!}{n!^2}\right)^{1/n}\le4\tag6
$$
and the Squeeze Theorem yields
$$
\lim_{n\to\infty}\left(\frac{(2n)!}{n!^2}\right)^{1/n}=4\tag7
$$
A: There is really nno meed to use the stirling fomula to calculate this integral, we can just use stolz formula to caluate 
$I_{n}=(\frac{(2n)!}{(n!)(n!)})^{\frac{1}{n}}$ Using stolz formula we can get $\quad$
$\lim log I_{n}=\lim log(\frac{2(2n+1)}{n+1})=log4 $
