TITLE: Calculating the limit of $[(2n)!/(n!)^2]^{1/n}$ as $n$ tends to $\infty$
QUESTION [7 upvotes]: 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!

REPLY [4 votes]: 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 $$\lim_{n\to\infty} \left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=4.$$