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

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 ${{2n} \choose n} ^ {1/n} \to 4$.

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.

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 $$ \lim_{n\to\infty}\left(\frac{(2n)!}{n!^2}\right)^{1/n}=4\tag7 $$