Compute $\lim\limits_{n\rightarrow\infty}\left(\frac{\left(2n\right)!}{\left(n!\right)^{2}}\right)^{\frac{1}{n}}$ Compute
$$\lim_{n\rightarrow\infty}\left(\frac{\left(2n\right)!}{\left(n!\right)^{2}}\right)^{\frac{1}{n}}$$
If you have some nice proofs and you're willing to share them, then I thank you and you definitely have my upvote!
 A: Note that
$$
\frac{(2n)!}{(n!)^2}=\binom{2n}{n}
$$
Then, using the simple identity
$$
\binom{2n+2}{n+1}=\binom{2n}{n}\frac{4n+2}{n+1}
$$
by induction, we get that
$$
\binom{2n}{n}=4^n\prod_{k=0}^{n-1}\frac{k+{\small1/2}}{k+1}
$$
Since
$$
\frac1{2n}=\underbrace{\frac12}_{k=0}\underbrace{\left(\frac12\cdot\frac23\cdot\frac34\cdots\frac{n-1}n\right)}_{1\le k\le n-1}\le\prod_{k=0}^{n-1}\frac{k+{\small1/2}}{k+1}\le1
$$
Therefore,
$$
\frac1{2n}4^n\le\binom{2n}{n}\le4^n
$$
Thus, by the Squeeze Theorem
$$
\lim_{n\to\infty}\binom{2n}{n}^{1/n}=4
$$
A: You can use the Stirling approximation,
$$n!\sim n^n e^{-n}\sqrt{2\pi n}$$
from which we have
$$\left(\frac{(2n)!}{(n!)^2}\right)^{1/n}\sim \left(\frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{n^{2n} e^{-2n}2\pi n}\right)^{1/n}=4\cdot\frac{1}{(\pi n)^{1/(2n)}}.$$
The latter term can be written
$$\frac{1}{(\pi n)^{1/(2n)}}=\exp\left(-\frac{1}{2n}\ln(\pi n)\right)\to 1,$$
so the limit is $4$.
A: You could use the fact that for a sequence  $(a_n)$ of positive numbers, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} {\root n\of{a_n}}$ and the two limits are equal. For a proof of this, see these notes of Pete L. Clark.
Apply this result to $a_n={(2n)!\over (n!)^2}$. One easily evaluates  $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=4$. Then $\lim\limits_{n\rightarrow\infty} {a_n^{1/n}}=4$.
