TITLE: Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$
QUESTION [13 upvotes]: I know that
$$
\lim_{n\to\infty}{{2n}\choose{n}}^\frac{1}{n} = 4
$$
but I have no Idea how to show that; I think it has something to do with reducing ${n}!$ to $n^n$ in the limit, but don't know how to get there. How might I prove that the limit is four?

REPLY [3 votes]: If we know1 that for a sequence $(a_n)$ of positive real numbers we have $$\lim_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} \frac{a_{n+1}}{a_n},$$
providing the limit on the right hand side exists, we can use this for $a_n=\binom{2n}n$. We get
$$\lim_{n\to\infty} \frac{\binom{2n+2}{n+1}}{\binom{2n}n} = \lim_{n\to\infty} \frac{(2n+1)(2n+2)}{(n+1)^2} = 4.$$
Notice that this approach is not that different from taking a logarithm and applying Stolz-Cesaro theorem. (Which is actually a possible way the result about $\sqrt[n]{a_n}$.)

1This result is proved in some other posts on this site, for example: Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$, Convergence of Ratio Test implies Convergence of the Root Test, How to show that $\lim_{n \to \infty} a_n^{1/n} = l$?, Elementary way to show $\lim_{n \rightarrow \infty} \sqrt[n]{a_n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$?.