Show that that $\lim_{n\to\infty}\sqrt[n]{\binom{2n}{n}} = 4$ 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?

The term ${2n\choose n}$ occurs as biggest of $2n+1$ positive summands in expanding $(1+1)^{2n}$, which directly shows $\frac{4^n}{2n+1}\le {2n\choose n}\le 4^n$.
From this the claim follows by using $\sqrt[n]n\to 1$.
