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?

Use Stirling formula
$$n! \sim \sqrt{2 \pi n} \left(\dfrac{n}e \right)^n$$
This gives us that
$$\dbinom{2n}n \sim \dfrac{4^n}{\sqrt{\pi n}}$$
Now conclude that
$$\lim_{n \to \infty} \dbinom{2n}n^{1/n} = 4$$
