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?
 A: Hint:
$$
\begin{align}
\binom{2n}{n}
&=\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\frac{(2n-4)(2n-5)}{(n-2)^2}\cdots\frac{4\cdot3}{2^2}\frac{2\cdot1}{1^2}\\
&=2^n\frac{2n-1}{n}\frac{2n-3}{n-1}\frac{2n-5}{n-2}\cdots\frac{3}{2}\frac{1}{1}\\
&=4^n\frac{n-1/2}{n}\frac{n-3/2}{n-1}\frac{n-5/2}{n-2}\cdots\frac{3/2}{2}\frac{1/2}{1}\tag{1}\\
&\ge4^n\frac{n-1}{n}\frac{n-2}{n-1}\frac{n-3}{n-2}\cdots\frac{1}{2}\cdot1/2\\
&=4^n\frac1{2n}\tag{2}
\end{align}
$$
$(1)$ and $(2)$ show that
$$
\frac1{2n}4^n\le\binom{2n}{n}\le4^n\tag{3}
$$
A: 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 $\lim_{n \rightarrow \infty} \sqrt[n]{a_n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$?.
A: You could use Stirling’s approximation:
$$\lim_{n\to\infty}\frac{n!}{\sqrt{2n\pi}(n/e)^n}=1\;.$$
Then
$$\binom{2n}n=\frac{(2n)!}{n!^2}\approx\frac{\sqrt{4n\pi}(2n/e)^{2n}}{2n\pi(n/e)^{2n}}=\frac{4^n}{\sqrt{n\pi}}\;,$$
so $$\binom{2n}n^{1/n}\approx\frac4{(\pi n)^{1/n}}\;.$$
I’ll let you fill in the details to show that $\approx$ here can be replaced by $\sim$, meaning that the ratio tends to $1$.
A: Hint: By induction, show that for $n\geq 2$ $$\frac{4^n}{n+1} < \binom{2n}{n} < 4^n.$$
A: 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 $\sqrt[n]n\to 1$.
A: 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
$$\lim_{n \to \infty} \dbinom{2n}n^{1/n} = 4$$
A: \begin{align}
\lim_{n\to\infty}\log\left(\sqrt[n]{2n \choose n}\right)&=\lim_{n\to\infty}\log\left(\sqrt[n]{\frac{2n!}{n!n!}}\right)\\
&=\lim_{n\to\infty}\log\left(\sqrt[n]{\frac{\prod_{k=1}^n(n+k)}{\prod_{k=1}^n k}}\right)\\
&=\lim_{n\to\infty}\frac{\sum_{k=1}^n\log\left(\cfrac{n+k}{k}\right)}{n}\\
&=\lim_{n\to\infty}\frac{\sum_{k=1}^n\log\left(1+\cfrac{n}{k}\right)}{n}\\
&=\int_0^1 \log\left(1+\frac1x\right)dx\\
&=\int_0^1 (\log(1+x)-\log x)dx\\
&=\left[(1+x)\log x(1+x)-x\log x\right]_0^1\\
&=2\log2=\log4\\
\therefore \lim_{n\to\infty}\sqrt[n]{2n \choose n}&=4
\end{align}
