TITLE: Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$
QUESTION [5 upvotes]: $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$  
I could not approach it beyond these simple steps,  
$\lim_{n\to \infty} \frac{1}{2n} \log(\frac{2n!}{(n!)^2})$
$=\lim_{n\to \infty} \frac{1}{2n} [\log(2n)+\cdots +\log(n+1)-\log(n)-\cdots-\log1]$
$=\lim_{n\to \infty} (\log(2n)^{1/2n}+\cdots+\log(n+1)^{1/2n}-\log(n)^{1/2n}-\cdots-\log1^{1/2n})$  
Now,I understand that I have to create a sum of limit and produce an integration or use the formula $\lim_{n\to \infty} \log(1+\frac1x)^x=e$ but I cannot do it. Please help!

REPLY [11 votes]: Stirling's Asympotic Approximation yields
$$
\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{1}
$$
and
$$
\begin{align}
\lim_{n\to\infty}\frac1{2n}\log\left(\frac{4^n}{\sqrt{\pi n}}\right)
&=\lim_{n\to\infty}\left(\frac{n\log(4)}{2n}-\frac{\log(\pi n)}{4n}\right)\\
&=\log(2)\tag{2}
\end{align}
$$

We can also approach this in a more elementary manner:
$$
\begin{align}
\binom{2n}{n}
&=\frac{2^n}{n!}\frac{2n(2n-1)(2n-2)(2n-3)\cdots4\cdot3\cdot2\cdot1}{2n(2n-2)(2n-4)(2n-6)\cdots8\cdot6\cdot4\cdot2}\\
&=\frac{2^n}{n!}(2n-1)(2n-3)\cdots3\cdot1\\
&=4^n\frac{(2n-1)(2n-3)\cdots3\cdot1}{2n(2n-2)\cdots4\cdot2}\tag{3}
\end{align}
$$
To bound $(3)$, we have
$$
\begin{align}
\frac{4^n}{2}
&=4^n\frac{2n(2n-2)\cdots6\cdot4}{2n(2n-2)\cdots4\cdot2}\\
&\ge\color{#C00000}{4^n\frac{(2n-1)(2n-3)\cdots5\cdot3}{2n(2n-2)\cdots4\cdot2}}\\
&\ge4^n\frac{(2n-2)(2n-4)\cdots4\cdot2}{2n(2n-2)\cdots4\cdot2}\\
&=\frac{4^n}{2n}\tag{4}
\end{align}
$$
Combining $(3)$ and $(4)$ yields
$$
\frac1{2n}\log\left(\frac{4^n}{2n}\right)
\le\frac1{2n}\log\binom{2n}{n}
\le\frac1{2n}\log\left(\frac{4^n}{2}\right)\tag{5}
$$
that is
$$
\log(2)-\frac{\log(2n)}{2n}\le\frac1{2n}\log\binom{2n}{n}\le\log(2)-\frac{\log(2)}{2n}\tag{6}
$$
and by the Squeeze Theorem, we get
$$
\lim_{n\to\infty}\frac1{2n}\log\binom{2n}{n}=\log(2)\tag{7}
$$

More Accuracy
Stirling's Approximation actually gives
$$
\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi n}}\tag{8}
$$
and inequality $(4)$ combined with the AM-GM yields
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\tag{9}
$$
So we have the bounds
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi n}}\tag{10}
$$