Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$ $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$  
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!
 A: HINT:
As $\displaystyle \binom{2n}n=\frac{(2n)!}{n! n!}=\prod_{1\le r\le n}\frac{n+r}r$
$$\ln\binom{2n}n=\sum_{1\le r\le n}\ln\left(\frac{n+r}r\right)=\sum_{1\le r\le n}\ln\left(\frac{1+\frac rn}{\frac rn}\right)$$
As 
 $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\lim_{n\to\infty}\frac1n\ln\binom{2n}n=\int_0^1\ln\left(\frac{1+x}x\right)dx$$
A: $\newcommand{\+}{^{\dagger}}%
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For $N \gg 1$, ${{\rm d}\ln\left(N!\right) \over {\rm d}N} \approx \ln\left(N\right)$. Consider $N$ and/or $n$ as real variables !!!.
$$
\color{#0000ff}{\large\lim_{n \to \infty}{1 \over 2n}\,\left\{\ln\left(\left[2n\right]!\right) - 2\ln\left(n!\right)\right\}}
=
\lim_{n \to \infty}{2\ln\left(2n\right) -2\ln\left(n\right)\over 2} = \color{#0000ff}{\large\ln\left(2\right)}
$$
Otherwise,
\begin{align}
&\lim_{x \to \infty}{\ln\Gamma\pars{2x + 1} - 2\ln\Gamma\pars{x + 1} \over 2x}
=
\lim_{x \to \infty}{2\Psi\pars{2x + 1} - 2\Psi\pars{x + 1} \over 2}
\\[3mm]&=
\lim_{x \to \infty}\bracks{\Psi\pars{2x} + {1 \over 2x} - \Psi\pars{x} + {1 \over x}}
=
\lim_{x \to \infty}\bracks{\Psi\pars{2x} - \Psi\pars{x}}
\end{align}
Since $\Psi\pars{z} \sim \ln\pars{z}$ when $\verts{z} \gg 1$, we'll have $\Psi\pars{2x} - \Psi\pars{x} \sim \ln\pars{2x} - \ln\pars{x} = \ln\pars{2}$.
$\Gamma$ and $\Psi$ are the ${\it Gamma}$ and ${\it Digamma}$ functions, respectively: $\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}$.
${\large\tt ADENDUM:}$
Since $\ds{\totald{\ln\pars{x}}{x} = {1 \over x}}$, the ${\large\tt\ln}$ function varies slowly when $x \gg 1$. Then, for large $N$:
$$
\left.\totald{\ln\pars{x}}{x}\right\vert_{x\ =\ N} \approx
{\ln\pars{\bracks{N + 1}!} - \ln\pars{N!} \over \pars{N + 1} - N} = \ln\pars{N + 1}
\approx \ln\pars{N}\,,\qquad N \gg 1
$$
A: As 
$$
\log\binom{2n}{n}=\log\left(\frac{(2n)!}{(n!)^2}\right)=\log((2n)!)-2\log(n!)
$$
using Stirling's approximation $\log(k!)\sim k\log k -k$
$$
\log((2n)!)-2\log(n!)\sim 2n\log(2n)-2n\log n=2n\log 2
$$
So 
$$
\lim_{n\to\infty}\frac{1}{2n}\log\binom{2n}{n}=
\lim_{n\to\infty}\frac{1}{2n}2n\log 2=\log 2
$$
A: With Stolz-Cesaro (${\rm L}=\log$, obviously):
$$\lim_{n\to\infty}{{\rm L}(2n)+\cdots +{\rm L}(n+1)-{\rm L}(n)-\cdots-{\rm L}1\over 2n}=$$
$$\lim_{n\to\infty}{({\rm L}(2n+2)+\cdots +{\rm L}(n+2)-{\rm L}(n+1)-\cdots-{\rm L}1)-({\rm L}(2n)+\cdots +{\rm L}(n+1)-{\rm L}(n)-\cdots-{\rm L}1)\over 2(n+1)-2n}$$
$$=\lim_{n\to\infty}{{\rm L}(2n+2)+{\rm L}(2n+1)-2{\rm L}(n+1)\over 2}=
{1\over2}\lim_{n\to\infty}{\rm L}\left({(2n+2)(2n+1)\over(n+1)^2}\right)=
{1\over2}{\rm L}(4)={\rm L}(2).$$
A: 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
$$
\frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi n}}\tag{10}
$$
A: $Let \lim_{n\to \infty} \frac{1}{2n} log{2n\choose n}=L$
This limit can be solved by Squeeze Theorem, by the following inequality
$\lim_{n\to \infty} \frac{1}{2n} log{\frac{4^n}{n+1}}\le L\le\lim_{n\to \infty} \frac{1}{2n} log{4^n}$
Now,upon a little manipulation,left limit tends to $log2$ and right limit tends to $log2$, so $L=log2$
Hence solved.
Also, the proof of inequality can be obtained by induction hypothesis.
Sorry, could not provide full solution but it is also and idea!
