What is $\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k} $? It was originally asked on another website but nobody has been able to prove the numerical result. The attempts usually go by Stirling's approximation or try to use the Silverman-Toeplitz theorem.
 A: As already noted by @SimpleArt, the expression equals
$$\tag 1 (n+1)\ln n!-2\sum_{k=1}^n\ln k!.$$
The sum on the right equals
$$\sum_{k=1}^n\sum_{j=1}^{k}\ln j = \sum_{j=1}^{n}(n—j+1)\ln j= (n+1)\ln n! - \sum_{j=1}^{n} j\ln j.$$
Therefore $(1)$ equals
$$\tag 2 -(n+1)\ln n! +2 \sum_{j=1}^{n} j\ln j.$$
Now $\ln n! = \sum_{k=1}^{n} \ln k,$ and this sum can be compared nicely with $\int_1^n \ln x \, dx$ to get
$$\ln n! = n\ln n - n +O(\ln n).$$
Similarly, $\sum_{j=1}^{n} j\ln j $ can be compared to $\int_1^\infty x \ln x \, dx$ to give
$$\sum_{j=1}^{n} j\ln j = \frac{n^2\ln n}{2}-\frac{n^2}{4} +O(n\ln n).$$
Putting this altogether, we see some nice cancellation shows $(2)$ equals $n^2/2 + O(n\ln n).$ Dividing by $n^2$ then gives $1/2$ for the desired limit.
A: By Stolz Cezaro
$$\lim\limits_{n\to\infty}\frac {1}{n^2}\sum\limits_{k=0}^{n}\ln\binom{n}{k}=\lim\limits_{n\to\infty}\frac {1}{2n+1} \left(\sum\limits_{k=0}^{n+1}\ln\binom{n+1}{k}-  \sum\limits_{k=0}^{n}\ln\binom{n}{k} \right)\\
=\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\left(\ln\binom{n+1}{k}- \ln\binom{n}{k} \right)\\
=\lim\limits_{n\to\infty}\frac {1}{2n+1} \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)!}{k!(n+1-k)!} \frac{k!(n-k)!}{n!} \right)\\
=\lim\limits_{n\to\infty}\frac { \sum\limits_{k=0}^{n}\ln\left(\frac{(n+1)}{(n+1-k)}  \right)}{2n+1}\\
=\lim\limits_{n\to\infty}\frac { \ln\left(\prod\limits_{k=0}^{n}\frac{(n+1)}{(n+1-k)}  \right)}{2n+1}\\
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+1)^n}{(n+1)!}  \right)}{2n+1}\\
$$
Applying stolz Cezaro again, we get
$$
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!}  \right)-\ln\left(\frac{(n+1)^n}{(n+1)!}  \right)}{2} \\
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!}  \frac{(n+1)!}{(n+1)^n}  \right)}{2} \\
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n}}  {(n+1)^n}  \right)}{2}\\
=\lim\limits_{n\to\infty}\frac { \ln\left(1+\frac{1}{n+1} \right)^n}{2}\\
=\frac{\ln(e)}{2}=\frac{1}{2}
$$
A: Here is as exact an answer as you want,
from my answer here:
Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$
$$\prod_{k=0}^n \binom{n}{k} \sim
C^{-1}\frac{e^{n(n+2)/2}}{n^{(3n+2)/6}(2\pi)^{(2n+1)/4}}
\exp\big\{-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\big\}\text{ as }n \to \infty
$$
where
$$\begin{align}
C
&= \lim_{n \to \infty} 
\frac1{n^{1/12}}
\prod_{k=1}^n \big\{k!\big/\sqrt{2\pi k}\big(\frac{k}{e}\big)^k\big\}\\
&\approx 1.04633506677...\\
\end{align}
$$
and the $\{B_p\}$
are the Bernoulli numbers,
defined by
$$\sum_{p \ge 0} B_p\frac{x^p}{p!} = \frac{x}{e^x-1}
.$$
Taking logs,
$$\sum_{k=0}^n \ln\binom{n}{k} \sim
\ln C+n(n+2)/2-(3n+2)\ln n/6-(2n+1)\ln(2\pi)/4
-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\\
\text{ as }n \to \infty
$$
Dividing by $n^2$,
$$\dfrac1{n^2}\sum_{k=0}^n \ln\binom{n}{k} 
\to \dfrac12
\text{ as }n \to \infty
$$
