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.

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 $n^2$,
$$\dfrac1{n^2}\sum_{k=0}^n \ln\binom{n}{k} 
\to \dfrac12
\text{ as }n \to \infty
$$
