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.

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
$$
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!}  \right)-\ln\left(\frac{(n+1)^n}{(n+1)!}  \right)}{2} \\
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n+1}}{(n+2)!}  \frac{(n+1)!}{(n+1)^n}  \right)}{2} \\
=\lim\limits_{n\to\infty}\frac { \ln\left(\frac{(n+2)^{n}}  {(n+1)^n}  \right)}{2}\\
=\lim\limits_{n\to\infty}\frac { \ln\left(1+\frac{1}{n+1} \right)^n}{2}\\
=\frac{\ln(e)}{2}=\frac{1}{2}
$$
