A limit on binomial coefficients Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.
What I can do is just use Stolz formula. But I could not proceed.

The limit is $\frac{1}{2}$. 
We have 
$$\begin{eqnarray*}
\sum_{k=0}^n \log {n\choose k} 
&=& \sum_{k=0}^n \log n! - \sum_{k=0}^n \log k! - \sum_{k=0}^n \log (n-k)! \\
&=& (n+1)\log n! - 2\sum_{k=1}^n \log k!. 
\end{eqnarray*}$$
But 
$$\begin{eqnarray*}
\sum_{k=1}^n \log k! 
&=& \sum_{k=1}^n \sum_{j=1}^k \log j \\
&=& \sum_{j=1}^n \sum_{k=j}^n \log j \\
&=& \sum_{j=1}^n (n-j+1)\log j \\
&=& (n+1)\sum_{j=1}^n \log j - \sum_{j=1}^n j\log j \\
&=& (n+1)\log n! - n^2 \frac{1}{n}\sum_{j=1}^n \frac{j}{n}\log \frac{j}{n} - \sum_{j=1}^n j\log n \\
&=& (n+1)\log n! - n^2\int_0^1 dx\ x\log x - \frac{n(n+1)}{2}\log n + O(n\log n) \\
&=& (n+1)\log n! +\frac{n^2}{4} - \frac{n(n+1)}{2}\log n + O(n\log n). 
\end{eqnarray*}$$
(The error estimate above can probably be tightened.)
Using Stirling's approximation we find 
$$\begin{eqnarray*}
\frac{1}{n^2}\sum_{k=0}^n \log {n\choose k}
&=& -\frac{n+1}{n^2}(\log n! - n\log n) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& -\frac{n+1}{n^2}(-n + O(\log n)) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& \frac{1}{2} + O\left(\frac{\log n}{n}\right).
\end{eqnarray*}$$
