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.

$x_n=\frac{1}{n^2}\sum_{k=0}^{n}\ln{n\choose k}=\frac{1}{n^2}\ln(\prod {n\choose k})=\frac{1}{n^2}\ln\left(\frac{n!^n}{n!^2.(n-1)!^2(n-2)!^2...0!^2}\right)$ since ${n\choose k}=\frac{n!}{k!(n-k)!}$
$e^{n^2x_n}=\left(\frac{n^n(n-1)!}{n!^2}\right)e^{(n-1)^2x_{n-1}}=\left(\frac{n^{n-1}}{n!}\right)e^{(n-1)^2x_{n-1}}$
By Stirling's approximation, $n! \sim n^ne^{-n}\sqrt{2\pi n}$
$e^{n^2x_n}\sim \left(\frac{e^n}{n\sqrt{2\pi n}}\right)e^{(n-1)^2x_{n-1}}$
$x_n \sim \frac{(n-1)^2}{n^2}x_{n-1}+\frac{1}{n}-\frac{1}{n^2}\ln(n\sqrt{2\pi n})$
The $\frac{1}{n}$ term forces $x_n$ to tend to infinity, because $\frac{1}{n^2}\ln(n\sqrt{2\pi n})$ does not grow fast enough to stop it diverging.
