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.
 A: 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 
$$\begin{eqnarray*}
\frac{1}{n^2}\sum_{k=0}^n \log {n\choose k}
&=& -\frac{n+1}{n^2}(\log n! - n\log n) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& -\frac{n+1}{n^2}(-n + O(\log n)) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& \frac{1}{2} + O\left(\frac{\log n}{n}\right).
\end{eqnarray*}$$
A: The sum does not diverge, but converges. I suspect it converges to $1/2$ based on the value at $n=5000$ [link], but I've only managed to bound it by $3/4$.
From Wikipedia, 
$$\binom{n}{k} \leq \left(\frac{n \cdot e}{k}\right)^k$$
Then,
$$\begin{align} x_n=\frac{1}{n^2}\sum_{k=0}^n\ln \binom{n}{k}&\leq \frac{1}{n^2}\sum_{k=0}^n\ln \left(\frac{n \cdot e}{k}\right)^k\\
&=-\frac{1}{n^2}\sum_{k=0}^n k\ln \left(\frac{k}{n \cdot e}\right)\\
&=-\frac{1}{n}\sum_{k=0}^n \left(\frac{k}{n}\right)\ln \left(\frac{k}{n \cdot e}\right)\end{align}$$
Looking at this as a Riemann Sum,
$$\begin{align}
\lim_{n \to \infty} x_n &\leq \lim_{n \to \infty}-\frac{1}{n}\sum_{k=0}^n \left(\frac{k}{n}\right)\ln \left(\frac{k}{n \cdot e}\right) \\
&= \int_0^1 -x\ln\left(\frac{x}{e}\right)\mathrm{d}x
\\
&=\left.\frac{x^2}{4}-\frac{x^2}{2}\ln x+\frac{x^2}{2}\ln e\right|_0^1
\\
&=\frac{3}{4}
\end{align}$$
You can also bound it from below by $\frac{1}{4}$ by taking $\binom{n}{k} \geq  \left(\frac{n}{k}\right)^k$.
A: I've managed to reduce it somewhat.
$$x_n=\frac{1}{n^2}\ln(\prod_{k=0}^n \binom{n}{k})$$
$$x_n=\frac{1}{n^2}\ln \left ( \frac{n!}{(n-0)!0!}\frac{n!}{(n-1)!1!}\frac{n!}{(n-2)!2!}...\frac{n!}{(n-n)!n!}\right )$$
$$x_n=\frac{1}{n^2}\ln \left ( \frac{n!^{(n+1)}}{(0!1!2!...n!)^2} \right )$$
$$x_n=\frac{n+1}{n^2}\ln \left ( n! \right )-\frac{2}{n^2}\ln \left ( 0!1!2!...n! \right )$$
$$x_n=\frac{n+1}{n^2}\ln \left ( n! \right )-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$
Using Stirling's approximation for large $n$ ($ \ln(n!)=n \ln(n)-n$):
$$x_n=\frac{n+1}{n^2}(n \ln(n)-n)-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$
$$x_n=\frac{n \ln(n)-n}{n}+\frac{n \ln(n)-n}{n^2}-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$
$$x_n=\ln(n)-1+\frac{ \ln(n)-1}{n}-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$
Noting that $\lim_{n \rightarrow \infty} \frac{\ln(n)}{n}=0$, for $n \rightarrow \infty$:
$$x_n=\ln(n)-1-\frac{2}{n^2}\sum_{k=0}^n \ln(k!)$$
A: $x_n=\frac{1}{n^2}\sum_{k=0}^{n}\ln{n\choose k}=\frac{1}{n^2}\ln(\prod {n\choose k})=\frac{1}{n^2}\ln\left(\frac{n!^n}{n!^2.(n-1)!^2(n-2)!^2...0!^2}\right)$ 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.
A: $$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{1}{n^2}\sum_{k=0}^{n}ln\frac{n!}{k!(n-k)!}
=\frac{1}{n^2}\sum_{k=0}^{n}(ln(n!)-ln(k!)-ln((n-k)!))
=\frac{1}{n^2}\sum_{k=0}^{n}ln(n!)-\frac{2}{n^2}\sum_{k=1}^{n}ln(k!)
=\frac{n+1}{n^2}ln(n!)-\frac{2}{n^2}\sum_{k=0}^{n}ln(k!)
$$
because ln(n!)~nln(n)-n,so
$$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{n+1}{n^2}(nln(n)-n)-\frac{2}{n^2}\sum_{k=1}^{n}(kln(k)-k)
$$
then I am going to prove $\sum_{k=1}^{n}(kln(k)) \rightarrow  \frac{n^2}{2}(ln(n)-1/2)$
$$
\int_{i=x-1}^{x}(iln(i))dx<xln(x)<\int_{i=x}^{x+1}(iln(i))dx
$$
and
$$
\int(xln(x))dx=x^2/2(lnx-1/2)
$$
so sum it up, we got
$$
\sum_{k=1}^{n}(kln(k)) \rightarrow  \frac{n^2}{2}(ln(n)-\frac{1}{2})
$$
then
$$
\frac{1}{n^2}\sum_{k=0}^{n}ln\binom n k
=\frac{n+1}{n^2}(nln(n)-n)-\frac{2}{n^2}\sum_{k=1}^{n}(kln(k)-k)
=\frac{n+1}{n}(ln(n)-1)-\frac{2}{n^2}(\frac{n^2}{2}(lnn-\frac{1}{2})-\frac{n(n+1)}{2})
=ln(n)-1-(ln(n)-1/2-1)=1/2
$$
A: For a two terms estimate of the behaviour of $\displaystyle\prod_{k=0}^{n}\binom{n}{k}$ one can check out March 2013 volume of Asymmetry problem V2-4 here (http://akotronismaths.blogspot.gr/p/asymmetry-electronic-mathematical.html)
