Calculate $\lim\limits_{n\to\infty}\dfrac{\sum\limits_{k=0}^n\log\binom{n}{k}}{n^2}$ How to prove if the following limit exists?
If it exists, what's the value?
$$\lim\limits_{n\to\infty}\dfrac{\sum\limits_{k=0}^n\log\binom{n}{k}}{n^2}$$
Thanks!
 A: Notice that we can write
$$ \frac{1}{n^2} \sum_{k=1}^{n} \log \binom{n}{k} = \frac{1}{n}\sum_{k=1}^{n} \left( \frac{2k-1}{n} - 1 \right) \log\left(\frac{k}{n}\right). $$
Taking $n \to \infty$, this converges to the integral
$$ \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x = \frac{1}{2}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\lim_{n \to \infty}{\sum_{k = 0}^{n}\ln\pars{n \choose k} \over n^{2}} & =
\lim_{n \to \infty}
{\sum_{k = 1}^{n + 1}\ln\pars{n + 1 \choose k} -
\sum_{k = 1}^{n}\ln\pars{n \choose k} \over \pars{n + 1}^{2} - n^{2}} =
\lim_{n \to \infty}
{\sum_{k = 1}^{n}\ln\pars{{n + 1 \choose k}/{n \choose k}} \over 2n + 1}
\\[5mm] & =
{1 \over 2}\,\lim_{n \to \infty}
{\sum_{k = 1}^{n}\bracks{\ln\pars{n + 1} - \ln\pars{n - k + 1}} \over n + 1/2}
\\[5mm] & =
{1 \over 2}\,\lim_{n \to \infty}{n\ln\pars{n + 1} - \ln\pars{n!} \over n + 1/2}
\\[5mm] & =
{1 \over 2}\,\lim_{n \to \infty}\bracks{\vphantom{\Large A}%
\pars{n + 1}\ln\pars{n + 2} - n\ln\pars{n + 1} - \ln\pars{n + 1}}
\\[5mm] & =
{1 \over 2}
\lim_{n \to \infty}\bracks{\pars{n + 1}\ln\pars{1 + 2/n \over 1 + 1/n}} =
\bbx{1 \over 2}
\end{align}
