How to evaluate $\lim\limits_{n\to +\infty}\frac{1}{n^2}\sum\limits_{i=1}^n \log \binom{n}{i} $ I have to compute:
$$ \lim_{n\to +\infty}\frac{1}{n^2}\sum_{i=1}^{n}\log\binom{n}{i}.$$
I have tried this problem and hit the wall.

We have:
$$ \prod_{i=1}^{n}\binom{n}{i} = \frac{n!^n}{\left(\prod_{i=1}^{n}i!\right)\cdot\left(\prod_{i=1}^{n-1}i!\right)}\tag{1}$$
hence:
$$ \frac{1}{n^2}\sum_{i=1}^{n}\log\binom{n}{i}=\frac{1}{n^2}\left((n-1)\log\Gamma(n+1)-2\sum_{i=1}^{n-1}\log\Gamma(i+1)\right)\tag{2}$$
and by Stirling's approximation:
$$ \log\Gamma(z+1) = \left(z+\frac{1}{2}\right)\log z-z+O(1) \tag{3}$$
and partial summation we get:
$$ 2\sum_{i=1}^{n-1}\log\Gamma(i+1) = n^2\left(\log n-\frac{3}{2}\right)+O(n)\tag{4}$$
so:
$$ \sum_{i=1}^{n}\log\binom{n}{i} = \frac{n^2}{2}+O(n\log n)\tag{5} $$
and our limit is just $\color{red}{\frac{1}{2}}.$

A simpler approach is given by the identity:
$$\begin{eqnarray*}\sum_{i=1}^{n}\log\binom{n}{i}&=&(n-1)\log n!-2\sum_{i=1}^{n-1}\sum_{k=1}^{i}\log k \\ &=&(n-1)\log n+\sum_{k=1}^{n-1}(2k-n-1)\log k\tag{6}\end{eqnarray*}$$
hence partial summation gives:
$$\begin{eqnarray*}\sum_{i=1}^{n}\log\binom{n}{i}&=&(1-n)\log\left(1-\frac{1}{n}\right)-\sum_{k=1}^{n-2}(k^2-kn)\log\left(1+\frac{1}{k}\right)\\&=&O(1)+\sum_{k=1}^{n-2}(n-k)+O\left(\sum_{k=1}^{n-2}\frac{n-k}{k}\right)\\&=&\frac{n^2}{2}+O(n\log n).\tag{7}\end{eqnarray*} $$

Still another approach through summation by parts:
$$\begin{eqnarray*}\sum_{i=1}^{n}\log\binom{n}{i}&=&-\sum_{i=1}^{n-1}i\cdot\log(i+1)+\sum_{i=1}^{n-1}(n-i)\log(i)\\&=&O(n)+\sum_{i=1}^{n-1}(2i-n)\log(n-i)\\&=&O(n)+\sum_{i=1}^{n-1}(2i-n)\log\left(1-\frac{i}{n}\right)\tag{8}\end{eqnarray*}$$
gives that the limit is provided by a Riemann sum:
$$\begin{eqnarray*}\lim_{n\to +\infty}\frac{1}{n^2}\sum_{i=1}^{n}\log\binom{n}{i}&=&\int_{0}^{1}(2x-1)\log(1-x)\,dx\\&=&\int_{0}^{1}\frac{x^2-x}{x-1}\,dx\\&=&\int_{0}^{1}x\,dx=\color{red}{\frac{1}{2}}.\tag{9}\end{eqnarray*}$$
This limit can be related with the entropy of the binomial distribution through the Kullback-Leibler divergence, for instance.
