What is the limit of this sequence involving logs of binomials? Define  $$P_{n} = \frac{\log \binom{n}{0} + \log \binom{n}{1} + \dots +\log \binom{n}{n}}{n^{2}}$$ where $n = 1, 2, \dots$ and $\log$ is the natural log function. Find $$\lim_{n\to\infty} P_{n}$$ Using the property of the $\log$, I am thinking to find the product of all the binomial coefficients. But it gets really messy.

Let's see if I can come up
with a simple proof
that the limit is 1/2.
The log of the numerator
(of the product of binomial coefficients)
 is easy:
$\dfrac{(n+1)\ln n!}{n^2}
\sim \dfrac{(n+1)(n \ln n - n + O(\ln n))}{n^2}
=(\ln n) -1 + o(1)
 $.
For the log of the denominator
(of the product of binomial coefficients)
,
we need 
$\begin{array}\\
d_n
&=\sum_{k=0}^n \ln k!\\
&=\sum_{k=1}^n \sum_{i=1}^k \ln i\\
&=\sum_{i=1}^n \sum_{k=i}^n \ln i\\
&=\sum_{i=1}^n (n-i+1) \ln i\\
&=\sum_{i=1}^n (n+1) \ln i-\sum_{i=1}^n i \ln i\\
&=(n+1)\ln(n!)-\sum_{i=1}^n i (\ln (i/n)+\ln(n))\\
&=(n+1)\ln(n!)-\sum_{i=1}^n i \ln (i/n)-\sum_{i=1}^n i \ln(n)\\
&=(n+1)\ln(n!)-\sum_{i=1}^n i \ln (i/n)-\ln(n)n(n+1)/2\\
&\sim (n+1)(n \ln n - n + O(\ln n))-\sum_{i=1}^n i \ln (i/n)-(n^2/2)\ln(n)+O(n \ln n)\\
&\sim \frac12 n^2 \ln(n)-n^2+o(n^2)-\sum_{i=1}^n i \ln (i/n)\\
\end{array}
$
The denominator
is
$\dfrac{2d_n}{n^2}
\sim\ln(n)-2-\frac{2}{n}\sum_{i=1}^n \frac{i}{n} \ln (i/n) + o(1)
$.
Subtracting them,
we get
$1+2\frac1{n}\sum_{i=1}^n \frac{i}{n} \ln (i/n) + o(1)
$.
That sum is a
Riemann approximation
(which is, of course,
what I was trying for)
to
$\int_0^1 x \ln(x) dx
=\frac12 x^2 \ln x - \frac{x^2}{4}|_0^1
=-\frac14
$.
The final result is
$1-2\frac14
=\frac12
$
as it should be.
