Q:

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. 

A:

$$Q_n=\prod_{k=1}^{n-1}\binom{n}{k} = \frac{(n!)^{n-1}}{\left(\prod_{k=1}^{n-1} k!\right)^2}=\prod_{k=1}^{n-1}\frac{n!}{(k!)^2}$$
So $$Q_{n+1} =\frac{(n+1)^{n-1}(n+1)!}{(n!)^2}Q_n=\frac{(n+1)^n}{n!}Q_n$$
So $$\log Q_{n+1} =\log Q_n + n\log(n+1)-\log(n!)$$
So:
$$\log Q_n = \sum_{k=1}^n \left((k-1)\log k -\log\left((k-1)!\right)\right)$$
Now $\log (k-1)!  = (k-1)\log (k-1) - (k-1) + O(\log k)$ so:
$$\log Q_n = \sum_{k=2}^n (k-1)\left(\log\left(1-\frac{1}{k-1}\right)\right) +\frac{n(n-1)}{2} + O(\log(n!))$$
So $$P_n =\frac{\log Q_n}{n^2}=\frac{1}{n^2}\sum_{k=1}^n \log \left(\left(1-\frac{1}{k-1}\right)^{k-1}\right)+ \frac{n(n-1)}{2n^2} + O\left(\frac{\log n}{n}\right)$$
But $\left(1-\frac{1}{k-1}\right)^{k-1}=O(1)$, so you are done. 
So I get $P_n\to\frac{1}{2}$ as the limit. 

A:

The limit is $1/2$.
Here is a proof by authority:
This is copied from here:
Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$
$$\prod_{k=0}^n \binom{n}{k} \sim
C^{-1}\frac{e^{n(n+2)/2}}{n^{(3n+2)/6}(2\pi)^{(2n+1)/4}}
\exp\big\{-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\big\}\text{ as }n \to \infty
$$
where
$$\begin{align}
C
&= \lim_{n \to \infty} 
\frac1{n^{1/12}}
\prod_{k=1}^n \big\{k!\big/\sqrt{2\pi k}\big(\frac{k}{e}\big)^k\big\}\\
&\approx 1.04633506677...\\
\end{align}
$$
and the $\{B_p\}$
are the Bernoulli numbers,
defined by
$$\sum_{p \ge 0} B_p\frac{x^p}{p!}  = \frac{x}{e^x-1}
.$$
If the $n^2$ root is taken,
the only term that
does not go to $1$
is
$e^{1/2}$
(from
$e^{n(n+2)/2}$)
so that
$\left(\prod_{k=0}^n \binom{n}{k}\right)^{1/n^2}
\sim e^{1/2}
$
or
$\dfrac{\sum_{k=0}^n \ln \binom{n}{k}}{n^2}
\sim 1/2
$. 
There is probably a relatively simple proof. 
It might be easier to find 
knowing the answer.