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: 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.
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 $\left(1-\frac{1}{k-1}\right)^{k-1}\to e^{-1}$, so you are done.
So I get $P_n\to\frac{1}{2}$ as the limit.
A: 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.
A: Here's an upper bound
(probably not good enough)
for a start:
From the arithmetic-geometric mean inequality,
$\prod_{i=0}^n \binom{n}{i}
<\left(\dfrac{\sum_{i=0}^n \binom{n}{i}}{n}\right)^n
=\left(\dfrac{2^n}{n}\right)^n
=\dfrac{2^{n^2}}{n^n}
$
so
$(\prod_{i=0}^n \binom{n}{i})^{1/n^2}
<\dfrac{2}{n^{1/n}}
< 2
$
or
$\dfrac{\sum_{i=0}^n \ln \binom{n}{i}}{n^2}
< \ln 2
$.
