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.
$$Q_n=\prod_{k=1}^{n-1}\binom{n}{k} = \frac{(n!)^{n-1}}{\left(\prod_{k=1}^{n-1} k!\right)^2}=\prod_{k=1}^{n-1}\frac{n!}{(k!)^2}$$

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.
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}
&lt;\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}
&lt;\dfrac{2}{n^{1/n}}
&lt; 2
$
or
$\dfrac{\sum_{i=0}^n \ln \binom{n}{i}}{n^2}
&lt; \ln 2
$.