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 $\left(1-\frac{1}{k-1}\right)^{k-1}\to e^{-1}$, so you are done.
So I get $P_n\to\frac{1}{2}$ as the limit.
