TITLE: Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$
QUESTION [6 upvotes]: Evaluating $$\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$$

I'm not quite sure where to start in evaluating this. Some pointers, or a solution, would greatly be appreciated.

REPLY [8 votes]: Since
$$
\int_1^n\log(x)\,\mathrm{d}x\le\log(n!)\le\int_1^{n+1}\log(x)\,\mathrm{d}x\tag{1}
$$
we have that
$$
\log(n!)=n\log(n)-n+O(\log(n))\tag{2}
$$
therefore,
$$
\log\left(\frac{n^{n-1}}{n!}\right)=n+O(\log(n))\tag{3}
$$

Note that
$$
\left.\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\middle/\frac{((n-1)!)^{n-1}}{(0!1!2!\cdots(n-1)!)^2}\right.
=\frac{n^{n-1}}{n!}\tag{4}
$$
Applying $(3)$, we have
$$
\begin{align}
\log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right)
&=\sum_{k=1}^n\log\left(\frac{k^{k-1}}{k!}\right)\\
&=\sum_{k=1}^n(k+O(\log(k)))\\
&=\frac{n^2}2+O(n\log(n))\tag{5}
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}\frac1{n^2}\log\left(\frac{(n!)^n}{(0!1!2!\cdots n!)^2}\right)=\frac12\tag{6}
$$