calculus - Compute the limit of $frac1{sqrt{n}}left(1^1 cdot 2^2 cdot3^3cdots n^nright)^{1/n^2}$

Compute the following limit:

$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}} $$

I'm interested in almost any approaching way for this limit. Thanks.

Answer

Let's begin
$$
\lim\limits_{n\to\infty}\frac{\left(\prod\limits_{k=1}^n k^k\right)^{\frac{1}{n^2}}}{\sqrt{n}}=
\lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log k - \frac{1}{2}\log n\right)=
$$
$$
\lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log\left(\frac{k}{n}\right)+\frac{1}{n^2}\sum\limits_{k=1}^n k\log n - \frac{1}{2}\log n\right)=
$$
$$

\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\log n\left(\frac{n^2+n}{n^2}-1\right)\right)=
$$
$$
\exp\left(\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\lim\limits_{n\to\infty}\frac{\log n}{n}\right)=
$$
$$
\exp\left(\int\limits_{0}^1 x\log x dx\right)=\exp\left(-1/4\right)
$$
And now we are done!