Compute the limit of $\frac1{\sqrt{n}}\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)^{1/n^2}$ 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.
 A: 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!
A: Another way: by Stolz-Cesaro theorem, we have
$$\begin{align}
L&=\lim_{n\to\infty} \frac{1}{n^2} \left(\sum_{k=1}^{n} k\log (k) - \frac{n^2\log(n) }{2}\right)\\
&=\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}\\
&=\lim_{n\to\infty} \frac{1}{n^2-(n-1)^2} \left(n\log (n) - \frac{n^2\log(n) }{2}+\frac{(n-1)^2\log(n-1)}{2}\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(2n\log (n) -n^2\log(n)+(n-1)^2(\log(n)+\log(1-1/n))\right)\\
&=\lim_{n\to\infty} \frac{1}{4n-2} \left(-n+o(n)\right)=-\frac{1}{4}
\end{align}$$
and therefore
$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}}=e^{-1/4}.$$
A: $$
\frac1{n^2}\sum_{k=1}^nk\log(k)-\frac12\log(n)=\frac1{n}\sum_{k=1}^n\frac{k}n\log\left(\frac{k}n\right)+\frac12\frac{\log(n)}n=\int_0^1x\log(x)\mathrm dx+o(1)
$$

Edit: Per request, a solution without integrals, using only the elementary version of Stirling's approximation. Let $s_n=\displaystyle\sum_{k=1}^nk\log(k)$. Then,
$$
\sum_{k=1}^n\log(k!)=\sum_{k=1}^n\sum_{i=1}^k\log(i)=\sum_{i=1}^n(n-i+1)\log(i)=(n+1)\log(n!)-s_n.
$$
On the other hand, Stirling's approximation in its simplest form is $\log(k!)=k\log(k)-k+r_k$ with $r_k=O(\log k)$, which yields
$$
\sum_{k=1}^n\log(k!)=\sum_{k=1}^nk\log(k)-\sum_{k=1}^nk+\sum_{k=1}^nr_k=s_n-\frac12n(n+1)+t_n,\quad t_n=\sum_{k=1}^nr_k.
$$
Comparing these two expressions and using once more $\log(n!)=n\log(n)-n+r_n$, one gets
$$
2s_n=n(n+1)\log n-n(n+1)+(n+1)r_n+\frac12n(n+1)-t_n,
$$
hence $2s_n-n^2\log n=-\frac12n^2+u_n$ with
$$
u_n=n\log(n)-\frac12n+(n+1)r_n-t_n.
$$
Since $r_n=O(\log n)$, $t_n=O(n\log n)$ and each term in $u_n$ is $O(n\log n)$, hence
$$
\frac{s_n}{n^2}-\frac{\log n}2=-\frac14+\frac{u_n}{2n^2}=-\frac14+O\left(\frac{\log n}{n}\right)\to-\frac14.
$$
