TITLE: What Is The Limit Of The Sequence: $\frac{n^3}{{((3n)!)^\frac{1}{n}}}$
QUESTION [5 upvotes]: What (if exists) the $\lim \limits_{n\to \infty}\dfrac{n^3}{{((3n)!)^\frac{1}{n}}}$?
I have no idea where to begin. Maybe I could use the ratio test? Please try to keep it as elementary as possible because we are only in the beginning of the course.
Thanks a lot.

REPLY [6 votes]: Stirling's approximation gives
$$
\lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e\tag{1}
$$
Using $(1)$ yields
$$
\begin{align}
\lim_{n\to\infty}\frac{n^3}{{((3n)!)^{\frac1n}}}
&=\lim_{n\to\infty}\frac{1}{27}\left(\frac{3n}{((3n)!)^{\frac{1}{3n}}}\right)^3\\
&=\frac{1}{27}\left(\lim_{n\to\infty}\frac{1}{3n}((3n)!)^{\frac{1}{3n}}\right)^{-3}\\
&=\frac{1}{27}\left(\frac1e\right)^{-3}\\
&=\frac{e^3}{27}\tag{2}
\end{align}
$$

Alternate Proof of $\boldsymbol{(1)}$
$$
\begin{align}
\log\left(\lim_{n\to\infty}\frac1n(n!)^{1/n}\right)
&=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac kn\right)\\
&=\int_0^1\log(x)\,\mathrm{d}x\\[9pt]
&=-1\tag3
\end{align}
$$
which is equivalent to $(1)$.