What Is The Limit Of The Sequence: $\frac{n^3}{{((3n)!)^\frac{1}{n}}}$ 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.
 A: 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 $(1)$.
A: We use the fact that if $(a_n)$ is a sequence of positive numbers and if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}}$ and the two limits are equal.
We apply this  to the sequence with terms $a_n={n^{3n}\over (3n)!}$.
We have:
$$
{a_{n+1}\over a_n} = {(n+1)^{3n+3}\over (3n+3)!}\cdot {(3n)!\over n^{3n}}
={(n+1)^3\over (3n+3)(3n+2)(3n+1) }\cdot {\Bigl({n+1\over n}\Bigr)^{3n} }
\quad\buildrel{n\rightarrow\infty}\over\longrightarrow\quad{e^3\over27}.
$$ 
A: Ok, since David posted the approach I was thinking about, I wanted to see (just for fun) if there are other ways to find the limit in a completely basic fashion.
There is! but this is a bit long, so I might skip some details.
First we state a few Lemmas:
Lemma A: 
If $x_n$ is a monotonically increasing sequence, and converges to $L$, then for all $n$, $L \ge x_n$. 
Similarly if $y_n$ is a monotonically decreasing sequence, and $y_n$ converges to $R$, then for all $n$, $R \le y_n$.
Proof: Left to the reader.
Lemma B:
$x_n = \left(1 + \frac{1}{n}\right)^n$ is a monotonically increasing sequence.
Proof: Consider $n$ copies of $1 + \frac{1}{n}$ and $1$ copy of $1$ and apply $\text{AM} \ge \text{GM}$.
$\square$
Lemma C:
$y_n = \left(1 + \frac{1}{n}\right)^{n+1}$ is monotonically decreasing.
Proof: Consider $n$ copies of $1 - \frac{1}{n}$ and $1$ copy of $1$ and apply $\text{AM} \ge \text{GM}$.
$\square$
Note that Lemmas B and C can also be proven using Bernoulli's inequality, though that might not be considered basic (but surely can be considered elementary). These proofs (including the above AM,GM proofs) have appeared on this site before, if I remember correctly.
Combining the above Lemmas, leads us to:
Proposition D:
$$ \left(1 + \frac{1}{n}\right)^{n+1} \ge e \ge \left(1 + \frac{1}{n}\right)^n$$
for all $n$.
Proof:  $x_n$ and $y_n$ of Lemmas B and C, both converge to $e$ and we now have the inequality by Lemma A.
Note that, you can define $e$ this way too. First, $y_n$ is convergent as it is monotonically decreasing and bounded below. That means that $x_n$ is convergent too and to the same limit. Call that limit $e$.
Using binomial theorem, one can show that $x_n \lt 3$, and so $e \le 3$ (this can probably be found in most textbooks which define $e$ as the limit of $x_n$).
$\square$
Now we come to the main result:
Proposition E
For all $n \ge 4$, we have that
$$ \left(\frac{n}{e}\right)^{n+8} \ge n! \ge \left(\frac{n}{e}\right)^{n-8}$$
Proof:
By induction on $n$.
The base case is easily verifed using $e \le 3$.
Let $A_n = \left(\frac{n}{e}\right)^{n+8}$. Then
$$\frac{A_{n+1}}{A_n} = (n+1)\frac{\left(1 + \frac{1}{n}\right)^{n+8}}{e} \ge n+1$$
using Proposition D.
Thus if $A_n \ge n!$, then $A_{n+1} \ge (n+1)!$.
Similary, if $B_n = \left(\frac{n}{e}\right)^{n-8}$, the we can show that
$B_{n+1} \le (n+1)B_n$ and by induction, $B_{n+1} \le (n+1)!$.
$\square$
Now we can apply Proposition E to your sequence to squeeze your term and get that the limit is $\frac{e^3}{27}$.
I will leave that to the reader.
