TITLE: Find $\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$
QUESTION [5 upvotes]: Find $$\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$$
I don't know what method to use, if  we divide numerator and denominator with $3^{3n}$, I don't see that we win something. I can't find two sequences, to use than the squeeze theorem. I'm stuck.

REPLY [3 votes]: The Stolz-Cesàro Theorem says that if
$$
\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=L
$$
then
$$
\lim_{n\to\infty}\frac{a_n}{b_n}=L
$$
This is the discrete, pre-calculus analog of L'Hôpital.

Suppose we let
$$
a_n=\log\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)
$$
and
$$
b_n=n
$$
Then
$$
a_{n+1}-a_n=\log\left(\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)
$$
and
$$
b_{n+1}-b_n=1
$$
From this, we get
$$
\begin{align}
\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}
&=\lim_{n\to\infty}\log\left(\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)\\
&=\log\left(\lim_{n\to\infty}\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)\\[3pt]
&=\log(1)\\[9pt]
&=0
\end{align}
$$
Therefore,
$$
\begin{align}
\log\left(\lim_{n\to\infty}\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}\right)
&=\lim_{n\to\infty}\frac{\log\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)}n\\
&=\lim_{n\to\infty}\frac{a_n}{b_n}\\
&=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\\[3pt]
&=0
\end{align}
$$
From which we get
$$
\lim_{n\to\infty}\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}=1
$$