Find $\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$ 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.

Let $a_n = \dfrac{3^{3n}(n!)^3}{(3n)!}$.
The limit $\lim a_n^{1/n}$ is related to the root test for series convergence.
In this context, it is well known that if $\lim \dfrac{a_{n+1}}{a_n}$ exists, then so does $\lim a_n^{1/n}$ and they are equal. Let's compute the ratios then:
$$
\lim \frac{a_{n+1}}{a_n} = \lim 3^3 \dfrac{(n+1)^3}{(3n+3)(3n+2)(3n+1)} = 1
$$
