A more concise way of showing the limit of $a_n := (n^{3n})/(n3^n)$ I am trying to show the limit of the sequence
$a_n:=\frac{n^{3n}}{n3^n}$
I have done it using the ratio test for limits as I will illustrate below, but I feel like there is an easier method but cannot find one; it would sate my curiosity to see it done in less lines:
Let $a_{n+1}:=\frac{(n+1)^{3n+3}}{(n+1)3^{n+1}}$
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{3n+3}}{(n+1)3^{n+1}}\times\frac{n3^n}{n^{3n}}=\frac{n}{3}(n+1)^2(1+\frac{1}{n})^{3n}\ge \frac{n}{3}(n+1)$$
We are utilising the comparison test to observe that $\frac{a_{n+1}}{a_n}$ is greater than a sequence which clearly tends to infinity, and thus by the ratio test $lim\frac{a_{n+1}}{a_n}>1$ as n tends to infinity, so the limit of $a_n=+\infty$ as $n \rightarrow+\infty$
Could anyone provide a shorter method which doesn't use the ratio test maybe?
Thanks!

$$
a_n=\frac{n^{3n}}{n3^n}=\frac{n^{3n-1}}{3^n}=(\frac{n}{3})^{n}*n^{2n-1}
$$
Do you see why this tends to infinity?
