Using the Cauchy root test I've got the next series: $$\sum_{n=1}^{\infty} \frac{3^n n!}{n^n} $$ I must to solve it to see if converges or not. So I've got the next process:  
Sol (attempt): Let $a_n=\frac{3^n n!}{n^n}$. By the Cauchy root test we've got that: 
$$\lim_{n\to\infty}\sqrt[n]{a_n} =\lim_{n\to\infty}\sqrt[n]\frac{3^n n!}{n^n} =\lim_{n\to\infty}\sqrt[n] {\left(\frac{3}{n}\right)^n n!} = \lim_{n\to\infty}\frac{3}{n}\sqrt[n]{n!} = \lim_{n\to\infty}\frac{3}{n}(n!)^\frac{1}{n} = 0$$ 
If $n$ tends to infinity we've got that $\frac{3}{n}=0$ and $ (n!)^\frac{1}{n}=(n!)^0=1$ 
Thus, the series converges absolutely.  
My question is about if I'm using the test in a good way, and if the limit is correct. If there's any error on my process I'd be grateful to know. 

We have $$\frac {a_{n+1}}{a_n}=   \frac {3(n+1)n^n}   {(n+1)^{n+1}}  =\frac {3}{ 
    (1+\frac {1}{n})^n   }\to \frac {3}{e}>1 \text { as } n\to \infty.$$
Or we can  apply the root test using Stirling's Formula: For $n>0$ we have $1-\frac {1}{6n}<n!^{-1}(\frac {n}{e})^n\sqrt {2\pi n}\;<1+\frac {1}{6n}$.
Although for this Q it is sufficient to use a  consequence of it that's much easier to prove: $\lim_{n\to \infty}n!\cdot \left(\frac {n}{e}\right)^{-n}=1, $ which implies that $$\lim_{n\to \infty}(a_n)^{1/n}=\frac {3}{e}>1.$$
