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. 
 A: 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 $$\lim_{n\to \infty}(a_n)^{1/n}=\frac {3}{e}>1.$$
A: The ratio test works readily:
$$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}(n+1)\not!}{(n+1)^{n+1}}\,\frac{n^{n}}{3^{n}\not{n\not!}}=\frac{3}{\Bigl(\cfrac{n+1}n\Bigr)^n}=\frac{3}{\Bigl(1+\cfrac{1}n\Bigr)^n}\to\frac 3{\mathrm e}>1.$$
A: 
I thought that it might be instructive to present a way forward that does not rely on Stirling's Formula.  In the following, we will use the inequality
$$\begin{align}
\frac1n \sum_{k=1}^n\log(k/n)&>\int_0^1 \log(x)\,dx\\\\
&=-1\tag1
\end{align}$$
since $\log(x)\le 0$ and concave for $x\in (0,1]$.  We now proceed.


Note that we have
$$\begin{align}
(n!)^{1/n}&=e^{\frac1n \log(n!)}\\\\
&=e^{\frac1n \sum_{k=1}^n\log(k)}\\\\
&=ne^{\frac1n \sum_{k=1}^n\log(k/n)}\tag2\\\\
&>ne^{-1}\tag3
\end{align}$$
where we used $(1)$ in going from $(2)$ to $(3)$.

Finally, applying $(3)$ yields
$$\begin{align}
\sqrt[n]{\frac{3^n\,n!}{n^n}}&>\frac3e\\\\
&>1\tag4
\end{align}$$
Letting $n\to \infty$ in $(4)$, we find that 
$$\lim_{n\to \infty}\sqrt[n]{\frac{3^n\,n!}{n^n}}>1$$
which implies from the root test that the series $\sum_{n=1}^\infty \frac{3^n\,n!}{n^n}$ diverges.

NOTE:  Given $(4)$, we see that the general terms of the series do not approach $0$.  Therefore, the root test is unnecessary.
A: Up to this point it is OK:
$$\lim_{n\to\infty}\frac{3}{n}(n!)^\frac{1}{n}$$
However, $(n!)^\frac{1}{n}=(n!)^0=1$ is not true. It is $\infty^0$ type of indeterminate form, so you must be careful. You can use the estimate: $e^n\ge \frac{n^n}{n!} \Rightarrow n!\ge \left(\frac{n}{e}\right)^n$:
$$\frac{3}{n}(n!)^\frac{1}{n}\ge \frac3n\cdot \frac{n}{e}=\frac3e>1.$$
A: Using what you already did, consider
$$A=\frac{3}{n}(n!)^\frac{1}{n}\implies \log(A)=\log(3)-\log(n)+\frac{1}{n}\log(n!)$$ and use Stirling approximation
$$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({n}\right)\right)+\frac{1}{12
   n}+O\left(\frac{1}{n^3}\right)$$ Replace and simplify to get
$$\log(A)=-1+\log(3)+\frac{\log(2\pi n)}{2n}+O\left(\frac{1}{n^2}\right)$$
A: Here is an answer which uses Stirling's approximation, which differs slightly by the other answers. Originally intended to answer a duplicate of this question, I noticed that none of the answers exactly matched mine, and as the question is still unanswered it may be of sight help... I used the formula the straightforward way without any log simplifications as per the other answer....
By Stirling's approximation, we have
$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$
Hence, at the limit to infinity, we can write
$\lim_{n \to \infty} \frac{3^nn!}{n^n}=\lim_{n \to \infty} \frac{3^n}{n^n}\frac{\sqrt{2\pi n}~n^n}{e^n}=\lim_{n \to \infty}\left(\frac{3}{e}\right)^n\sqrt{2\pi n}$
As 3>e, the limit at infinity clearly diverges, hence the entire sum diverges......
