Does $\sum\limits_{n=1}^\infty \frac{n^n}{3^n n!}$ converge? 
Test the convergence of the series $$\sum_{n=1}^\infty \frac{n^n}{3^n n!}$$

I know that if the nth term tends to $\infty$ then the series is divergent and if it is tends to 0 it is convergent . Also I'm familiar with some test e.g. Ratio test, d'Alembert test, comparison test etc. But I could not solve it in proper way.
I know as $n$ increases $n^n$ increases more rapidly than $n!$ or $3^n$ but no idea when they both are multiplied
 A: I don't see a problem in solving it by using D'Alembert's Ratio-Test. 
You need to find the convergence of $$\sum_{n=1}^\infty \dfrac{n^n}{3^n n!}$$
So let $u_n= \dfrac{n^n}{3^n n!} $ which implies  $u_{n+1}= \dfrac{(n+1)^{n+1}}{3^{n+1} (n+1)!}$. 
Hence $$\dfrac{u_n}{u_{n+1}}=\dfrac{n^n}{3^n n!} . \dfrac{3^{n+1} (n+1)!}{(n+1)^{n+1}}$$
 $$\dfrac{u_n}{u_{n+1}}=\dfrac{n^n}{3^n n!} . \dfrac{3^{n}.\ 3 \ .(n+1).n!}{(n+1)^{n}.(n+1)}$$
$$\dfrac{u_n}{u_{n+1}}=\dfrac{3.n^n}{(n+1)^n}$$
$$\dfrac{u_n}{u_{n+1}}=\dfrac{3}{(1+\dfrac{1}{n})^n}$$
$$\lim_{n\to \infty} \dfrac{u_n}{u_{n+1}}= \dfrac{3}{e} > 1 $$
$$( \ \rm{ Since } \ \lim_{n \to \infty } (1+\dfrac{1}{n})^n = e  ) $$ See this for the proof of the term $e$.
Hence by D'Alemberts ratio test we have $l>1 ( l=\lim_{n\to \infty} \dfrac{u_n}{u_{n+1}})$ , $\sum_{n=1}^\infty \dfrac{n^n}{3^n n!}$ converges. 
Thank you.
A: The ratio test should work, but
1. you have to do some clever manipulation to $(n+1)^{n+1}/n^n$, and
2. you have to know the value of $\lim_{n\to\infty}(1+n^{-1})^n$. 
Does that help?
A: Using D'alambert's ratio test we can get that the series $$\sum \frac{n^n}{x^n\cdot n!}$$ converges for all $x>e$ and diverges for $x<e$.
It worth mentioning that at $x=e$ the sum diverges (can be seen by using Stirling's approximation).
