Convergence or Divergence $\sum_{n=0}^{\infty} \frac{n^n}{3^{1+3n}}$
I have, by Cauchy Criterion
$\lim_{n \rightarrow \infty} \sqrt[n]{|a_n|}= \frac{1}{27} \lim_{n \rightarrow \infty} \frac{n}{3^{1/n}}= ?$
How a finish?
Cheers!
 A: Note that the $n^{th}$ terms is $a_n = \dfrac13 \left(\dfrac{n}{27} \right)^n$. 
For $n > 54$, we have that $a_n > \dfrac{2^n}3$.
Hence, what is $\lim_{n \to \infty} a_n$? What can you say about the series after finding what $\lim_{n \to \infty} a_n$ is?
A: Maybe you can tidy things up first
$$\sum\limits_{n = 0}^\infty  {\frac{{{n^n}}}{{{3^{1 + 3n}}}}}  = \frac{1}{3}\sum\limits_{n = 0}^\infty  {\frac{{{n^n}}}{{{3^{3n}}}}}  = \frac{1}{3}\sum\limits_{n = 0}^\infty  {\frac{{{n^n}}}{{{27^n}}}}  = \frac{1}{3}\sum\limits_{n = 0}^\infty  {{{\left( {\frac{n}{27}} \right)}^n}} $$
Cauchy then says $$\lim \; a_n^{1/n}=\lim\; \frac n {27}\to\infty$$
So the sequence isn't summable. Maybe we can shed some light on this.
The sequence of the form $$a_n=\left(\frac{n}{\lambda}\right)^n$$
has the particular problem that $a_n\not \to0$. Indeed, Whenever $n>2\lambda $, 
$$a_n=\left(\frac{n}{\lambda}\right)^n>2^n$$
which means that $a_n\to \infty$. 
A: If all is about convergence or divergence of the series, you can also use the D'Alembert criterion (ratio test):
$$ \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} 
= \lim_{n \rightarrow \infty}  \frac{\frac{(n+1)^{n+1}}{3\cdot 3^{3(n+1)}}}{\frac{n^{n}}{3\cdot 3^{3n}}} 
= \lim_{n \rightarrow \infty} \frac{(n+1)^{n+1}\cdot 3\cdot 3^{3n}}{n^{n} \cdot 3\cdot 3^{3(n+1)}} 
= \lim_{n \rightarrow \infty} \frac{(n+1)}{3^3} \left(1+\frac{1}{n}\right)^{n}$$
You should know that:
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{n}=e,$$
so:
$$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{27}\lim_{n \rightarrow \infty}(n+1) = \infty. $$
Thus, the series diverges.
