<s>[INST] More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:  
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alembert's ratio test) is:
$$u_n=\frac{n^3}{3^n}\;,$$  
so  
$$\begin{align*}
\frac{|u_{n+1}|}{|u_n|} &=\frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3}\;.
\end{align*}$$
How do we get from there to...  
$$=\frac{n^3+3n^2+3n+1}{3n^3}$$  
What happens with $3^n$ in the numerator and power of $n+1$ in the denominator? How do they cancel out?
Also, in the very next step that all goes to being equal to  
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\frac{1}{3}<1\;,$$  
which means the series is convergent.
But how do we get to $\dfrac{1}{3}$?
 [/INST] 
What happens with the $3^n$ in the numerator and the $3^{n+1}$ in the denominator?

Recall the following laws of exponents:
$$a^{b}a^c = a^{b+c}, \quad \text{ and } \quad \frac{a^b}{a^c} = a^{b-c} = \frac{1}{a^{c-b}},$$
for any $a>0$, and any real numbers $b,c$. In particular, if $a=3$, $b=n$, and $c=n+1$, then:
$$\frac{3^n}{3^{n+1}}=\frac{1}{3^{(n+1)-n}} = \frac{1}{3^1}=\frac{1}{3}.$$ 
</s>