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 $\dfrac{1}{3}$?

Here is one way to simplify the limit and arrive at the answer. Hopefully, it will let you see how terms cancel out.
\begin{align*}
\left|\frac{u_{n+1}}{u_{n}}\right| &= \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3} \\
&= \frac{(n+1)^3}{n^3}\cdot \frac{3^n}{3^{n+1}} \\
&= \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{3} \\
&= \left(1 + \frac{1}{n}\right)^3 \cdot \frac{1}{3} \\
\Rightarrow \lim_{n \rightarrow \infty} \left|\frac{u_{n+1}}{u_{n}}\right| &= 1^3 \cdot \frac{1}{3} = \frac{1}{3}
\end{align*}
