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 $\dfrac{1}{3}$?

*

*You have a factor of $3^n$ in the numerator, and a factor of $3^{n+1}$ in the denominator. So
$$\frac{3^n(\text{stuff})}{3^{n+1}(\text{other stuff})} = \frac{3^n(\text{stuff})}{3\times 3^{n}\text{(other stuff)}} = \frac{\text{stuff}}{3(\text{other stuff})}.$$
Since $3^{n+1}=3\times 3^n$.

*Dividing numerator and denominator by $n^3$, we have
$$\begin{align*}
\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{3n^3} &= \lim_{n\to\infty}\frac{\frac{1}{n^3}(n^3+3n^2+3n+1)}{\frac{1}{n^3}(3n^3)}\\
&= \lim_{n\to\infty}\frac{1 + \frac{3}{n} + \frac{3}{n^2}+\frac{1}{n^3}}{3}\\
&= \frac{\lim\limits_{n\to\infty}(1 + \frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3})}{\lim\limits_{n\to\infty}3}\\
&= \frac{1 + 0 + 0 + 0}{3} = \frac{1}{3}.
\end{align*}$$
