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 $\dfrac{1}{3}$?