Problem: Find the radius of convergence of the series
\[
\sum_{n=1}^{\infty} \frac{n^3}{3^n}z^n,~ z \in \mathbb{C} .
\]
Solution: We recall that if $R$ is the radius of convergence, then
\begin{align*}
\frac{1}{R} & = \lim_{n \to \infty} \left\vert \frac{\frac{(n+1)^3}{3^{n+1}}}{\frac{n^3}{3^n}} \right\vert = \lim_{n \to \infty} \left\vert \frac{(n+1)^3}{3n^3} \right\vert = \frac{1}{3}.
\end{align*}
Therefore, the radius of convergence will be $3$.