$\sum_{n=1}^{\infty} \frac{n^3}{3^n} z^n$ As a power series in $z$, it has a radius of convergence, which is what I believe you are looking for.
Regroup it as $\displaystyle n^3 \left( \dfrac{z}{3} \right)^n$.
Use the ratio test: In a power series $\displaystyle \sum_{n=1}^{\infty} a(n) x^n$, if $\displaystyle \lim_{n \rightarrow \infty} \left \lvert \dfrac{a(n+1)}{a(n)} \right \rvert = c$, then the series converges for $\displaystyle \lvert x \rvert < \frac1c$.
Taking $x = z/3$, you see that $a(n) = n^3$.
$\lim_{n \rightarrow \infty} \left \lvert \dfrac{a(n+1)}{a(n)} \right \rvert = \lim_{n \rightarrow \infty} \left( 1 + \frac1n\right)^3 = 1$
So convergence occurs for $|z| < 3$.
The question of convergence remains open for the endpoints, $z = \pm 3$ of the convergence interval. This is settled by realizing that at both those endpoints, the magnitudes of the terms are all greater than those of the corresponding terms of $\displaystyle \sum_{n=1}^{\infty} \left( \pm x \right)^n$ and that both of these are non-convergent. So by the comparison test, convergence fails at the endpoints, and is strictly confined to the interior of the interval $|z| < 3$.
Is it right to my procedure?
Is there another easier way to do it?