$\displaystyle \sum \frac{n^2}{3^nz^n}\\$
$\displaystyle \frac{1}{R}=\lim_{n->\infty}\sqrt[n]{|\sum \frac{n^2}{3^nz^n}|}\\$
what to do now?
I don't know why you have the summation inside the square root. Correct application of the nth root test means that the given series will converge for values of z that satisfy $\displaystyle \lim_{n \rightarrow +\infty} \left| \frac{n^{2/n}}{3z}\right| < 1 \Rightarrow \frac{1}{3|z|} \lim_{n \rightarrow +\infty} n^{2/n} < 1$.
Then you have to test each endpoint.
Since power series and stuff belongs at least to calculus II , I assume you've already had calculus I: limits, continuity and stuff...right?
Then you must know that n^(1/n) --> 1 when n --> oo, but then
n^(2/n) = [n^(1/n)]^2 --> 1*1 = 1 by arithmetic of limits.
Tonio
Hmmm...rechecking your question I realize that what you posted is NOT a power series in z since z appears in the denominator, so this is at most a Laurent series, a very different stuff belonging to complex analysis.
What did you really mean to ask here? Is z a variable or just a parameter?
Tonio
True, because as I pointed out before what you ngave is not a power series! What you gave is a palin series with a parameter (constant) z, as you said. There's no convergence radius here, then.
For something to be a proper power series the powers of the VARIABLE have to be positive integers.
Tonio
Instead of blindly applying formulas you would be well advised to understand the principles of where those formulas come from and then applying those same principles to the questions you get. The question you have posted and the subsequent exchanges of replies is a classic example of why I give that advice.