Radius of Convergence for (-1)^n(i^n)(n^2)(Z^n)/3^n in Complex Analysis

[Mattofix]

Homework Statement

Find the radius of convergence of

(-1)^n(i^n)(n^2)(Z^n)/3^n

The Attempt at a Solution

i have got to lZl i (n+1)^2/3n^2

but am unsure how to complete it...

 

[NateTG]

For the sum to converge, the norm of the terms must go to zero.
So, what is:
$$\lim_{n \rightarrow \infty} \left | n^2 \left(\frac{-iZ}{3}\right)^n\right|$$

 

[HallsofIvy]

How and why did you "get" that? At first I thought you were using the "root test" but that won't work with the n².

(Am I correct that "n" is the index of summation and "i" is the complex base? If so |i|= |-1|= 1)

I would be inclined to use the "ratio test": a sequence $\sum a_n$ converges if the ratio $|a_{n+1}/a_n|$ converges to a number less than 1. Here, $|a_{n+1}|= (n+1)^2 Z^{n+1}/3^{n+1}$ so the ratio becomes $((n+1)/n)^2 Z/3$. Since (n+1)/n goes to 1, so does ((n+1)/n)^2 and we have |Z|/3< 1. The radius of convergence is 3.