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

Mattofix
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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...
 
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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|
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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