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

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SUMMARY

The radius of convergence for the series defined by the expression (-1)^n(i^n)(n^2)(Z^n)/3^n is determined to be 3. The analysis employs the ratio test, where the terms of the series are evaluated as |a_{n+1}/a_n|, leading to the conclusion that |Z|/3 must be less than 1 for convergence. The limit of the ratio simplifies to ((n+1)/n)^2 Z/3, which approaches 1 as n approaches infinity. Thus, the series converges when |Z| < 3.

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  • Understanding of complex numbers and their properties, specifically the modulus of complex numbers.
  • Familiarity with series convergence tests, particularly the ratio test.
  • Knowledge of limits and their application in evaluating series.
  • Basic understanding of sequences and summation indices.
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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...
 
Last edited:
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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.
 

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