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Radius of convergence

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the radius of convergence of


    3. 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: Jan 29, 2008
  2. jcsd
  3. Jan 29, 2008 #2


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    For the sum to converge, the norm of the terms must go to zero.
    So, what is:
    [tex]\lim_{n \rightarrow \infty} \left | n^2 \left(\frac{-iZ}{3}\right)^n\right|[/tex]
  4. Jan 29, 2008 #3


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    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 [itex]\sum a_n[/itex] converges if the ratio [itex]|a_{n+1}/a_n|[/itex] converges to a number less than 1. Here, [itex]|a_{n+1}|= (n+1)^2 Z^{n+1}/3^{n+1}[/itex] so the ratio becomes [itex]((n+1)/n)^2 Z/3[/itex]. 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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