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Radius of convergence of complex power series

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the radius of convergence of the series:

    ∑ n^-1.z^n
    n=1

    Use the following lemma:
    ∞ ∞
    If |z_1 - w| < |z_2 - w| and if ∑a_n.(z_2 - w)^n converges, then ∑a_n.(z_1 - w)^n also
    n=1 n=1
    converges.
    The contrapositive is also true.

    2. Relevant equations



    3. The attempt at a solution
    Hi, here's what I've done:

    Let r be the radius of convergence. The series converges when |z-w| < r and diverges for |z-w|> r.

    In this question, z = z, w = 0.

    Use the ratio test for convergence:

    lim |a_n+1 / a_n| = lim |zn / n+1 |
    n->∞
    = |z|

    Thus the series converges when |z| < 1
    diverges when |z| > 1

    So let |z - 0| = |z_2 - w|
    Then, by the lemma, all |z_1 - 0| converges, where |z_1 - 0| < |z - 0|
    and the converse is true for a diverging series.

    Thus the radius of convergence = |z|

    ---

    I think I've covered everything, but I don't think I've made very satisfactory use of the lemma. Can anyone please point me in the right direction?
    Thanks for any help.
     
  2. jcsd
  3. Oct 7, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your final answer appears to be "Thus the radius of convergence is |z|" which is meaningless. the radius of convergence is a real number, not a variable.

    You had already said "Thus the series converges when |z| < 1
    diverges when |z| > 1". What more do you need?
     
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