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Showing a (complex) series is (conditionally) convergent.

  1. Sep 20, 2010 #1
    I've been reading a complex analysis book which had an example showing [tex]\sum^\infty_{n=1}1/n \cdot z^n[/tex] is convergent in the open unit ball.

    I'm now looking at the case when [tex]|z| = 1[/tex]. Clearly [tex]z = 1[/tex] is the divergent harmonic series, but i know this series is in fact convergent for all other [tex]|z| = 1[/tex].

    In order to prove this, i need to be able to show that the related series [tex]\sum^\infty_{n=1}z^n[/tex] is bounded, whenever [tex]|z| = 1[/tex] and [tex]z \neq 1 [/tex],.

    I can solve this problem when ever the argument of z is a rational multiple of pi, but other than that I'm stuck. Any help proving that this related series is bounded would be very helpful.

    Thanks!
     
  2. jcsd
  3. Sep 20, 2010 #2

    mathman

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    If you sum the truncated (at N) series, you have (1-zN+1)/(1-z). For z=1, you have 0/0 (no good). For zā‰ 1, the expression is well defined, so see what happens as N -> āˆž.
     
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