On the radius of convergence of a power series

  1. Hi, I'm new here. I am curious that why a power series must have a radius of convergence? I mean, even in a complex plane, there is always a so-called convergent radius for a power series. Is it possible that a power series is convergent for a certain range in one direction, and for an apparent shorter/longer range in some other direction? So far all the text books I read do not give lessons over this question. Any answer or hint or instruction will be much appreciated.
  2. jcsd
  3. LCKurtz

    LCKurtz 8,454
    Homework Helper
    Gold Member

    The answer is no, it can't have a larger range of convergence in a different direction. The relevant theorem is the Cauchy-Hadamard theorem. Lots of links on the internet, one of which is:

  4. This is a really great question, though. Its answer is part of the beauty of complex analysis.
  5. HallsofIvy

    HallsofIvy 41,265
    Staff Emeritus
    Science Advisor

    It is, basically, an application of the "ratio test".

    If [itex]f(z)= \sum a_n(z- z_0)^n[/itex]] then the series converges, absolutely, as long as
    [tex]\lim_{n\to\infty}\frac{|a_{n+1}(z- z_0)^{n+1}|}{|a_n (z- z_0)^n|}[/tex][tex]= |z- z_0|\lim_{n\to\infty}\frac{a_{n+1}{a_n}|< 1[/tex]
    and diverges if that limit is larger than 1.

    As long as
    [tex]\lim_{n\to\infty}\frac{a_{n+1}}{a_n}= A[/tex]
    exists, then we have that the power series converges for
    [tex]|z- z_0|< \frac{1}{A}[/tex]
    and diverges for
    [tex]|z- z_0|> \frac{1}{A}[/tex]

    You can get the same result by using the root test instead of the ratio test:
    [itex]\sum a_n (z- z_0)^n[/itex] converges absolutely as long as
    [tex]\lim_{n\to\infty}\left(a_n(z- z_0)^n)^{1/n}= \left(\lim_{n\to\infty}\sqrt[n]{a_n}\right)|z- z_0|[/tex]
    is less than 1.
  6. thank you all for the replies. That does help.
  7. OK, I think I get it. Thank you.
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