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On the radius of convergence of a power series

by piggees
Tags: convergence, power series, radius
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Aug16-10, 08:42 AM
P: 4
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.
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Aug16-10, 09:14 PM
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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:
Aug17-10, 02:44 AM
P: 355
This is a really great question, though. Its answer is part of the beauty of complex analysis.

Aug17-10, 07:22 AM
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On the radius of convergence of a power series

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.
Aug17-10, 07:41 AM
P: 4
thank you all for the replies. That does help.
Aug17-10, 11:00 AM
P: 4
OK, I think I get it. Thank you.

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