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On the radius of convergence of a power series 
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#1
Aug1610, 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 socalled 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
Aug1610, 09:14 PM

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Thanks
PF Gold
P: 7,583

The answer is no, it can't have a larger range of convergence in a different direction. The relevant theorem is the CauchyHadamard theorem. Lots of links on the internet, one of which is:
http://eom.springer.de/c/c020870.htm 


#3
Aug1710, 02:44 AM

P: 355

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



#4
Aug1710, 07:22 AM

Math
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Thanks
PF Gold
P: 39,348

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. 


#5
Aug1710, 07:41 AM

P: 4

thank you all for the replies. That does help.



#6
Aug1710, 11:00 AM

P: 4

OK, I think I get it. Thank you.



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