On the radius of convergence of a power series

[piggees]

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 textbooks I read do not give lessons over this question. Any answer or hint or instruction will be much appreciated.

 

[LCKurtz]

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:

http://eom.springer.de/c/c020870.htm

 

[foxjwill]

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

 

[HallsofIvy]

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

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

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

You can get the same result by using the root test instead of the ratio test:
\sum a_n (z- z_0)^n converges absolutely as long as
\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|
is less than 1.

 

[piggees]

thank you all for the replies. That does help.

 

[piggees]

OK, I think I get it. Thank you.