# On the radius of convergence of a power series

1. Aug 16, 2010

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

2. Aug 16, 2010

### 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

3. Aug 17, 2010

### foxjwill

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

4. Aug 17, 2010

### HallsofIvy

Staff Emeritus
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.

5. Aug 17, 2010

### piggees

thank you all for the replies. That does help.

6. Aug 17, 2010

### piggees

OK, I think I get it. Thank you.