Radius of Convergence for Series from Rudin

  • Thread starter Thread starter Poopsilon
  • Start date Start date
  • Tags Tags
    Series
Click For Summary

Homework Help Overview

The discussion revolves around determining the radius of convergence for the series \( \sum \frac{1}{1+z^n} \). Participants are exploring the behavior of this series for various values of the complex variable \( z \).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to identify the convergence of the series for different magnitudes of \( z \), with suggestions that it converges for \( |z| > 1 \). Others express difficulty in applying the ratio or root tests and consider bounding the series for convergence analysis.

Discussion Status

Participants are actively engaging with the problem, raising questions about the applicability of various convergence tests and discussing the implications of bounding the series. There is an acknowledgment of the complexity involved in determining convergence for values of \( z \) within specific ranges.

Contextual Notes

There is a mention that the series is not a power series, which influences the approach to finding the radius of convergence. Additionally, the nature of \( z \) as a complex variable is noted, which adds complexity to the analysis.

Poopsilon
Messages
288
Reaction score
1
I can't seem to find the radius of convergence for this series, or even a suitable function that bounds it for certain values of z. The series is £ 1/(1+z^n) sorry for the pound sign I'm on an iPod touch. Thanks.
 
Physics news on Phys.org
Looks to me like every z whose magnitude is greater than 1.
 
Poopsilon said:
I can't seem to find the radius of convergence for this series, or even a suitable function that bounds it for certain values of z. The series is £ 1/(1+z^n) sorry for the pound sign I'm on an iPod touch. Thanks.

I've always thought of 'radius of convergence' in terms of power series. This is not a power series.

EDIT: This answer is wrong see my below post. Upon inspection, it looks like it converges for values greater than 1 as you say (by looking at its limits for different values of z).
 
Last edited:
How would I go about proving that, I can't seem to get the ratio or root test to work. I could bound it by 1/z^n which does converge outside the circle of radius 1 but bounding it is only true when Re(z^n) >= -1/2, I think, and that seems to depend on z and n in a very complicated way.
 
Couldn't you just use the fact that |z^n + 1| = |z^n - (-1)| \geq |z^n| - 1 > 0?
And then a simple application of the comparison test should work I think.
 
Poopsilon said:
How would I go about proving that, I can't seem to get the ratio or root test to work. I could bound it by 1/z^n which does converge outside the circle of radius 1 but bounding it is only true when Re(z^n) >= -1/2, I think, and that seems to depend on z and n in a very complicated way.

It seems complex but one can break it into pieces to simplify the problem.

\sum^{\infty}_{n=0} \frac{1}{1+z^{n}}

We are looking for the convergence of the series for differing values of z where possible values of z are from (-inf, +inf). Keep also in mind that n is always a positive integer (n > 0) and since we are interested in convergence, we are determining the behavior of this infinite series for very large values of n. For some arbitrarily large element n, is that element infinitesimally negligible as to converge?

Use intuition/reasoning to generalize a piece-wise solution for convergence. ie: for z between (-inf, -1), at z = 0, at z = 1, at z > 1 etc to find convergence of the series.

Basically, ask yourself this:
  1. At an arbitrary z value, z = number, does the series converge?
  2. Can I generalize this solution for z = someNumber to a range of numbers?
  3. Have I exhausted all possible z values?

From this, although tedious, you can cover all of your bases. These are where your convergence/divergence tests come in. One can use Direct Comparison Test (hint: z < -1 and z > 1) to come to a conclusion that it does converge for some z. The numbers in between (-1 < z < 1) you have to figure out if it converges. Note that these are boundary cases that must be explicitly treated as a 'smaller' problem.
 
Last edited:
carlodelmundo said:
We are looking for the convergence of the series for differing values of z where possible values of z are from (-inf, +inf).

Although I don't think the OP explicitly said this, z is complex. I just verified by checking the problem in Baby Rudin.
 
My apologies. Please take my responses with a grain of salt.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 22 ·
Replies
22
Views
5K