Radius of Convergence for Series from Rudin

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The radius of convergence for the series \( \sum^{\infty}_{n=0} \frac{1}{1+z^{n}} \) is determined to be greater than 1 for values of \( z \) where \( |z| > 1 \). The discussion highlights the challenges in applying the ratio and root tests, suggesting the use of the Direct Comparison Test for convergence analysis. It emphasizes the need to evaluate the series for various ranges of \( z \), particularly for complex values, and to treat boundary cases explicitly.

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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.
 
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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).
 
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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.
 
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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.
 

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