What is the radius of convergence of

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Homework Help Overview

The discussion revolves around determining the radius of convergence for the series Σ anzn, where a_n = (n + (-1)^n) / n^2. Participants are exploring the application of the Cauchy-Hadamard Theorem in the context of complex analysis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply the Cauchy-Hadamard Theorem and are discussing the limit superior of the sequence. There is consideration of the behavior of the series at specific points on the boundary of the convergence disk, particularly at z=1 and z=(-1).

Discussion Status

Some participants have provided informal reasoning regarding the radius of convergence being one, while others are questioning the completeness of the arguments presented. There is an acknowledgment of the need for a more formal proof and exploration of boundary behavior.

Contextual Notes

Participants note that some information may be missing in their proofs and are considering the implications of singularities outside the disk of convergence. There is a focus on the oscillating term (-1)^n and its impact on the convergence analysis.

Shackleford
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Homework Statement



z ∈ ℂ

What is the radius of convergence of (n=0 to ∞) Σ anzn?

Homework Equations



I used the Cauchy-Hardamard Theorem and found the lim sup of the convergent subsequences.

a_n = \frac{n+(-1)^n}{n^2}

limn→∞ |an|1/n

The Attempt at a Solution



I think that the radius of convergence is one, i.e. |z| < 1. I figured that the numerator would tend to n with the oscillating 1 and so you'd get \frac{n^{1/n}}{n^{2/n}} = 1

R = 1/limn→∞ |an|1/n
 
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Shackleford said:

Homework Statement



z ∈ ℂ

What is the radius of convergence of (n=0 to ∞) Σ anzn?

Homework Equations



I used the Cauchy-Hardamard Theorem and found the lim sup of the convergent subsequences.

a_n = \frac{n+(-1)^n}{n^2}

limn→∞ |an|1/n

The Attempt at a Solution



I think that the radius of convergence is one, i.e. |z| < 1. I figured that the numerator would tend to n with the oscillating 1 and so you'd get \frac{n^{1/n}}{n^{2/n}} = 1

R = 1/limn→∞ |an|1/n

That's a little informal but it looks fine. You might want worry about what happens at z=1 and z=(-1) if you are concerned about the boundary cases.
 
Dick said:
That's a little informal but it looks fine. You might want worry about what happens at z=1 and z=(-1) if you are concerned about the boundary cases.

I assume a singularity is there and outside of the disk the series is divergent. Well, I'm leaving out a few bits of information in my "proof" here. I'll state a bit more of the background. What would a more formal proof look like?
 
Shackleford said:
I assume a singularity is there and outside of the disk the series is divergent. Well, I'm leaving out a few bits of information in my "proof" here. I'll state a bit more of the background. What would a more formal proof look like?

Rather than just ignoring the (-1)^n handle it with a squeeze type thing. E.g. n/2<=n+(-1)^n<=2n. You can easily find the outer limits. Discuss what happens when z=1 or z=(-1).
 

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