Conditional Convergence in a Power Series

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Discussion Overview

The discussion revolves around the concept of conditional convergence in power series, specifically exploring whether there exists a power series \(\sum_n^\infty c_n (z-a)^n\) with a radius of convergence \(R\) such that all \(z\) for which \(|z-a| = R\) exhibit purely conditional convergence, meaning neither divergence nor absolute convergence occurs.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the existence of a power series that conditionally converges on the boundary of its radius of convergence.
  • Another participant suggests modifying a conditionally convergent series by incorporating a term \((z-a)^n\) to explore its convergence behavior.
  • A different participant points out that for \(z = 2\), the series diverges, indicating that the proposed modification does not achieve the desired conditional convergence.
  • Another participant proposes a series \(\sum_{n=1}^{\infty} \frac{(-|z|)^n}{n}\) and argues that it converges for \(|z| < 1\) and conditionally converges at \(|z| = 1\), although they later acknowledge that this does not meet the requirement of being a power series.

Areas of Agreement / Disagreement

Participants express differing views on the possibility of achieving purely conditional convergence in power series, with no consensus reached on whether such a series exists.

Contextual Notes

Participants reference specific series and convergence behaviors, but the discussion remains open-ended with no definitive conclusions or proofs provided regarding the existence of the desired power series.

nonequilibrium
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I was wondering if there's an example of a power series [tex]\sum_n^\infty c_n (z-a)^n[/tex] with radius of convergence R so that all z for which |z-a| = R there is purely conditional convergence? (no divergence but also no absolute convergence) Or perhaps a reason why that's impossible?
 
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I'm not sure, but I might try taking a conditionally convergent series and put a (z-a)^n in the summand. For example, try

[tex]f(z) \equiv -\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(1-z)^n[/tex]

Without the -(1-z)^n this is a conditionally convergent series. I picked -(1-z)^n to tack on because without the (-1)^(n+1) this series gives the logarithm when |1-z| < 1. For the logarithm, the series diverges beyond this radius. I would guess that the modified series I proposed might conditionally converge beyond that radius, instead of diverging, but I haven't checked for sure.
 
Hm, take z = 2, then you get the harmonic series and it diverges. Thanks for the try though.

Apparently one can proof that [tex]\sum \frac{z^n}{n}[/tex] converges conditionally FOR ALL |z| = 1 except for z = 1. That's awfully close, but sadly not enough :(
 
What about...
[tex] f(z) = \sum_{n=1}^{\infty} \frac{(-|z|)^n}{n}[/tex]

Looks to me like it converges for all |z| < 1. Consider the consider you care about: |z| = 1 = R.

Of course, since |z| = |-z| the sign change won't affect convergence. And of course, if plugging |z|= 1, you get conditional convergence automatically!

Does that work for you?

Edit: Oops, misread. You wanted a power series. My bad. : (
 
Yeah a power series, sorry :( thanks for the effort though :)
 

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