Convergence on the unit circle

  • #1

Homework Statement



Determine the behavior of convergence on the unit circle, ie |z| = 1 of:

Ʃ [itex]\frac{z^{n}}{n^{2}(1 - z^{n})}[/itex]

Homework Equations



Obviously this is divergent then z is a root of unity. The question is what happens when z is not a root of unity.

The Attempt at a Solution


My thought was originally that it would converge. Now I think it may diverge. To show this I look at
| [itex]\frac{z^{n}}{n^{2}(1 - z^{n})}[/itex] | = [itex]\frac{1}{n^{2}|1 - z^{n}|}[/itex], since |z| = 1, and show this does not converge to 0 and hence the series must diverge.

now for ε = 1. I want to show that z[itex]^{n}[/itex] will land close enough to 1(for infintely many n) so that
[itex]\frac{1}{n^{2}|1 - z^{n}|}[/itex] > 1.

or there exists infinitely many n such that [itex]\frac{1}{(|1 - z^{n}|}[/itex] [itex]\geq[/itex] n[itex]^{2}[/itex]

Homework Statement



We know that with |z|=1, z[itex]^{n}[/itex] is dense on the unit circle. Hence for any n there exsists an m that would make
[itex]\frac{1}{(|1 - z^{m}|}[/itex] [itex]\geq[/itex] n[itex]^{2}[/itex] true.
Now can we show the stars will allign and get m to equal n infinitely many times.

Thanks for any help.
 
Last edited:

Answers and Replies

Related Threads on Convergence on the unit circle

Replies
1
Views
1K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
5
Views
958
Top