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Convergence on the unit circle
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[QUOTE="Trevor Vadas, post: 3532727, member: 359658"] [h2]Homework Statement [/h2] Determine the behavior of convergence on the unit circle, ie |z| = 1 of: Ʃ [itex]\frac{z^{n}}{n^{2}(1 - z^{n})}[/itex] [h2]Homework Equations[/h2] Obviously this is divergent then z is a root of unity. The question is what happens when z is not a root of unity. [h2]The Attempt at a Solution[/h2] 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 infinitely 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] [h2]Homework Statement [/h2] 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. [/QUOTE]
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Convergence on the unit circle
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