Math 321 - The Dirichlet Test

What if z r zn well start with the series n0 r then

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Unformatted text preview: So all three series converge for all complex numbers z with |z | < R and diverge for all complex numbers with |z | > R. What if |z | = R? ∞ zn We’ll start with the series n=0 R . Then we can compute exactly the partial sum 1 − z n+1 n R if z = R zm z = Fn (z ) = (2) 1− R R m=0 n+1 if z = R 1 z 1− R As expected, if |z | < R this converges to for |z | > R, because z R n+1 = z n+1 R as n → ∞. Also as expected, this diverges → ∞. I claim that this also diverges whenever |z | = R. For z = R, it is obvious because n + 1 → ∞. For |z | = R with z = R, does not blow up as n → ∞, but it cannot converge either, because z n+2 R − z n+1 R = z n+1 R z...
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This document was uploaded on 02/24/2014 for the course MATH 321 at University of British Columbia.

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