Math 321 - The Dirichlet Test

# Fix any 0 and set x fn z zc z r n zm r fn z

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Unformatted text preview: What if ∞ ∞ 1 1zn = n=0 n which diverges. So |z | = R? Well, if z = R, then the series is n=0 n R that leaves |z | = R but with z = R. This is where the Dirichlet test comes in handy. Fix any ε > 0 and set X= fn ( z ) = z∈C z R n zm R Fn (z ) = gn = February 3, 2008 m=0 1 n |z | = R, |z − R| ≥ ε X n as in (2) |z − R | = ε |z | = R The Dirichlet Test 2 For z ∈ X n+1 z 1− R Fn (z ) = z 1− R n+1 1+ z 2R ≤ ≤1R ε |R − z | R so that the hypotheses of the Dirichlet test are satisﬁed and the series converges uniformly ∞ 1zn on X . We conclude that n=0 n R converges for |z | < R and for |z | = R, z = R and diverges for |z | > R and for z = R. February 3, 2008 The Dirichlet Test 3...
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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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