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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.
 Winter '08
 JoelFeldman
 Math

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