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Unformatted text preview: (x) ≥ 0, gn+1 (x) ≥ 0 and every gk (x) − gk+1 (x) ≥ 0. For each ﬁxed x,
lim gn+1 (x) = 0. So (1) guarantees that {sn (x)} is a Cauchy sequence and hence conn→∞ verges. Call the limit s(x). Taking the limit of (1) as m → ∞ gives
s(x) − sn (x) ≤ 2M gn+1 (x)
February 3, 2008 The Dirichlet Test 1 Since gn+1 (x) converges uniformly to zero as n → ∞, we have that sn (x) converges uniformly to s(x) as n → ∞. n ∞ ∞ n 1z
z
, n=0 n R and
Example. We shall consider three diﬀerent power series:
n=0 R
∞
1 zn
, for some ﬁxed R > 0. For all three series, the radius of convergence is
n=0 n2 R
exactly R since, for ℓ ∈ {0, 1, 2},
ℓ
n lim sup
n→∞ 11
nℓ Rn = 1
R n lim sup
n→∞ 1
n = 1
R...
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 Winter '08
 JoelFeldman
 Math

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