The Real And Complex Number Systems

have, by (*)
2n−1
∑
2n−1
(ε >)
a k sin kx
∣
∣ = ∑
a k sin kx
k=n
≥
=
k=n
2n−1
∑
a 2n sin 1 2 since a k > 0 and a k ↘
k=n
( 1
2 sin 1 2)
(2na 2n ) .
That is, we have proved that 2na 2n → 0 as n → ∞. Similarly, we also have
(2n − 1) a 2n−1 → 0 as n → ∞. So, we have proved that na n → 0 as n → ∞.
(⇐) Suppose that na n → 0 as n → ∞, then given ε > 0, there exists a
positive integer n 0 such that as n ≥ n 0 , we have
ε
|na n | = na n <
2 (π + 1) . (*)
In order to show the uniform convergence of ∑ ∞
n=1 a n sin nx on R, it suffices
to show the uniform convergence of ∑ ∞
n=1 a n sin nx on [0, π] . So, if we can
show that as n ≥ n 0
n+p
∑
a k sin kx
< ε for all x ∈ [0, π] , and all p ∈ N
∣
∣
k=n+1
then we complete [ ] it. We consider two cases as follows. (n ≥ n 0 )
π
As x ∈ 0, , then
n+p
∣
∣
∣
n+p
∑
k=n+1
n+p
a k sin kx
∣ = ∑
≤
=
k=n+1
n+p
∑
k=n+1
n+p
∑
k=n+1
a k sin kx
a k kx by sin kx ≤ kx if x ≥ 0
(ka k ) x
ε pπ
≤
by (*)
2 (π + 1) n + p
< ε.
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