The Real And Complex Number Systems

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π
And as x ∈ , π , then
n+p
∣
n+p
∑
k=n+1
a k sin kx
∣ ≤
≤
≤
≤
≤
m
∑
k=n+1
a k sin kx +
∣
n+p
∑
k=m+1
[ π
]
a k sin kx
∣ , where m = x
m∑
a k kx + 2a m+1
sin x by Summation by parts
k=n+1
2
ε
2 (π + 1) (m − n) x + 2a m+1
sin x 2
ε
π
2 (π + 1) mx + 2a m+1
x by 2x [0,
π ≤ sin x if x ∈ π 2
ε
2 (π + 1) π + 2a m+1 (m + 1)
< ε 2 + 2 ε
2 (π + 1)
< ε.
Hence, ∑ ∞
n=1 a n sin nx converges uniformly on R.
Remark: (1) In the proof (⇐), if we can make sure that na n ↘ 0, then
we can use the supplement on the convergnce of series in Ch8, (C)-
(6) to show the uniform convergence of ∑ ∞
n=1 a n sin nx = ∑ ∞
n=1 (na n) ( )
sin nx
n
by Dirichlet’s test for uniform convergence.
(2)There are similar results; we write it as references.
(a) Suppose a n ↘ 0, then for each α ∈ ( 0, π 2
)
,
∑ ∞
n=1 a n cos nx and
∑ ∞
n=1 a n sin nx converges uniformly on [α, 2π − α] .
Proof: The proof follows from (12) and (13) in Theorem 8.30 and
Dirichlet’s test for uniform convergence. So, we omit it. The reader
can see the textbook, example in pp 231.
(b) Let {a n } be a decreasing sequence of positive terms. ∑ ∞
n=1 a n cos nx
uniformly converges on R if and only if ∑ ∞
n=1 a n converges.
Proof: (⇒) Suppose that ∑ ∞
n=1 a n cos nx uniformly converges on R, then
let x = 0, then we have ∑ ∞
n=1 a n converges.
(⇐) Suppose that ∑ ∞
n=1 a n converges, then by Weierstrass M-test, we
have proved that ∑ ∞
n=1 a n cos nx uniformly converges on R.
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