The Real And Complex Number Systems

So, by Weierstrass M-test, we have proved that ∑ ∞
n=1 x/nα (1 + nx 2 ) converges
uniformly on R if α > 1/2. Hence, ∑ ∞
n=1 x/nα (1 + nx 2 ) converges
uniformly on every finite interval in R if α > 1/2.
9.20 Prove that the series ∑ ∞
n=1 ((−1)n / √ n) sin (1 + (x/n)) converges
uniformly on every compact subset of R.
Proof: It suffices to show that the series ∑ ∞
n=1 ((−1)n / √ n) sin (1 + (x/n))
converges uniformly on [0, a] . Choose n large enough so that a/n ≤ 1/2, and
therefore sin ( 1 + ( )) (
x
n+1 ≤ sin 1 +
x
n)
for all x ∈ [0, a] . So, if we let fn (x) =
(−1) n / √ n and g n (x) = sin ( 1 + n) x , then by Abel’s test for uniform convergence,
we have proved that the series ∑ ∞
n=1 ((−1)n / √ n) sin (1 + (x/n))
converges uniformly on [0, a] .
Remark: In the proof, we metion something to make the reader get
more. (1) since a compact set K is a bounded set, say K ⊆ [−a, a] , if we can
show the series converges uniformly on [−a, a] , then we have proved it. (2)
The interval that we consider is [0, a] since [−a, 0] is similar. (3) Abel’s test
for uniform convergence holds for n ≥ N, where N is a fixed positive
integer.
9.21 Prove that the series ∑ ∞
n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) converges
pointwise but not uniformly on [0, 1] .
Proof: We show that the series converges pointwise on [0, 1] by considering
two cases: (1) x ∈ [0, 1) and (2) x = 1. Hence, it is trivial. Define
f (x) = ∑ ∞
n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) , if the series converges
uniformly on [0, 1] , then by Theorem 9.2, f (x) is continuous on [0, 1] .
However,
f (x) =
{ 1
2
log (1 + x) if x ∈ [0, 1)
log 2 if x = 1
Hence, the series converges not uniformly on [0, 1] .
Remark: The function f (x) is found by the following. Given x ∈ [0, 1),
then both
∞∑
t 2n = 1
1 − t and 1 ∞∑
t n 1
=
2 2 2 (1 − t)
n=0
converges uniformly on [0, x] by Theorem 9.14. So, by Theorem 9.8, we
n=0
.
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