The Real And Complex Number Systems

lim n→
|sinn 1a b − sinna b|
lim n→
sinn 1a b − sinna b
lim n→
sup 2 cosna b cos a
2
− sinna b sin a 2
sin a 2
lim n→
sup 2 cosna b cos a sin a 2 2
|sin a| ≠ 0
which is impossible. So, ∑ sinna b diverges.
Remark: (1) By the same method, we can show the divergence of ∑ cosna b if
a ≠ n for all n ∈ Z and b ∈ R.
(2) The reader may give it a try to show that,
and
p
by considering ∑ n0
(**).
(3) The series ∑ sink
k
p
∑
n0
cosna b
Proof: First, it is clear that ∑ sink
k
|∑ sin k| ≤
1
sin 1 2
partial sums as follows: Since
sin
p1
2 b
sin b 2
sin a p 2 b *
p
p1
sin
∑ sinna b b 2
cos a p sin b 2 b **
n0
2
e inab . However, it is not easy to show the divergence by (*) and
converges conditionally.
converges by Dirichlet’s Test since
, we consider its
. In order to show that the divergence of ∑ sink
k
3n3
n
∑ sin k ∑
sin 3k 1 sin 3k 2 sin 3k 3
k
3k 1 3k 2 3k 3
k1
k0
and note that there is one value is bigger than 1/2 among three values |sin 3k 1|,
|sin 3k 2|, and|sin 3k 3|. So,
3n3
∑
k1
sin k
k
n 1
2
≥ ∑
3k 3
k0
which implies the divergence of ∑ sink .
k
Remark: The series is like Dirichlet Integral sinx
0
x
Integral converges conditionally.
(4) The series ∑ |sink|r diverges for any r ∈ R.
k
Proof: We prove it by three cases as follows.
(a) As r ≤ 0, we have
So, ∑ |sink|r diverges in this case.
k
(b) As 0 r ≤ 1, we have
∑
|sin k|r
k
≥ ∑ 1 k .
dx. Also, we know that Dirichlet