The Real And Complex Number Systems

Remark: In order to show the series ∑ sin 1 n diverges, we consider Cauchy Criterion
as follows.
n sin 1 ≤ sin 1 ... sin 1
2n n 1
n n
and given x ∈ R, forn 0,1,2,..., we have
|sin nx| ≤ n|sin x|.
So,
sin 1 2 ≤ sin 1 ... sin 1
n 1
n n
for all n. Hence, ∑ sin 1 n diverges.
Note: There are many methods to show the divergence of the series ∑ sin 1 n . We can
use Cauchy Condensation Theorem to prove it. Besides, by (11), it also owrks.
(9) O-Stolz’s Theorem.
n 1
Proof: LetS n ∑ j1 j
and X n log n. Then by O-Stolz’s Theorem, it is easy to see
lim n→
S n .
(10) Since n k1 1 1 k
diverges, the series ∑ 1/k diverges by Theorem 8.52.
(11) Lemma: If a n is a decreasing sequence and ∑ a n converges. Then
lim n→ na n 0.
Proof: Sincea n → 0anda n is a decreasing sequence, we conclude that a n ≥ 0.
Since ∑ a n converges, given 0, there exists a positive integer N such that as n ≥ N,
we have
a n ..a nk /2 for all k ∈ N
which implies that
k 1a nk /2 since a n ↘.
Let k n, then as n ≥ N, wehave
n 1a 2n /2
which implies that as n ≥ N
2n 1a 2n
which implies that
lim n→
2na 2n 0 since lim n→
a n 0. *
Similarly, we can show that
lim n→
2n 1a 2n1 0. **
So,by(*)adn(**),wehaveprovedthatlim n→ na n 0.
Remark: From this, it is clear that ∑ 1 n diverges. In addition, we have the convergence
of ∑ na n − a n1 . We give it a proof as follows.
Proof: Write