The Real And Complex Number Systems

lim n→
sin ... sin n
1 ... 1 n
.
Proof: Write
sin ... sin n
1 1sin ... 1 1 n n sin n
,
1 ... 1 n
1 ... 1 n
the by Toeplitz’s Theorem, we have proved it.
E Theorem 8.16 emphasizes the decrease of the sequence a n , we may ask if we
remove the condition of decrease, is it true The answer is NOT necessary. For example,
let
a n 1 n −1n1 . 0
2n
F Some questions on series.
(1) Show the convergence of the series ∑ n1
log n sin 1 n .
Proof: Since n sin 1 n 1 for all n, logn sin 1 n 0 for all n. Hence, we consider the
new series
∑
n1
− log n sin 1
n ∑ log
n1
sin 1/n
1/n
as follows. Let a n log sin1/n and b
1/n n log 1 1 , then
n 2
lim
a n
n→
b 1 n 6 .
In addition,
∑ b n ≤ ∑ 1 n 2
by e x ≥ 1 x for all x ∈ R. From the convergence of ∑ b n , we have proved that the
convergence of ∑ a n by Limit Comparison Test.
∑ n1
(2) Suppose that a n ∈ R, and the series ∑
n1
converges absolutely.
a n
n
Proof: By A. P. ≥ G. P. , we have
a n2 1 n 2
2
which implies that ∑ n1
a n
n
≥
converges absolutely.
a n2 converges. Prove that the series
Remark: We metion that there is another proof by using Cauchy-Schwarz inequality.
the difference of two proofs is that one considers a n , and another considers the partial
sums S n .
Proof: ByCauchy-Schwarz inequality,
which implies that ∑ n1
a n
n
n
∑
k1
|a n|
k
2
≤
a n
n
∑ a2
k
k1
converges absolutely.
Double sequences and series
8.28 Investigate the existence of the two iterated limits and the double limit of the
n
∑
k1
1
k 2