The Real And Complex Number Systems

n ∑
kn
a k ∑
kn
a k b k
b n
b k
∑ C k d k − d k1 C d − C n−1 d n
kn
∑
kn
C k d k − d k1 − C n−1 d n
by C lim k→ C k and lim k→ d k 0. In order to show the existence of lim n→ b n ∑ kn
a k ,
it suffices to show the existence of lim n→ ∑
kn
C k d k − d k1 . Since the series
∑
kn
C k d k − d k1 exists, lim n→ ∑
kn
C k d k − d k1 0. From above results, we have
proved the convergence of lim n→ b n ∑ kn
a k .
Note: We also show that lim n→ b n ∑
kn
a k 0 by preceding sayings.
Supplement on the convergence of series.
A Show the divergence of ∑ 1/k. We will give some methods listed below. If the
proof is easy, we will omit the details.
(1) Use Cauchy Criterion for series. Since it is easy, we omit the proof.
(2) Just consider
1 1 2 1 3 1 4 ... 2 1 n ≥ 1 1 2 2 1 4 1 ...2n−1 2 n
1 n 2 → .
Remark: We can consider
1 1 ... 1 1 ... 2 10
1 ...≥ 1 11 100
10 9 100 90 ...
Note: The proof comes from Jing Yu.
(3) Use Mathematical Induction to show that
1
k − 1 1 k 1
k 1 ≥ 3 if k ≥ 3.
k
Then
1 1 2 1 3 1 4 1 5 1 6 ....≥ 1 3 3 3 6 3 9 ...
Remark: The proof comes from Bernoulli.
(4) Use Integral Test. Since the proof is easy, we omit it.
(5) Use Cauchy condensation Theorem. Since the proof is easy, we omit it.
(6) Euler Summation Formula, the reader can give it a try. We omit the proof.
(7) The reader can see the book, Princilpes of Mathematical Analysis by Walter
Rudin, Exercise 11-(b) pp 79.
Suppose a n 0, S n a 1 ...a n ,and∑ a n diverges.
(a) Prove that ∑ an
1a n
diverges.
Proof: Ifa n → 0asn → , then by Limit Comparison Theorem, we know that
diverges. If a n does not tend to zero. Claim that does not tend to zero.
∑ an
1a n
a n
1a n