The Real And Complex Number Systems

In addition,
n
(2). ∑
k1
(1). lim n→
A n b n1 exists.
n
|A k b k1 − b k | ≤ M ∑
k1
|b k1 − b k | ≤ M ∑
k1
|b k1 − b k |.
(2) implies that
n
(3). ∑ A k b k1 − b k converges.
k1
n
By (1) and (3), we have shown that ∑ k1
a k b k converges.
Remark: (1) The result is first discovered by Richard Dedekind (1831-1916).
(2) There is an exercise by (b), we write it as a reference. Show the convergence of the
k
−1
series ∑ k1
.
k
Proof: Leta k −1 k
, then in order to show the convergence of
1/3
∑ k1
−1
k
and b
k 2/3 k 1
k
, it suffices to show that ∑
k k1
a k : S n
n ∈ N, there exists j ∈ N such that j 2 ≤ N j 1 2 . Consider
S n a 1 a 2 a 3 a 4 ...a 8 a 9 ....a 15 ...a j
2 ...a n
n
is bounded sequence. Given
≤ 3a 3 5a 4 7a 15 9a 16 ..4k − 1a 2k 2 −1 4k 1a 2
2k
if j 2k, k ≥ 2
3a 3 5a 4 7a 15 9a 16 ..4k − 3a 2k−2
2 if j 2k − 1, k ≥ 3
then as n large enough,
S n ≤ −3a 4 5a 4 −7a 16 9a 16 ... −4k − 1a 2k
2 4k 1a 2k
2
−3a 4 5a 4 −7a 16 9a 16 ... −4k − 5a 2k−2
2 4k − 3a 2k−2
2
which implies that as n large enough,
S n ≤ 2 ∑ a 2j
2 2 ∑
1
j2
j2 2j 4/3 : M 1 *
Similarly, we have
M 2 ≤ S n for all n **
By (*) and (**), we have shown that
n
a k : S n is bounded sequence.
∑ k1
Note: (1) By above method, it is easy to show that
−1
k
∑
k1
converges for p 1/2. For 0 p ≤ 1/2, the series diverges by
1
n 2 p ... 1
n 2 2n p ≥ 2n 1
n 2 n p ≥ 2n 1
n 2 n p ≥ 2n 1
n 1 2p ≥ 2n n 1 1 1.
k p
log k
−1
.
k
(2) There is a similar question, show the divergence of the series ∑ k1
Proof: WeuseTheorem 8.13 to show it by inserting parentheses as follows. We insert
k
−1log
parentheses such that the series ∑ forms ∑−1 k b k . If we can show ∑−1 k b k
k