The Real And Complex Number Systems

Show that k1
a 2n2
1 a 2n2
a 2n1
a 2n
1 a 2n
for n 1, 2, . . .
1 −1 k a k converges if, and only if, ∑ k1
−1 k a k converges.
a
1c
Proof: First, we note that if b, then 1 a1 − b 1, and if b
1a
c , then
1 1 − b1 c. Hence, by hypothesis, we have
1 1 a 2n 1 − a 2n1 *
and
1 1 a 2n2 1 − a 2n1 . **
()Suppose that ∑
k1
−1 k a k converges, then lim k→ a k 0. Consider Cauchy
Condition for product,
1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1 for q 1, 2, 3, . . . .
If p 1 2m, andq 2l, then
1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1
|1 a 2m 1 − a 2m1 1 a 2m2l − 1|
≤ 1 a 2m − 1by(*)and(**)
a 2m → 0.
Similarly for other cases, so we have proved that k1
1 −1 k a k converges by
Cauchy Condition for product.
()This is a counterexample as follows. Let a n −1 n
n, thenitiseasytoshowthat
a 2n2
a
1 a 2n1
2n2
In addition,
n
k1
1 −1 k a k
k1
However, consider
n
∑
k1
n
∑
k1
n
∑
k1
n
n
a 2k − a 2k−1
exp 1
2k
exp −1k
k
− exp
exp
expb k 1
2k 1
2k − 1
exp −1n
n
a 2n
1 a 2n
for n 1, 2, . . .
−1
2k − 1
n
∑
k1
−1 k
k
,whereb k ∈
≥ ∑ exp−1 1
k1
2k 1
2k − 1
So, by Theorem 8.13, we proved the divergence of ∑ k1
→ as n → .
−1 k a k .
− 1
≥ 0 for all
→ exp− log 2 as n → .
−1
2k − 1 , 1
2k
8.45 A complex-valued sequence fn is called multiplicative if f1 1andif
fmn fmfn whenever m and n are relatively prime. (See Section 1.7) It is called
completely multiplicative if
f1 1andiffmn fmfn for all m and n.