The Real And Complex Number Systems

Consider
which implies that
which implies that
b n1 b 1 −1
2
n
→ 1asn → .
n1 j2 b j n
j1 b j −1/2
a 1 1/2 a 2
−2/3
an1 1
b n1
2/3
lim n→
a n1 a 1 a 22 1/3 .
Remark: There is another proof. We write it as a reference.
Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that
a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Leta 2 ≥ a 1 .Sincea n2 a n a n1 1/2 ,
then inductively, we have
a 1 ≤ a 3 ≤...≤ a 2n−1 ≤...≤ a 2n ≤...≤ a 4 ≤ a 2 .
So, both of a 2n and a 2n−1 converge. Say
lim n→
a 2n x and lim n→
a 2n−1 y.
Note that a 1 ≠ 0anda 2 ≠ 0, so x ≠ 0, and y ≠ 0. In addition, x y by
a n2 a n a n1 1/2 . Hence, a n converges to x.
By a n2 a n a n1 1/2 , and thus
n
j1 a2 j2 n
j1 a j a j1 a 1 a 22 a n1 n−2 j1 a2
j2
which implies that
which implies that
a n1 a2 n2 a 1 a2
2
lim n→
a n x a 1 a 22 1/3 .
8.11 a 1 2, a 2 8, a 2n1 1 2 a 2n a 2n−1 , a 2n2 a 2na 2n−1
a 2n1
, L 4.
Proof: First, we note that
a 2n1 a 2n a 2n−1
2
≥ a 2n a 2n−1 by A. P. ≥ G. P. *
for n ∈ N. So, by a 2n2 a 2na 2n−1
a 2n1
and (*),
a 2n2 a 2na 2n−1
a 2n1
≤ a 2n a 2n−1 ≤ a 2n1 for all n ∈ N.
Hence, by Mathematical Induction, itiseasytoshowthat
a 4 ≤ a 6 ≤...≤ a 2n2 ≤...≤ a 2n1 ≤...≤ a 5 ≤ a 3
for all n ∈ N. It implies that both of a 2n and a 2n−1 converge, say
lim n→
a 2n x and lim n→
a 2n−1 y.
With help of a 2n1 1 a 2 2n a 2n−1 , we know that x y. In addition, by a 2n2 a 2na 2n−1
a 2n1
,
a 1 2, and a 2 8, we know that x 4.
8.12 a 1 −3 2 n1 2 a n3 , L 1. Modify a 1 to make L −2.
Proof: ByMathematical Induction, itiseasytoshowthat
− 2 ≤ a n ≤ 1 for all n. *