The Real And Complex Number Systems

So,
3a n1 − a n a n3 − 3a n 2 ≥ 0
by (*) and fx x 3 − 3x 2 x − 1 2 x 2 ≥ 0on−2, 1. Hence, a n is an
increasing sequence with a upper bound 1. So, a n is a convergent sequence with limit L.
So,by3a n1 2 a n3 ,
L 3 − 3L 2 0
which implies that
L 1or − 2.
So, L 1sinca n ↗ and a 1 −3/2.
In order to make L −2, it suffices to let a 1 −2, then a n −2 for all n.
8.13 a 1 3, a n1 31an
3a n
, L 3 .
Proof: By Mathematical Induction, itiseasytoshowthat
a n ≥ 3 for all n. *
So,
a n1 − a n 3 − a n 2
≤ 0
3 a n
which implies that a n is a decreasing sequence. So, a n is a convergent sequence with
limit L by (*). Hence,
31 L
L
3 L
which implies that
L 3 .
So, L 3 since a n ≥ 3 for all n.
and
8.14 a n b n1
b n
,whereb 1 b 2 1, b n2 b n b n1 , L 1 5
Hint. Show that b n2 b n − b2 n1 −1 n1 and deduce that |a n − a n1 | n −2 ,ifn 4.
Proof: ByMathematical Induction, itiseasytoshowthat
Thus, (Note that b n ≠ 0 for all n)
|a n1 − a n| b n2
b n1
b n2 b n − b2 n1 −1 n1 for all n
b n ≥ n if n 4
2
.
− b n1
−1n1 ≤ 1
b n b n b n1 nn 1 1 if n 4.
n2 So, a n is a Cauchy sequence. In other words, a n is a convergent sequence, say
lim n→ b n L. Then by b n2 b n b n1 ,wehave
b n2
b n
1
b n1 b n1
which implies that (Note that 0 ≠L ≥ 1sincea n ≥ 1 for all n)
L
L 1 1
which implies that
L 1 5
2
.