Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.1 SEQUENCES ¤ 459
(b)
From the first graph, it seems that the smallest possible value of N corresponding to ε =0.1 is 9,sincen 5 /n! < 0.1
whenever n ≥ 10,but9 5 /9! > 0.1. From the second graph, it seems that for ε =0.001, the smallest possible value for N
is 11 since n 5 /n! < 0.001 whenever n ≥ 12.
75. Theorem 6: If lim |an| =0then lim − |an| =0, and since − |an| ≤ an ≤ |an|,wehavethat lim an =0by the
n→∞ n→∞ n→∞
Squeeze Theorem.
77. To Prove: If lim
n→∞ a n =0and {b n } is bounded, then lim
n→∞ (a nb n )=0.
Proof: Since {b n} is bounded, there is a positive number M such that |b n| ≤ M and hence, |a n||b n| ≤ |a n| M for
all n ≥ 1. Letε>0 be given. Since lim a n =0, there is an integer N such that |a n − 0| < ε if n>N.Then
n→∞ M
|a nb n − 0| = |a nb n| = |a n||b n| ≤ |a n| M = |a n − 0| M< ε M
· M = ε for all n>N.Sinceε was arbitrary,
lim
n→∞ (a nb n )=0.
79. (a) First we show that a>a 1 >b 1 >b.
a 1 − b 1 = a + b − √
ab = 1 a − 2 √ √a √ 2
ab + b = 1 2 2
2 − b > 0 [since a>b] ⇒ a1 >b 1 .Also
a − a 1 = a − 1 (a + b) = 1 (a − b) > 0 and b − b 2 2 1 = b − √ ab = √ √b √
b − a < 0,soa>a 1 >b 1 >b.Inthesame
way we can show that a 1 >a 2 >b 2 >b 1 and so the given assertion is true for n =1. Suppose it is true for n = k,thatis,
a k >a k+1 >b k+1 >b k .Then
a k+2 − b k+2 = 1 (a 2 k+1 + b k+1 ) − a k+1 b k+1 =
a 1 2 k+1 − 2
a k+1 b k+1 + b k+1 = 1 √ak+1
− 2
b
2 k+1 > 0,
a k+1 − a k+2 = a k+1 − 1 2 (a k+1 + b k+1 )= 1 2 (a k+1 − b k+1 ) > 0,and
b k+1 − b k+2 = b k+1 − a k+1 b k+1 = b k+1
bk+1
− √ a k+1
< 0 ⇒ a k+1 >a k+2 >b k+2 >b k+1 ,
so the assertion is true for n = k +1. Thus, it is true for all n by mathematical induction.
(b) From part (a) we have a>a n >a n+1 >b n+1 >b n >b, which shows that both sequences, {a n } and {b n },are
monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem.
(c) Let lim a n = α and lim b n = β. Then lim a a n + b n
n+1 = lim
n→∞ n→∞ n→∞ n→∞ 2
2α = α + β ⇒ α = β.
⇒
α = α + β
2
⇒