Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

50 D CHAPTER 11 IN_ FINITE SEQUENCES AND SERIES
85. (a)
50
From the graph, it appears that the sequence { ~~ }
5
converges to 0, that is, lim n 1
= 0.
n-c::o n.
'-~~----~----~---L----~
0 10
(b)
O.Or3 _____________ "'
y= 0.1
7.5 "'--'~--~-----'-----~-------' 12.5
0
y = 0.001
9.5 ' 0.1. From the second graph, it seems that for c = 0.001, the smallest pos~ible value for N
is 11 si n~e n 5 fn! < 0.001 whenever n ~ 12.
87. Theorem 6: If lim lanl = 0 then lim -lanl = 0, and since -ian I ~an ~ Ia,. I, we have that lim an= 0 by the
n -+oo n-+oo n-oo
Squeeze Theorem.
89. To Prove: If lim an = 0 and {bn} is bounded, then lim (a.,b,.) = 0.
tl. -tOO
n -+oo
Proof: Since {bn} is bounded, there is a posi!ive number M such that lb,.l ~ M and hence, lanl lbnl ~ lanl M for
all n ~ 1. Let f: > 0 be given. Since lim an = 0, there is an integer N such that ian - 01 < Me if n > N . Then.
n~oo
la,.bn - Ol = lanbnl = lanl lbnl ~ lanl M = Ia,.- Ol M < ~ · M = f: for all n > N. Since f: was arbitrary,
lim (anbn) = 0.
n -too
91. (a) First we show that a > a1 > b1 > b.
a1 - b1 = ai b -.Jab = ~ ( ~- 2Vllb +b) = ~ ( y'a - -/bf > 0 [since a > b] => a1 > b1 . Also
a- a1 = a-Ha+ b)= Ha - ~) > 0 and b- b1 = b- .Jab= Vii( Vb - 01) < 0, so a >. a1 > b1 >b. In the same
way we can show that a1 > a2 > b2 > b1 and so .the given assertion is true for n = 1. Suppose it is true for n = k, that is,
ak > ak+l > bk+ l > bk. Then
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