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

11 . D INFINITE SEQUENCES AND SERIES
11.1 Sequences
1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers,
(b) The terms a,. approach 8 as n becomes large. In fact, we can make an as close to 8 as we like by taking n sufficiently
large.
(c) The terms an become large as n becomes large. In fact, we can make an as large as we like by taking n sufficiently large.
2n . { 2 4 6 8 10 } { 4 3 8 5 }
3. an = n2 + 1, so the sequence IS 1 + 1' 4 + 1 ' 9 + 1' 16 + 1' 25 + 1' . . . = 1, 5' 5' 17' 13 ' . . . .
(-1)n- l . . { 1 -1 1 -1 1 } { 1 1 1 1 1 } ·
5. an = 5n , so the sequence IS 51' 52' 53 ' 54' 55 '... = 5' - 25' 125'-625' 3125'... .
1 . {1 1 1 1 1 } {11 1 1 1 }
7. ~ = (n + 1)!, so the sequence IS 2!'"3!' 4f• 5!' 6!'... = 2' 6' 24 , 120, 720'... .
9. a 1 = 1, a,,+ 1 = 5an- 3. Each term is defined in terms of the preceding term. a2 = 5a1 - 3 = 5(1)- 3 = 2.
a 3 = 5a2 - 3 = 5(2) - 3 = 7. a4 = 5a3 - 3 = 5(7) - 3 = 32. as = 5a4 - 3 = 5(32) - 3 = 157.
The sequence is {1, 2, 7, 32, 157, ... }.
a3
Q.4 = --=
2/5 2
=
1 + a3 1 + 2/5 7•
a 5 = 1
: 4 a
4
=:' 1
~;/7 = ~- ,The sequence is {2, ~. ~. ~' ~ •... }.
The denominator of the nth term is the nth positive odd integer, so an= - - 1 -.
2n-1
15. { - 3, 2, -1, ~,- ;~, ... }. The first term is -3 and each term is - ~ times the preceding one, so a,. = -3 ( -~ f - 1 .
17. { ~, -~,%,- 1 :, 2 5~, ... }. The numerator of the nth term is n 2 and its denominator is n + 1. Including the alternating signs,
n2
wegetan = (-1)n+l __ _
n+1
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