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

SECTION 11.1 SEQUENCES 0 47
35. a,. = cos( n /2). This sequence diverges since the tenns don't approach any particular real number as n---> oo.
I
The terms take on val4es between - 1 and I.
_ (2n - 1)! = (2n - 1}! · = 1 · ___. 0 as n---> oo. C
37· an - (2n + 1)! (2n + 1}(2n}(2n- 1)! (2n + 1}(2n) onverges
1 + e - 2n
--'------ ---> 0 as n---> oo because 1 + e- 2 n ---> 1 and e" - e-n -+ oo.
en- e n
Converges
2 -n n 2 • • x 2 H 2x 11 2
41. an = n e = -. Smce lim - = lim - = lim - = 0, it follows from Theorem 3 that lim an= 0. Converges
en X--+00 e X X-tCO eX X--tOO e:t n-oo
2
43. 0 < cos n < _..!:.._ · [since 0 :s; cos
2 1
n :s; 1], so since lim 2
- 2n - 2" n->oe> " n ·
= 0, { 'c 0 82
n } converges to 0 by the Squeeze Theorem.
2
. ( / ) sin(1/
/ n
n)
.
s·
mce
li
m
sin(1/x)
/ = 1 . tm sin t [ w h ere t = 1 x = 1, tt •O ows · om Theorem 3
45. an = n sm 1 n =
1
that {an } converges to I .
x ..... oo 1 X t-o+ t
/ ] . ., II fr
(
2)"' . ( 2)"
2
= lim = 2 ::::}
x ->oo 1 + 2/x
lim 1 + - = lim e 1 n Y = e 2 , so by Theorem 3, lim 1 + - = e 2 • Converges
!t-oo X :r-oo n -+oo n
( 2n2 1) (2 1/n 2 )
49. an = In(2n 2 + 1)- ln(n 2 + 1) =In n 2 + = In . + / n 1 1 1 2
-+ In 2 as n ---> oo. Converges
51. a.,, = arctan(ln n). Let f(x) = arctan(ln x). Then lim f(x) = ~s ince In x -+ oo as x---> oo and arctan is continuous.
x - oo
Thus, lim a,= lim f(n) = ~-
n-+oo
n -+oo
Converges
53. {0, 1, 0, 0, 1, 0, 0, 0, 1, ... } diverges since the sequence takes on only two values, 0 and I, and never stays arbitrarily close to
either one (or any other value) for n sufficiently large.
n ! 1 2 3 (n- 1) n 1 n n
55. an= 2
n = 2 · 2 · 2 · · · · · - - 2
- · 2 2': 2 · 2 (for n > 1] = 4--> oo as n--+ oo, so {an} diverges.
57. 2~------------~ From the graph, it appears.that the sequence converges to 1.
{ ( -2/e)n} converges to 0 by (7), and hence {1 + ( - 2/e)"}
.. . . .. . .
converges to 1 + 0 = 1.
0
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