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

100 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
l nsinnl n 1 · .
5. !ani = - 2
- -
1
::; ~ 1
{ (
< - ,so Jan!-> 0 as n-> oo. Thus, lim a,.= 0. The sequence.{an} is convergent.
n + n + n n --+oo •
3)4"} ( 3)4:t
7. 1 + ~ is convergent. Let y ~ 1 +; . Then
lim lny = lim 4xln(1 + 3/x) = lim ln( 1 + 3 /x) ~ lim ~ ( - ~)
:r:-oo :z:--+oo x --+ oo 1/ (4x} x--+oo - 1/ (4x2)
limy= lim
( 1+ - 3)"" =e 12 .
.:c-oo n---.oo n
I . 12 2
liD = 1 ,so
x - oo 1 + 3/x
9. We use induction, hypothesizing that an- 1 < an < 2. Note first that 1 < a2 = t (1 + 4) = 1 < 2, so the hypothesis holds
fo r n = 2. Now assume that ak-1 < ak < 2. Then ak = ~(ak - 1 + 4) < ~(a~.:+ 4) < ~(2 + 4) = 2. Soak < ak+l < 2,
and the induction is complete. To find the limit of the sequence, we note that L = lim an = lim an+l =>
'ft.-oo n-oo
L = HL + 4) => L = 2.
n n 1
00
n · ·
00
1
11. --g--- 1
< 3
= 2 , so L; --g--- 1
converges by the Comparison Test with the convergent p-series L; 2
[ p = 2 > 1].
n + n n n=l n + n=l n
13 ). lan+ll 1' [ (n+1)3 5n ] l' (1 1)3 1 1 1 ~ na .b h R . .,.,
. 1m -- = 1m
Tl.___.OO an n - 00 n n n-oo n 5 5 n=l 511.
5 + 1 · 3 = rm +- ·- =- < , so LJ - converges y t e at1o ,est. ·
15. Let f ( x) = ~. Then f is continuous, positive, and decreasing on [2, oo), so the Integral Test applies.
xvlnx
00
1
1t
1 !.In L !n t
f (x)dx= lim ~ dx [u = ln x. du=_: 0, {bn} is decreasing, and lim bn = 0, so the series 1
E (- 1)n- 1 Vn converges by the Alternating
n + n--+oo n=l n + 1
Series Test.
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