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

48 D CHAPTER 11 INFINITE SEQUENCES AND SERIES
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59.
From the graph, it appears that the sequence converges to t.
As n--+ oo,
3/ n
.. ...
2 +2
=>
8 + 1/ n
so lim an =~ ·
n -+oo
0 '----~--~--"--'----./ 21
61. 2.~-------------.. From the. graph, it appears 'that the sequence {an} ,;_ { ~ 2 ~o~: } is
..
..
divergent, since it osci llates between 1 and -1 (approximately). To
n2
prove this, suppose that {a,.} converges to L. If bn = --, 2
then
1+n
{bn} converges to 1, and lim abn =!::. = L. But an = cosn, so
n-+oo n 1 bn
63.
- 2
0~-~-~~-----~-J IO
lim abn does not exist. This contradiction shows that {an} diverges.
-oon
•
From the graph, it appears that the sequence approaches 0.
1 · 3 · 5 · · · · · (2n- 1) 1 3 5 2n- 1
O < a,.= n = - · - · - · · .. ·--
(2n) 2n 2n 2n 2n
1 1
: at = 1060, a2 = 1123.60, a3 = 1191.02, a4 = 1262.48, and as = 1338.23.
(b) lim a,. = 1000 lim (1.06)", so the sequence diverges by (9) with r = 1.06 > 1.
n - oo n-+oo
67. (a) We are given that the init!al population is 5000, so Po = 5000. The number of catfish increases by 8% per month and is
decreased by 300 per month, so Pt = Po + 8%Po - 300 = 1.08Po - 300, H = l.08P1 - 300, and so on. Thus,
P,. = 1.08Pn-t - 300.
(b) Using the recursive formula with Po = 5000, we get P 1 = 5100, P 2 = 520.8, P 3 = 5325 (rounding any portion of a
catfish), P,. = 5451, P 5 = 5587, and P6 = 5734, which is lhe number of catfish in the pond after six monlhs.
69. If lrl ~ 1, then {r"} diverges by (9), so {nrn} diverges also, since lnr''l = n lr"l ~ lr''l· If lrl < 1 then
= lim ___!__I ."'
:::: 0, so lim nr" = 0, and hence { nr"} converges.
:C-tOO :c-oo r-:Z: X-+00 - T T-% %-+00 - n T n-oo .
lim xr"' = lim -=..._ :lb ~ ( In\
whenever lrl < 1.
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