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

SECTION 11.2 SERIES 0 53
13.
n s.,
1 0.29289
2 0.42265
3 0.50000
4 0.55279
5 0.59175
6 0.62204 o'-~~~~--~~~~~L-~~ u
7 0.64645
From the graph and the table, it seems that the series converges.
8 0.66667
9 0.68377
10 0.69849
so I: 00 (1- - --- 1) = lim ( 1 - - -- 1) = 1
n=l Vn v'n + 1 k-oo y'kTI ·
15. (a) lim an = lim ~ = ~.so the seqitence {an} is convergent by (11.1.1).
n-oo n-oo 3n + 1 3
(b) Since lim an = ~ =I 0, the series f: an is divergent by the Test for Divergence.
n -+oo
n=l
17. 3 - 4 + 1 ; -
64 9
+ ··· is a geometric series with ratio r = -l Since lrl = ~ > 1, the series diverges.
19. 10 - 2 + 0.4 - 0.08 + · · · is a geometric series with ratio - 1
20 = -i· Since lr\ = k < 1, the series converges to
21. f: 6(0.9)n-l is a geometric series with first term a = 6 and ratio r = 0.9. Since lrl = 0.9 < 1, the series converges to
n=l
a 6 6
-- = --=- = 60.
1- 7· 1-0.9 0.1
23. f: ( ~ 3 ~n - l = ~ f: · (-~ ) n- 1 . The latter series is geometric w.ith a= 1 and ratio r =-~.Si nce lrl = ~ < 1, it
n = l 4 4 n=l 4
converges to ( 1 / ) = ~. Thus, tbe given series converges to (-!) ( ~) = t.
1 - - 3 4
oo ?Tn 1 oo (?T)n· . . . . ?T . . .
25. I: n+l = - I: - IS a geometnc senes w1th rat1o r = - 3
. Smce \rl > 1, the senes dtverges.
n=O 3 3 n=O 3
1 1 1 1 1 ~ 1 1 ~ 1 Th' . I . I f h d' h . .
27. - + - + - + - + - + · · · = L- -- = - 3
L- - . IS ts a constant mu t1p e o t e tvergent armoruc senes, so
3 6 9 12 15 n = l 3n n=l n
it diverges.
~ n - 1 . b h "' " o· . lim li n - 1 1 ....£ 0
29. L., ----- dtverges y t e .est .or 1vergence smce . an = m - 3 1
= - 3
r .
n=l 3n - 1 n-oo n-oo n -
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