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

68 D CHAPTER 11 INFINITE SEQUENCES AND SERIES
35. 2: b 2 n = 2: 1/(2n) 2 clearly converges (by comparison with ~e p-series for p = 2). So suppose that 2: (-1r-I bn
. .
converges. Then by Th~orem 11.2.8(ii), so does 2: [(-1 t- 1 1
bn + bn] ::: 2 ( 1 + l + i + · · · ) = 22: - . But 2
thi~
. n -1
diverges by comparison with the harmonic series, a contradiction. Therefore, 2: (....:1)"- 1 b,. must diverge. The Alternating
Series Test does not apply since {bn} is not. decreasing.
11.6 Absolute Convergence and the Ratio and Root Tests
1. (a) Siilce lim I an+l I = 8 > 1, part (b) of the Ratio Test tells us that the series 2: an is divergent.
n-oo an. . .
(b) Since lim I an+l I = 0.8 < 1: part (a) of the Ratio Test tells us that the series 2: an is absolutely convergent (and
n.-CX> an
therefore convergent).
(c) Since lim I an+l I = 1, the Ratio Test fails and the series 2: an might converge or it might diverge.
-~ an . .
3. lim lan+l l = lim In+ 1 · 5 " 1 = lim 1.!. · n+ 1 1 =.!.lim 1 + 1 /n = .!. (1) = .!. < 1,sotheseries E ~is
n-too an n-oo 5n+l n n-too 5 n 5 n-too 1 5 5 n=l 5n
absolutely co.nvergent by the Ratio Test.
1
5. bn = - 5 n + n-->oo . n=O 5n + 1
1 > 0 for n 2::' 0, { bn} is decreasing for n ~ 0, and lim bn = 0, so E ( -1 t converges by the Alternating
Series Test. To determine absolute convergence, choose an = ~ to get
- n
li m -b
an
=
1'
tm
1/n l' ' 5n + 1 5 0 1 d. . b .h L ' . C . ,.. . h th
1 / ( 5 1 ) = rm --= > , so ~ -- 1verges y t e mut ompar1son . est w1t e
n~~ n n-~ n + n-+oo n n=l 5n + 1
' .
hannonic series. Thus, the series E 5
( - 1 )" 1
is conditionally convergent.
n=O n+
7. lim lak+Il = lim [(k-+: 1 ) (~_)'"+ 1 ] = lim k +
1 (~)
1
= ~ lim (1
. k~oo a k k-oo k(~)k k -too k 3 3 k~ oo ' k
+ .!..) = ~'( 1 ) = 1 < ·1,sotheseries
E k ( ~) k is apsolutely convergent by the Ratio Test. since the terms of this series are positive, absolute convergence is the
n=1
same as convergence.
9. lim 1 an+ I I = lim [ (l.l)n+l . ~J = lim (1.1)n 4 1
·= (1.1) lim
n -- 1,
n = l
- 1)n (l.~)n diverges by the Ratio Test.
n
3
' 3
· = (1.1) lim ( ~ ~ )4
el/n e ( 1 ) oo 1 · ' 00 e l /n
11. Since 0 ~ - 3 - ~ 3 = e 3 and 2: 3 is a convergent p-series [p = 3 > 1 ], L: - 3
- converges, and so
n n n n = l n n = l n
oo (- 1)"el/n.
2::
3
is absolutely convergent.
n=l n
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