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

64 D CHAPTER 11 INFINITE SEQUENCES AND SERIES
21 . Use the Limit Comparison Test with an = 2
f+"2 and bn = ; :
n +n+ 1 n 12
. an . n 3 1 2 Jn+2 . (n 3 1 2 Jn+2)l(n 3 1 2 vn) . ) 1 +2ln /1 1
hm - = hm = lim = lim = - = - > 0
n--.oo bn n--.oo 2n 2 + n + 1 n--.oo (2n 2 + n + 1)ln 2 n--.oo 2 + 1ln + 1l n 2 2 2 ·
Since f: ; 12
is a convergent p-series [P = ~ > 1], the series f: 2
f+"2 also converges.
n=l n . n= l n + n + 1
5+ 2n 1
23. Use the Limit Compa'rison Test with a,. = ( 1
+ n 2
)2 and bn = n 3
:
lim an li n 3 (5 + 2n) l' 5n 3 + 2n 4 1ln 4 l' * + 2 . 2 0 s· ~ 1 .
n --.oo bn. = n--.~ (1 + n2)2 = n~ ( 1 + n2)2 · l l (n2 )2 = n!...~ (.l;,. + l )2 = > . mce n'S::l n3 IS a convergent
...
p-senes . [ p = 3 > 1 ], t h e senes . ~ 5 + 2n l
L.J (
2 ) a so converges.
2
n.=l 1 + n
n 2 1 oo ../n 4 + 1
2 ( ) = -- for all n ~ 1, so I: 3 2
diverges by comparison with
n n + 1 n + 1 · n=l n + n
f: - 1 - = f: ~.which diverges because it is a p-series with p = 1 ~ 1.
n.=ln+1 n=2n
27. Use the Limit Comparison Test with an = (1 + ~) 2 e- n and bn = e-": lim abn = lim .(1 + ~) 2 = 1 > 0. Since
n n-oo n n-oo n
f: e-n = f: · :. is a convergent geometric series [lrl = ~ < 1], the series f: (1 + ~) 2 e-n also converges.
n=l n=l e n=l n
29. Clearly n! = n(n - 1)(n - 2) · · · (3)(2) > 2 · 2 · 2 · · .. · 2 · 2 = 2''- 1 so..!_ < - 1 -. E - 1 - is a convergent geometric
. - ' n! - 2n-l n =1 2 n -1
series [lr I = ~ < 1], so E --\ converges by the Comparison Test.
n=l n .
31. Use the Limit Comparison Test with an == sin (;) and bn = ; . Then I: an and 2: bn are series with positive terms and
I . an . sin(1ln) . sinO . ~ . . . .
1m - = 1 rm _ = hm - - = 1 > 0. Smce L.J bn IS the divergent harmontc senes,
n--.oo bn n--.oo 1ln 0--+D 0 n=l
00
2: sin (1l n) also diverges. [Note that we could also use !'Hospital's Rule to ev~l uate the limit:
n = 1 ·
. sin(ll x) H • cos(l/x) · ( -l/x 2 ) . 1
hm I = lim I 2
= hm cos - =cosO= 1.)
x - oo 1 X %-oo -1 X x_____.oo X
lO 1 1 1 1 1 1 1 1 .
33. n~l ~ = .J2 + ..ff7 + v'82 + · · · + .JI0,05I ~ 1.24856. Now ~ < ..JTiJ = n 2 , so the error IS
00
1
Rw ~ Tw ~ 21 dx = lim [-- 1 ] t = lim ( -- 1 +- 1 ) = - 1 = 0.1.
t --.oo t 10 10
10 X t --.oo X
10
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