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

54 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
31. Converges.
00
00
00
1 + 2"
( 1 2")
[ (
L:--=L: - + - =L: - 1 ) " +- ( 2 ) " ]
n=1 3" n=1 3n 3" . n=1 3 3
-~ ~ - ! 2-~
- 1 - 1/ 3 + 1 - 2/3 - 2 + - 2
[sum of two convergent geometric series]
00
33. L: o/2 = 2 + v'2 + ?12 + ~ + · · · diverges by the Test for Divergence since
n=l
lim a,. = lim o/2 = lim 2 1 / n = 2° = 1 =/= 0.
n.-oo n-oo n-oo
2 ) .
35. E In ( 2
n 2
+ 1 diverges by the Test for Divergence since
n=l n + 1
37. E (~)" is a geometric series with ratio r = -I ~ 1.047. It diverges because lrl ~ 1.
k=O
00
39. L: arctan n diverges by the Test for Divergence since lim an = lim arctan n = % =/= 0.
n=l n.-+oo n-+oo
1
41 ~ ~ ( 1 1
1 1
)" · · · · h fi
d · s· I I
n = l e n=1 e e e e
th ·
. 6 ~ = 6 - 1s a geometric senes w1t rst term a = - an ratio r = - . mce r = - < 1, e senes converges
1/e 1/e e 1 oo 1 ·
) = 1. Thus, by Theorem 8(ii),
1 - 1 e 1 - 1 e e e - 1 n=1 n n + 1
to --
1
- = --
1
- · - = --.By Example 7, L: (
oo (1 1 · )
00
1 "" · 1 · 1 1 e - 1 e
n~1 en + n(n + 1) = n~l en + n~l n(n + 1) = e- 1 + 1 = e - 1 + e- 1 = e - 1·
43. Using partial fractions, the partial sums of the series E +are
' n =2 n -1
n
s,. = L: .
2 n (
. = L: -.
1
---.
1
-
)
i = 2 (t - 1)(t + 1) i=2 t- 1 t + 1
= (1-!) + (! -!) + (! -!) + ... + ( '-1 - _1) + (-1 - .!.)
3 2 4 3 5 n-3 n- 1 n-2 n
Th . . I . . d 1 1 1 1
n - n
1s sum 1s a te escopmg senes an sn = + - 2
~ -- 1
- - ·
Thus, E --i:-- 1
= l~ Sn = lim (1 +! - - 1 -- .!.) = ~-
n = 2 n - n-+oo n-+00 2 n- 1 n 2
oo 3 n 3 n (1 1 )
45. For the series :E , s,. = :E -. -.-- = L: -:- - -. -
n=1 n(n+3) i =1 t(1 + 3) i=1 \ t+3
[using partial fractions). The latter sum is
(1 - l) + (l ....: l) + (1 - l) + (l - l) + ... + (-1 - l) + (-1 - _1 ) + (-1 - _1 ) + (.!. - --4)
4 2 5 3 6 4 7 n-3 n n -2 n + 1 n-1 n+2 n n+3
= 1 + 2 1 + 3- 1 1 1 1 n+1 - n+ 2 - n+3 [tl. e escopmg senes ' ]
TI ~ 3 l' l' (1 1 1 1 1 1 ) 1 1 11 c
ms, 6 (
n = l n n + n-oo n-oo n 1l n
3 ) = tm Bn = 1m + 2 + 3 - - + 1 - - + 2 - ..,----+ 3
= 1 + 2 + 3 = 6 . onverges
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