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

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54 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES<br />

31. Converges.<br />

00<br />

00<br />

00<br />

1 + 2"<br />

( 1 2")<br />

[ (<br />

L:--=L: - + - =L: - 1 ) " +- ( 2 ) " ]<br />

n=1 3" n=1 3n 3" . n=1 3 3<br />

-~ ~ - ! 2-~<br />

- 1 - 1/ 3 + 1 - 2/3 - 2 + - 2<br />

[sum of two convergent geometric series]<br />

00<br />

33. L: o/2 = 2 + v'2 + ?12 + ~ + · · · diverges by the Test for Divergence since<br />

n=l<br />

lim a,. = lim o/2 = lim 2 1 / n = 2° = 1 =/= 0.<br />

n.-oo n-oo n-oo<br />

2 ) .<br />

35. E In ( 2<br />

n 2<br />

+ 1 diverges by the Test for Divergence since<br />

n=l n + 1<br />

37. E (~)" is a geometric series with ratio r = -I ~ 1.047. It diverges because lrl ~ 1.<br />

k=O<br />

00<br />

39. L: arctan n diverges by the Test for Divergence since lim an = lim arctan n = % =/= 0.<br />

n=l n.-+oo n-+oo<br />

1<br />

41 ~ ~ ( 1 1<br />

1 1<br />

)" · · · · h fi<br />

d · s· I I<br />

n = l e n=1 e e e e<br />

th ·<br />

. 6 ~ = 6 - 1s a geometric senes w1t rst term a = - an ratio r = - . mce r = - < 1, e senes converges<br />

1/e 1/e e 1 oo 1 ·<br />

) = 1. Thus, by Theorem 8(ii),<br />

1 - 1 e 1 - 1 e e e - 1 n=1 n n + 1<br />

to --<br />

1<br />

- = --<br />

1<br />

- · - = --.By Example 7, L: (<br />

oo (1 1 · )<br />

00<br />

1 "" · 1 · 1 1 e - 1 e<br />

n~1 en + n(n + 1) = n~l en + n~l n(n + 1) = e- 1 + 1 = e - 1 + e- 1 = e - 1·<br />

43. Using partial fractions, the partial sums of the series E +are<br />

' n =2 n -1<br />

n<br />

s,. = L: .<br />

2 n (<br />

. = L: -.<br />

1<br />

---.<br />

1<br />

-<br />

)<br />

i = 2 (t - 1)(t + 1) i=2 t- 1 t + 1<br />

= (1-!) + (! -!) + (! -!) + ... + ( '-1 - _1) + (-1 - .!.)<br />

3 2 4 3 5 n-3 n- 1 n-2 n<br />

Th . . I . . d 1 1 1 1<br />

n - n<br />

1s sum 1s a te escopmg senes an sn = + - 2<br />

~ -- 1<br />

- - ·<br />

Thus, E --i:-- 1<br />

= l~ Sn = lim (1 +! - - 1 -- .!.) = ~-<br />

n = 2 n - n-+oo n-+00 2 n- 1 n 2<br />

oo 3 n 3 n (1 1 )<br />

45. For the series :E , s,. = :E -. -.-- = L: -:- - -. -<br />

n=1 n(n+3) i =1 t(1 + 3) i=1 \ t+3<br />

[using partial fractions). The latter sum is<br />

(1 - l) + (l ....: l) + (1 - l) + (l - l) + ... + (-1 - l) + (-1 - _1 ) + (-1 - _1 ) + (.!. - --4)<br />

4 2 5 3 6 4 7 n-3 n n -2 n + 1 n-1 n+2 n n+3<br />

= 1 + 2 1 + 3- 1 1 1 1 n+1 - n+ 2 - n+3 [tl. e escopmg senes ' ]<br />

TI ~ 3 l' l' (1 1 1 1 1 1 ) 1 1 11 c<br />

ms, 6 (<br />

n = l n n + n-oo n-oo n 1l n<br />

3 ) = tm Bn = 1m + 2 + 3 - - + 1 - - + 2 - ..,----+ 3<br />

= 1 + 2 + 3 = 6 . onverges<br />

© 2012 Cengagc Learning. All Righ!S Reserv<strong>ed</strong>. Muy not be scann<strong>ed</strong>. copi<strong>ed</strong>. or duplicotcd, or po>l<strong>ed</strong> lo n publicly occcs:Jiblc wobsitc. In whole or ip part.

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