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

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62 0 CHAPTER 11 INFINITE SEQUENCES 1,\ND SERIES<br />

(c) The estimate in part (b) iss ~ 1.64522 with error ~ 0.005. The exact value given in Exercise ~4 is 1r 2 / 6 ~ 1.644934.<br />

The difference is less than 0.0003.<br />

(d) Rn ~ /.~ --;. dx = .!.. So Rn < 0.001 if.!. < - 1 - {::} n > 1000.<br />

,. x n n 1000<br />

r R,. ~ 1 oo (2x + 1) - 6 dx = t~ [ 10( ; 2<br />

~ ) 1 5 l0(<br />

39. f(x) = 1/{2x + 1) 6 is continuous, positive, and decreasing on [1, oo), so the Integral Test applies. Using (3),<br />

2<br />

n \<br />

1 ) 5<br />

. To be correct to five decimal places, we want<br />

10( 2<br />

n 1 + 1<br />

) 5<br />

~ 1 ~ 6<br />

¢:} (2n + 1) 5 ~ 20,000 {::} n ~ ~ ( .e-'20,000- 1) ~ 3.12, ;muse n = 4.<br />

4<br />

1 1 1 1 1<br />

S4 = n~ 1<br />

(2n + 1)6 = 36 + S6 + 7<br />

6 + ge ~ 0.001446 ~ 0.00145.<br />

41. f: n-1.oo 1 = f: 1 ~ 01<br />

is a convergent p-series with p = 1.001 > 1. Using (2), we get<br />

n = 1<br />

n=1 n<br />

Rn < !.oo x-l.001 dx = lim [ x - o.oo1 ] t = - 1000 lim [-1-]t = - 1000(--1-) = 1000.<br />

- n t->oo - 0.001 " t->oo x0.001 n n0.001 n0.001<br />

1000 - 1000<br />

We want Rn < 0.000 000 005 {::} n 0<br />

. 001 < 5 x 10 9 {::} n°·001 > 5<br />

x 10 9<br />

n > ( 2 X 10ll) 1000 = 2 1000 X 1011,000 ~ l.07 X 10301 X 1011.000 = l.07 X 1011.301.<br />

{::}<br />

43. (a) From the figure, a2 +as+···+ an ~ Jt f(x) dx, so with<br />

y<br />

1 1 1 1 1 1" 1<br />

f(x)=-,-+-+-+ .. ·+ - ~ -dx= ln n.<br />

x2 3 4 n 1 x<br />

1 1 1 1<br />

Thus, Bn = 1 + z + '3 + 4 + .. · + ;:;: ~ 1 + ln n.<br />

(b) By part (a), s 10 o ~ 1 + ln 10 6 ~ 14.82 < 15 and<br />

0 2 3 4·" n<br />

8109 ~ 1 + ln 10 9 ~ 21.72 < 22.<br />

45. bin" = ( e 1 " b) Inn = ( e 1 " ")In b = nln b = n _ 1 n b. This is a p-series, which converges for all b such that - ln b > 1 ¢:}<br />

1<br />

lnb < - 1 ¢:} b < e- 1 {::} b < 1/e [with b > 0].<br />

11.4 The Comparison Tests<br />

1. (a) We cannot say anything about I: an. If an > bn for all nand :L b,. is convergent, then :L an could be convergent or<br />

divergent. (See the note after Example 2.)<br />

(b) If an < bn for all n, then I; an is convergent. [This is part (i) of the Comparison.Test.)<br />

n n 1 1 " II ~ n b . 'th ~ 1 , w h' 1c I 1 converges<br />

n 2n n n=l n + 1 n = l n<br />

3. 2<br />

n 3 + 1<br />

< - 2 3<br />

= - 2<br />

< 2 .or a n ~ 1, so L..- -<br />

2<br />

- 3<br />

-- converges y companson WI L..-<br />

2<br />

because it is a p-series with p = 2 > 1.<br />

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