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

62 0 CHAPTER 11 INFINITE SEQUENCES 1,\ND SERIES
(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.
The difference is less than 0.0003.
(d) Rn ~ /.~ --;. dx = .!.. So Rn < 0.001 if.!. < - 1 - {::} n > 1000.
,. x n n 1000
r R,. ~ 1 oo (2x + 1) - 6 dx = t~ [ 10( ; 2
~ ) 1 5 l0(
39. f(x) = 1/{2x + 1) 6 is continuous, positive, and decreasing on [1, oo), so the Integral Test applies. Using (3),
2
n \
1 ) 5
. To be correct to five decimal places, we want
10( 2
n 1 + 1
) 5
~ 1 ~ 6
¢:} (2n + 1) 5 ~ 20,000 {::} n ~ ~ ( .e-'20,000- 1) ~ 3.12, ;muse n = 4.
4
1 1 1 1 1
S4 = n~ 1
(2n + 1)6 = 36 + S6 + 7
6 + ge ~ 0.001446 ~ 0.00145.
41. f: n-1.oo 1 = f: 1 ~ 01
is a convergent p-series with p = 1.001 > 1. Using (2), we get
n = 1
n=1 n
Rn < !.oo x-l.001 dx = lim [ x - o.oo1 ] t = - 1000 lim [-1-]t = - 1000(--1-) = 1000.
- n t->oo - 0.001 " t->oo x0.001 n n0.001 n0.001
1000 - 1000
We want Rn < 0.000 000 005 {::} n 0
. 001 < 5 x 10 9 {::} n°·001 > 5
x 10 9
n > ( 2 X 10ll) 1000 = 2 1000 X 1011,000 ~ l.07 X 10301 X 1011.000 = l.07 X 1011.301.
{::}
43. (a) From the figure, a2 +as+···+ an ~ Jt f(x) dx, so with
y
1 1 1 1 1 1" 1
f(x)=-,-+-+-+ .. ·+ - ~ -dx= ln n.
x2 3 4 n 1 x
1 1 1 1
Thus, Bn = 1 + z + '3 + 4 + .. · + ;:;: ~ 1 + ln n.
(b) By part (a), s 10 o ~ 1 + ln 10 6 ~ 14.82 < 15 and
0 2 3 4·" n
8109 ~ 1 + ln 10 9 ~ 21.72 < 22.
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 ¢:}
1
lnb < - 1 ¢:} b < e- 1 {::} b < 1/e [with b > 0].
11.4 The Comparison Tests
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
divergent. (See the note after Example 2.)
(b) If an < bn for all n, then I; an is convergent. [This is part (i) of the Comparison.Test.)
n n 1 1 " II ~ n b . 'th ~ 1 , w h' 1c I 1 converges
n 2n n n=l n + 1 n = l n
3. 2
n 3 + 1
< - 2 3
= - 2
< 2 .or a n ~ 1, so L..- -
2
- 3
-- converges y companson WI L..-
2
because it is a p-series with p = 2 > 1.
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