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

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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS 0 61
27. T he function f(x) = coFxx is neither positive nor decreasing on (1, oo), so the hypotheses of the Integral Test are not
. · oo COS7Tn
satisfied for the senes L: ~ .
n = l y n
00 1
29. We have already shown (in Exercise 21) that when p = 1 the series E (l ) diverges, so assume that p =1- L
. n=2 n n n p
f (x) = ( 1 ) is continuous and positive on [2, oo), and f'(x) = - ~~ ~ : 1 < 0 ifx > e - P, so that f is eventually
X ln x P · X X 'P
decreasing and we can use the Integral Test.
roo ---:-:--1-:-- d l' [ (ln X) 1-p] t
} 2 x(ln x)P x=t~ 1 - p
2
[forp :f- 1] = lim [ (ln t)1- p - (ln 2)1 - p]
h oo · 1-p 1-p
This limit exists whenever 1 - p < 0 ¢:? p > 1, so the series converges for p > L
31 . Clearly the series cannot converge if p 2': -~ . becau se then lim n(1 + n 2 )P =1- 0. So assume p. < -~ . Then ·
n-- 1. Unless specified otherwise, the domain of a function f is the
set of real numbers x such that the expression for f ( x) makes sense and defines a real number. So, in the case of a series, it's
the set of real numbers x such that the series is convergent.
00 (3)4
35. (a) n~l ;;:
E 4 = 81 I: 4 = 81 90 = -1o
00 81 00 1 (7T4) 97T4
n=l n
n = 1 n
00
1 1 1 1
00
1 . 7T 4 ( 1 1 ) 7r 4 17
(b) k~6 (k - 2)" = 34 + 44 + 54 + ... = k~3 k4 = 90 - 14 + 24 [subtract a I and a2] = 90 - 16
37. (a) j (x) = ...;- is positive and continuous and f'(x ) =- ~ is negative for x > 0, and so the Integral Test applies.
X X -
00 . 1 1 1 1 1
n~l n2 ~ sw = I2 + 22 + 32 + .. : + 102 ~ 1.549768.
Rw :::; roo ...;- dx = lim [- 1 ] t = lim (-.!. + 1
1
} 10
X t --.oo X
10
00
1 100
11 X 10 X
t --.oo
t
0 ) = 1
, so the error is at most 0.1.
10 (b) 8 10 + 2 dx :::; S :::; SlO + 21
dx =? SlQ + 1
\ :::; S :::; 8 10 + fa
1.549768 + 0 .090909 = 1.640677 :::; s :::; 1.549768 + 0.1 = 1.649768, so we gets ~ 1.64522 (the average of 1.640677
and 1.649768) with error :::; 0.005 (the maximum of 1.649768 - 1.64522 and 1.64522- 1.640677, rounded up).
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