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

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS 0 59
11.3 The Integral :rest and Estimates of Sums
1 r 2 ·1
1. The picture sh?WS that a 2 = 2
1. 3 < 1
1
x 1 . 3 dx,
y
1 131 00 1 [ 00 1
a 3 = "1.3 < "1.3 dx, and so on, so 2:: 1':3 < "1.3 dx. The
. 3 . 2 X n = 2 n . 1 X
integral converges by (7.8.2) withp = 1.3 > 1, so the series converges.
0 2 3 4
X
3. The function f(x) = 1/ ~ = x- 1 / 5 is continuous, positive, and decreasing on [1 , oo ), so the Integral Test applies.
f oo x- 1 1 5 dx = lim
1
1 . . t-.oo 1 t-.oo
J' x- 1 / 5 dx = lim [2x 4 1 5 ] t = lim (~t 4 1 5 - .§.) = oo, so E. 1/ ifrf diverges.
4
1 t--.oo ·. 4 n = 1
5. The function f(x) = ( 2
x ~ 1
) 3
is continuous, positi~e , and decreasing on [1, oo), so the lf1tegral Test applies .
. roo
1
dx ~ lim r 1 dx = lim [-.!
1
] t = lim ( -
1
+ _.!_) = _.!__
1 1
(2x + 1)3 t-+oo 1 1 (2x + 1) 3 t-.oo. 4 (2x + 1)2
1
t- oo 4(2t + 1)2 36 36
Since this improper integral is-convergent, the series E ( 2
1
1 ) 3
. n=l n +
is also convergent by the Integral Test.
7. The function f(x) =-/-- is continuous, positive, and decreasing on [1, oo), so the futegral Test applies.
X + 1 . .
roo ~ 'dx = lim r -/-- 1
dx = lim [- 2
1 ln( x
2
+ 1 )] t = - 2
1 lim [ln( e + 1) - ln 2] = oo. Sine~ this improper
} 1
X + 1 t-.oo } 1
X + t-->oo .
1
. t--+oo ·
integral is divergent, the series E ~ is also divergent by the Integral Test.
n=1 .n + 1
9. E ~ is a p-series with p = -/2 > 1, so it converges by (1 ).
n=1 n . .
1 1 1 1 ~ 1 Thi . . . 'th 3 1 . b ( )
11.1+-+-+ - + - + ···= 6 3 · s tsap-seneswL p = > ,so1tconverges y I.
8 27 64 125 n=l n
1 1 1 1
00
1 1
13. 1 +- +- + - + - + · · · = L: --. The function f(x) = 2
x _- 1· is
3 5 7 9 n =l 2n - 1
continuous, positive, and decreasing on [1, oo ), so the· Integral Test applies.
1 -·- dx = lim ft - 2
1
dx = lim a 1n 12x - 111 ~ = ~ lim (1n(2t - 1) - 0) = oo: so the series r: - 2
1
r oo -
} 1 2x- 1 t -+oo 1 X- 1 t -+oo t-->oo n=l n - 1
diverges.
oo vfn+ 4 'oo ( vfn 4 ) oo 1 oo 4
15. L: --2- = L: - 2 +2 = L: 3 / 2 + L: 2•
n =1 n n = l n n n=l n n =1 n
~ 1 . . 'th 3 .
6 3!2 IS a convergent p-senes w1 p = 2 > 1.
n = l n
E 4 = 4 E ~ is a constant multiple of a convergent p-series with p = 2 > 1, so it converges. The sum of two
2
n=l n . n = l n
convergent series_ is-convergent, so the original series is conv~rgent.
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