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

60 0 CI:IAPTER 11 INFINITE SEQUENCES AND SERIES
17. The function f(x) = ~
4
is continuous, positive, and decreasing on [1, oo), so· we can apply the Integral Test.
X + .
l
oo 1 · . ~t 1 · . [1 _1 x]t · ·1 . [ · _1 (t) _1 (1)]
--dx = lim - - dx = lim -tan - = - hm tan - - tan -
1 x2 + 4 t --+00 1 x 2 + 4 t --+oo 2 2 1 2 t--+oa 2 2
=.!. [~ - tan- 1 (.!.)]
. 2 2 . 2
Therefore, the series f: ~ converges.
n=l n +
19 ill n . ln n . ln 1 0 Tb fu . f( ) lnx . . d · · [ 2 )
. ~ - = ~ -
3 3
smce - = . e nchon . x = - 3
1s contmuous an postttve on , oo .
n = l n n=t n 1 X
I
f'(x) = x 3 (1/x)-(ln x)(3x 2 )=x 2 -3x 2 lnx=1 - 3lnx - 1 {::}
. (x3)2 x6 x4
3
x > e 1 1 3 ~ 1.4,' so f is decre~ing on [2, oo), and the Integral Test applies.
lnx dx ~ lim [- lnx - - 1 1
-] t = lim [-_!._(2 lnt + 1) + .!.] (~) - , so the series 4
f: In~
100
Inx dx = lim 1t
2
x3 t -oo 2
x3 t-+oo 2x 2 4x 2 1
converges.
t-+oo 4t 2 4 n= 2 n
(*): u = hl x, dv = x - 3 dx => du = (1/x) dx, v = -~x - 2 , so
I ln x d 1 -2 1 . ;· 1 -2 ( 1 / ) d 1 -21n · 1 ;· -3d 1 -2 1
1 -2 . C
~ x = - 2 x n x - - 2 x x x = - 2 x x + 2 x x = - 2 x nx- 4 x + .
(**):lim (~2lnt+ 1 ) J:!:_lim 2/ t =-l lim ]:_ =0.
t--+oo 4t 2 t --+oa 8t
4 t--+oo t2
21. f (x) = +is continuous and positive on [2, oo), and also decreasing since J'(x) =- 1 2 Tlnln) 2
< 0 for x > 2, so we can'
x = x . x x
use the lntegra1Test.1 ."" ln 1 1
dx= lim [ln(lnx)J; = lim [ln( lnt)~ln(ln2)]= oo,s otheseries f - -diverges.
. 2 X X t--+oo t--+oo . n=2 n 1 n n
23. The function f(x)'= e 1 1"jx 2 is continuous, positive, and decreasing on [1, oo), so the Integral Test applies.
[g(x) = e 1 1~ is decreasing and dividing by x 2 do es~'t change that fact.]
l
oo ~ t e1frz; · t · . 00 el/n
f(x) dx = lim - 2 - dx = lim [-e11 x] = - lim (e 1 / t - e) = -(1 - e) = e- 1, so the series 2:: - 2 -
1 t-oo 1 X t --t-oo . 1 t --+CXJ n=1 n
converges.
. 1 1 1 1 -
25. The funct10n f(x) = - - - 2
= - - -
3 2
+ -- [by partial fractions] is continuous, positive and decreasing on [1, oo),
q; +x x x x+ l
so the Integral Test applies.
00
1 It
f(x)dx = lim ( 21 -- 1 +-- 1 ) dx = lim [- -1 - ln x +ln(x +:t) . J t
1 t --+oo 1 X X X + 1 t--+oo X l
= hm . [-- 1 + ln-- t + 1 + 1-ln2 ] =0 + 0 + 1-ln2
t--+oo t t
Th e . mtegra ' I converges, so th e senes . 1 ~ - -- I
2 3
converges. .
n=l n +n
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