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

17
lim I an+l I=
SECTION 11.7 STRATEGY FOR TESTING SERIES 0 73
lim 11 · 3 · 5 .. · · · (2n- 1)(2n 1) . 2 · 5 · 8 · .... (3n - 1) I = lim 2n 1
· n-oo a n n--+oo 2 · 5 · 8 · · · · · (3n- 1)(3n + 2) 1 · 3 · 5 · · · · · (2n- 1) n--+oo 3n + 2
= lim 2 + 1/ n = ~ < 1 ,
n--+oo 3 + 2/ n 3
oo 1 . 3. 5 .. · .. (2n - 1)
so the series 2::: ( 3
) converges by the Ratio Test.
n=l 2 · 5 · 8 · · · · · n - 1 .
ln x
1
2 - ln x ? lrin. .
2
19. Let f(x) = r . Then f (x) = 2
x 312
< 0 when lnx > 2 or x > e-, so r ts decreasmg for n > e .
vx . . vn
. I R I lim In n I' 1/n 1' 2 0 h . ~ ( )"Inn b
By I'Hosptta 's u e, r = tm ( ) = tm r = , sot e senes n6-
1
- 1 r;;, converges y the
n--+oo vn n--+oo 1/ 2 Vn n - oo vn - vn
Alternating Series Test.
21 . lim lan. l = lim IC- l )ncos(l/n 2 )1 = lim icos(1/n 2 ).1 =cos O= 1, so the series E (-l)'' cos(l/ n 2 ) diverges by the
n-oo n-oo n -'oo . n =l
Test for Divergence.
23. Using the Limit Comparison Test with a n = tan(.!.) and b,. = .!., we have
n . n
lim an = lim tan(1/ n) = lim tan(l/ x) M: lim sec2(1/x). (- 1/x2) = lim sec2(1/ x) = 12 = 1 > 0. Since
n-oo bn n --+oo 1/ n :z:--+oo 1/x :r-oo -1/x 2 :z:--+oo
00 00
2::: bn is the divergent harmonic series, 2::: a,.. is also divergent.
n = l
n.=l
U
I
. n + 1 oo n!
f"L-H>O an .,._.00 e n n. ft. - 00 en + n+ln! r&.-oc e n n= l en
· ~ li I an+l I . 1' I (n + 1)! c" 2 1
1 . (n + 1)n! · e"2 1
25. se t 1e Ratto . est. m - - = un < +t) ~ · -
1
= rm 2 2 = 1m ~+t = 0 < 1, so :L ---:r
converges.
27. roo ln: dx = lim [- ln X - .!.] t . [using integration by parts] M: 1. So E ln: converges by the Integral Test, and since
./ 2 X t-oo X X 1 n=l n
k ln k k ln k ln k . . oo k In A: •
-----,-
3
< - k 3
= -k?, the gtven senes 2:::
3
converges by the Companson Test.
(k + l) .. ·- k=t(k+ l)
29. f a n = f (-1)"___!_h = f (
- 1)" b,,. Now b,. = - 1 - 1
- > 0, {b,..} is decreasing, and lim b, = 0, so the series
n =l n=l cos n "·=1 cos 1 n n -oo
converges by the Alternating Series Test.
0 W . 1 2 2 d ~ 1 '. . . ~ 1 . b h
r: rtte co~ lt n = < - an 6 - ts a convergent geometrtc senes, so 6 --
1
- ts convergent y t e
~ e" + e-" e" n = l e" n = l cos 1 n
Comparison Test. So E (
n=l
- 1)" ___!_h is absolutely convergent and therefore convergent.
cos n
31. lim a~.; = lim k 5 4
k--+oo k-oo 3 +
k
k = (divide by 4")
oo s"
Thus, 2::: ~ diverges by the T