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

102 0 ~HAPTER 11 INFINITE SEQUENCES AND SERIES
39. Use the Limi~ Comparison Test. lim I(~ )a,, I = lim n + 1 = Jim (1 + .!) = 1 > 0.
n--+00 an n - oo n n-+oo n
Since E Ia,. I is convergent, so is I: \ ( n : 1 ) a,. I, by the Limit Comparison Test.
41. lim I a,.+ll = lim [ lx + 21"+1 . n4" ,.] = lim [__2!:___ lx + 2[] = [x + 21 < 1 ¢:} lx + 21 < 4, so R = 4.
n - oo an !'-oo (n + 1) 4"+ 1 lx + 21 n-oo n + 1 4 4
oo (x + 2)n
lx + 21 < 4 ¢:? -4 < x + 2 < 4 ¢:? - 6 < x < 2. lfx = - 6, then the series E becomes
. • n =l n 4"
f: (- 4
4r = f: (- l )", the alternating harmonic series, which converges by the Alternating Series Test. When x = 2, the
n=l 71. n = l n
series becomes the harmonic series f: .!, which diverges. Thus, I= [- 6, 2).
n=l n
43. lim I Un+l l = lim 1 2 n+ 1 (x- 3 )"+1 · y'n+3 ,., = 2 jx- 31 lim Jn + 3 = 2lx- 31 < 1 ¢:} jx- 31 < ~.
n-oo a,. n-oo ../n + 4 2"(x - 3) n-oo n + 4
soR = ~- lx- 31 < ~ ¢:?
· · oo 2"{x - 3}"
-~ < x- 3 < ~ ¢:} ~ < x < ~ · For x =~. the series I: .;n+3 becomes
· n =l n +3
45.
I: ~ = I: ;/ 2
, which diverges [P = ~ $ 1), but for X = ~' we get I: ~ .which is a convergent
n=O y n + 3 n=3 n n=O v n + 3
alternating series, so I = [ ~ , ~).
n /("l(x) / (n)(:g)
1
0 sinx 2
1 cosx .il
2 - sin x
3 - cosx
4 sin x
•.
2
1
- 2
_ _il
2
1
2
"C) {3) (7r) {4}C)
sinx = f(~)+f'(~)(x-~)+!.____i_(x-~) 2 + / 6 (x-~) 3 +/ 6 (x - ~) 4 + ...
6 6 . 6 2! 6 3! 6 4! 6
47. - 1
+-
1 -X = 1 _ ~- X) = n~O (- x)" = n~O {-1)" x" for lxl < 1
2 00
=> ~ =I: {-1f'x"+ 2 withR= 1.
1+x n = O
© 2012 Ccngnge Learning. All RighiS R=n-.:d. May not be S