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

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102 0 ~HAPTER 11 INFINITE SEQUENCES AND SERIES<br />

39. Use the Limi~ Comparison Test. lim I(~ )a,, I = lim n + 1 = Jim (1 + .!) = 1 > 0.<br />

n--+00 an n - oo n n-+oo n<br />

Since E Ia,. I is convergent, so is I: \ ( n : 1 ) a,. I, by the Limit Comparison Test.<br />

41. lim I a,.+ll = lim [ lx + 21"+1 . n4" ,.] = lim [__2!:___ lx + 2[] = [x + 21 < 1 ¢:} lx + 21 < 4, so R = 4.<br />

n - oo an !'-oo (n + 1) 4"+ 1 lx + 21 n-oo n + 1 4 4<br />

oo (x + 2)n<br />

lx + 21 < 4 ¢:? -4 < x + 2 < 4 ¢:? - 6 < x < 2. lfx = - 6, then the series E becomes<br />

. • n =l n 4"<br />

f: (- 4<br />

4r = f: (- l )", the alternating harmonic series, which converges by the Alternating Series Test. When x = 2, the<br />

n=l 71. n = l n<br />

series becomes the harmonic series f: .!, which diverges. Thus, I= [- 6, 2).<br />

n=l n<br />

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 < ~.<br />

n-oo a,. n-oo ../n + 4 2"(x - 3) n-oo n + 4<br />

soR = ~- lx- 31 < ~ ¢:?<br />

· · oo 2"{x - 3}"<br />

-~ < x- 3 < ~ ¢:} ~ < x < ~ · For x =~. the series I: .;n+3 becomes<br />

· n =l n +3<br />

45.<br />

I: ~ = I: ;/ 2<br />

, which diverges [P = ~ $ 1), but for X = ~' we get I: ~ .which is a convergent<br />

n=O y n + 3 n=3 n n=O v n + 3<br />

alternating series, so I = [ ~ , ~).<br />

n /("l(x) / (n)(:g)<br />

1<br />

0 sinx 2<br />

1 cosx .il<br />

2 - sin x<br />

3 - cosx<br />

4 sin x<br />

•.<br />

2<br />

1<br />

- 2<br />

_ _il<br />

2<br />

1<br />

2<br />

"C) {3) (7r) {4}C)<br />

sinx = f(~)+f'(~)(x-~)+!.____i_(x-~) 2 + / 6 (x-~) 3 +/ 6 (x - ~) 4 + ...<br />

6 6 . 6 2! 6 3! 6 4! 6<br />

47. - 1<br />

+-<br />

1 -X = 1 _ ~- X) = n~O (- x)" = n~O {-1)" x" for lxl < 1<br />

2 00<br />

=> ~ =I: {-1f'x"+ 2 withR= 1.<br />

1+x n = O<br />

© 2012 Ccngnge Learning. All RighiS R=n-.:d. May not be S

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