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

74 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
"2 jn 1 1 1 oo n2
33. lim ~ = lim (~1 ) = lim [( 1 ) I t = li ( 1· 11
r = - < 1, so the series I: (·-n-)
n-+oo n-+00 n + n-+00 n + n m + n e . n=1 n + 1
converges by the Root Test.
35. an = n 1 ; 1 /n = n . ~ 1 /n, so let bn = ; and use the Limit Comparison Test. l . a, lim 1 1 0
n-->oo
un - = . --= >
n-+oo bn n-+oo n1/n
[see Exercise 4.4.61 ], so th~ series f: 1
1 11
,. diverges by comparison with the divergent harmonic series.
. n=l n
. 00
37. lim ~ = lim (2 1 /" - 1) = 1 - 1 = 0 < 1, so the series I: ( y/2 - 1)"' converges by the Root Test.
n-~oo n-too n-=1 .
11.8 Power Series
1. A power series is a series of the form :L:::'=o enx" = co + c1x + c2x 2 + c3x 3 + · ··,where x is a variable and the en's are
constants called the coefficients of the series.
More generally, a series of the form :L:::'=o c,,(x - a)" =eo+ c 1 (x- a)+ c2(x- a) 2 + · · · is called a power series in
( x - a) or a power series centered at a or a power series about a, where a is a constant.
I I a ~ (-1) "+ 1
lim n+ 1 = lim ( (n+1)x"-H ) I = lim I (-1)--x n+1 I = lim [( 1 + -1) JxJ ] = JxJ. By the Ratio Test, the
n--+00 Un n-tOO - 1 71. nxn n-H)O n n-00 n '
00 .
series I: ( -1)"nxn converges when !x i < 1, so the radius of convergence R = 1. Now we' II check !he endpoints, that is,
n=l
x = ± 1. Both series f: (- 1)"n(±1)" = f: (=F1)"n diverge by the Test for Divergence since lim J(=F1)"nl = oo. Thus,
n=l · n=l TI.--+OO
the interval of convergence is I = ( -1, 1).
x" ' . I a,.+1 1 . I x"+l 2n - 11 . (2n - 1 ) . (2- 1/n / lxJ ) = JxJ. By
n- 1 n-+oo an n-+oo n + x" n-+oo n + n-+oo 2 + 1 n
5. If an = - 2
--, then lim -- = hm - 2 1 · -- = !1m - 2 1
Jxl = hm
·oo ~ oo 1 ·
diverges by
n=l n - 1 n=l n - 1
the Ratio Test, the series I: - 2
--converges when Jx! < 1, so R = 1. When x = 1, the series I: - 2
. . I ~ 1 . 1 1 d 1 ~ 1 d' . . . I . I fth h . .
companson Wit 1 6 - srnce --- > - an - 6 - 1verges smce It ts a constant mu tip e o e armoruc senes.
n=1 2n 2n - 1 . 2n 2 n=l n
When x = - 1, the series f: ( - 1 ) n converges by t~e Alternating Series Test. Thus, the interval of converge~ce is [ -1, 1).
n=1 2n -1
7. If an= x",, then lim lan+l l = lim I ( x"+\; ·. n ! I= lim 1- x -1 = !xi lim -
n. n -+ oo an n-+oo n + 1 . X" n -+oo n + 1 n-+oo n + 1
So, by the Ratio Test, R = oo and I= ( - oo, oo).
1 - = lxl· 0 = 0 < 1 for all real x.
lim I an+1 I = lim J (n + 1)2 x"+l . ~~ = lim I x(n + 1)2 1 = lim [JxJ (1 + .!.)2] = EJ.(1)2 = ~ JxJ. By the
n - oo . an n-CXJ 2n+l n 2 x"· n.-oo 2n2 n-oo 2 n 2
. .
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