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

SECTION 11.8 POWER SERIES 0 75
l
00 n2xn
Ratio Test, the series 2:: ( -1)'" .- ,.- converges when t lx l < 1 ¢> lx l < 2, so the radius of convergence is R = 2.
n = l 2
lim I (:fl)n n 2 1 = oo. Thus, the interval of convergence is I = ( -2, 2).
n -oo
( - 3)"x"
11. If an = n 312
, then
00
2:: (=F1)"'n 2 diverge by the Test for Divergence since
n=l
( 3)n+l n+l 3/2 I I (. )3/21 ( )3/2
I I I
11m . -- a,.+t = 1' lffi - x · n = 1' Im - 3 X --
n = 3 1 X
I 1' tm . 1
n-•oo a., n~oo (n + 1)3/2 ( - 3)nxn n~oo n + 1 n-+oo 1 + 1/n .
= 3lx l (1) = 3 lxl
00 ( -3)"
By the Ratio Test, the series 2:: '- x "' converges when 3 lxl < 1 ¢> lx l < %, so R = %- When x = S· the series
,.= 1 nyn
· I: ~ converges by the Alternating Series Test. When x = - %, the series 2::
00 (- 1)" . 00 1
n=l n
( p = ~ > 1). Thus, the interval of convergence is [-S, t].
312
n=l n
is a convergent p-series
I
13.I f an = ( - 1 ) "~. then lim lan+t l = lim I xn+l . _4" lnn i= E11im 1n n =1.::1 · 1
4" ln n n~oo an n-+oo 4n+l ln(n + 1) x" 4 n~oo ln(n + 1) 4
[by !' Hospital's Rule] = 1: 1. By the Ratio Test, the series converges when 1:1 < 1 ¢:> lx l < 4, so R = 4. When
00 00 00
xn [(-1)(- 4)t 1 . 1 1
00
1
- > - an d 2:: - is tbe
n=2 n n = 2 n n =2 n n n n n n = 2 n
x = - 4, 2:(-1)" 4
n ln = 2::
4 " ln = 2: - 1 - . Smce lnn < n for n ~2 . - 1
divergent harmonic series (without then = 1 term), f - 1
1 ' is divergent by the Comparison Test. When x = 4,
n=2 n n .
00 xn 00 1
2:: ( - 1)n 4
n ln = 2:: ( -1)"-ln , which converges by the Alternating Series Test. Thus, I = ( - 4, 4] .
n =2 n n=2 n
(x - 2)"' . I an+t I . ·1(x - 2)"+1 n 2 + 1 I . n 2 + 1
15. If an = 2 1 , then lim - - = lim ( 1 ) 2 1 · ( 2 ) = lx- 21 lim ( 1 ) 2 = lx - 21. By the
n + n ~oo an n-+oo n + + X - " n~ oo n + + 1
oo (x- 2)"
Ratio Test, the series 2::
2
converges when lx- 21 < 1 [R = 1] ¢:> -1 < x- 2 < 1 ¢:> 1 < x < 3. When
n=O n + 1
x = 1, the series f: (- 1)" + converges by the Alternating Series Test; when x = 3, the series 1
f: - 1 - 2
- converges by
n = O n + n=O n + 1
comparison with the p -series f: ~ [p = 2 > 1]. Thus, the interval of convergence is I = [1, 3]. ·
n = l n .
00
3"'(x + 4)"
By the Ratio Test, the series .~ Vn converges when 3 lx + 41 < 1 ¢:> lx + 41 <
1
1 [ R = t ] ¢:>
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