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

80 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
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19. By Example 5, ( )2 = L: (n + 1}xn. Thus,
1- X n =O
!( ) _ 1 + X _ 1 X _ ~ ( } ,n ~ ( n+l
x - (1- x)2 - (1-x)2 + (1- x)2 - ~o n + 1 x + n~o n + 1}x
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= 2::(n+1}xn+ E nxn [make the starting values equal]
n=O
n=l
= 1 + f; [(n + 1) + n]xn = 1 + f; (2n + 1}xn = f; (2n + 1}xn with R = 1.
n=l n=l n = O
The series converges when l-x 2 / 16l < 1 x 2 < 16 lxl < 4, so R = 4. The partial sums are 9 1 = ~.
16
82 = s1 -
x 3 x 5 x 1 x 9
, .... Note that 8 1 corresponds to the first term of the infinite
162 , 83 = 82 + 163 , 84 = 83 - 164 , 8s = S4 + 165
sum, regardless of the value of the summation variable and the value of the exponent.
As n increases, 8n(x) approximates f better on the interval of convergence, which is ( - 4, 4}.
1 + x) I dx I dx I dx I dx
=I L~
0 (-1)nxn + n~o xn] dx = 1 [(1- x + x2 - x 3 +x 4 - • .. ) + (1 + x + x 2 + x 3 +x 4 + ... )] dx
23. f(x) = In ( - - = ln(1 + x) - ln(1 - x) = - - + -- = ( ) + - -
1 - X 1 +X 1 - X 1 - -X 1 - X
= 1 (2 + 2x 2 + 2x 4 + · · ·) dx =I f; 2x 2 " dx = C + f; 2 2 X 2n + l
n=O n=O n + 1
, oo zx2n+l oo 1
But /(0) = ln t = 0, soC = 0 and we have f(x) = E --with R = 1. Ifx = ±1, then f(x) = ± 2 E - 2 1
,
n = O 2n + 1 n =O n +
which both diverge by the Limit Comparison Test with b,. = .!. .
n
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