A7F08Sol-138

A7F08Sol-138 - Math 138 Assignment 7 Solutions Fall 2008 1....

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Unformatted text preview: Math 138 Assignment 7 Solutions Fall 2008 1. a) Use the Integral Test to show that the series X n =2 ln n n 3 2 converges. Solution Let f ( x ) = ln x x 3 2 for x 2. Then f ( x ) = x 3 2 ( 1 x )- ln x 3 2 x 1 2 x 3 = 2- 3 ln x 2 x 5 2 . Thus f ( x ) < 0 for ln x > 2 3 or x > e 2 3 1 . 95, and f ( x ) decreases for x 2. Since ln x > 0 for x > 1, f ( x ) > 0 for x > 1. Also f ( x ) is continuous for x > 1, and so we can apply the Integral Test. Z 2 ln x x 3 2 dx = lim b Z b 2 ln x 1 x 3 2 dx Int. by Parts: u = ln x,dv = 1 x 3 2 , so du = 1 x dx,v =- 2 x 1 2 = lim b - 2 ln x x 1 2 b 2 + 2 Z b 2 1 x 3 2 dx =- 2 lim b ln b b 1 2 + 2 ln 2 2 + 2 lim b - 2 x 1 2 b 2 =- 2 lim b 1 x 1 2 b + 2 ln 2 2 + 2 lim b - 2 b + 4 2 , by LH opitals rule =- 4 lim b 1 b + 2 ln 2 2- 4 lim b 1 2 + 4 2 = 0- 0 + 2(ln 2 + 2) . The integral converges and so the series also converges, by the integral test. b) Use the Corollary to the Integral Test (page 701 of the text) to determine an upper bound for the error if S 100 , the sum of the terms up to and including n = 100, is used to approximate the sum of the series in part (a). Solution The corollary of the Integral test tells us that if S = X n = k f ( n ), and S N = N n = k f ( n ), then R N = S- S N < Z N f ( x ) dx . So, if S = X n =2 ln n n 3 2 S 100 , then the error in this approximation is R N = S- S 100 < Z 100 ln x x 3 2 dx ....
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A7F08Sol-138 - Math 138 Assignment 7 Solutions Fall 2008 1....

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