A8F08Sol-138

# A8F08Sol-138 - Math 138 Assignment 8 Solutions Fall 2008 1...

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Unformatted text preview: Math 138 Assignment 8 Solutions Fall 2008 1. a) Determine the radius of convergence and the interval of convergence for the power series : ∞ X n =1 (2- 3 x ) n n . Note: the series is not in standard power series form, but can be treated in the same manner. Solution Consider lim n →∞ (2- 3 x ) n +1 ( n +1) (2- 3 x ) n n = lim n →∞ n n + 1 | 2- 3 x | = | 2- 3 x | . By the Ratio Test, the series converges absolutely for | 2- 3 x | < 1. This gives 3 2 3- x < 1 so 2 3- x < 1 3 , so the radius of convergence is R = 1 3 . 2 3- x < 1 3 = ⇒ - 1 3 < 2 3- x < 1 3- 1 3- 2 3 <- x < 1 3- 2 3- 1 <- x <- 1 3 1 >x > 1 3 , or 1 3 <x < 1 Check the endpoints: at x = 1 3 the series is ∞ X n =1 (2- 3 ( 1 3 ) ) n n = ∞ X n =1 (1) n n = ∞ X n =1 1 n , the Harmonic series which diverges. At x = 1 the series is ∞ X n =1 (2- 3(1)) n n = ∞ X n =1 (- 1) n n , the alternating Harmonic series, which con- verges (conditionally). The interval of convergence is 1 3 , 1 . b) Determine the radius of convergence and the interval of convergence for the power series : ∞ X n =0 (- 1) n n 4 2 2 n x n . Solution Consider lim n →∞ (- 1) n +1 ( n +1) 4 2 2( n +1) x n +1 (- 1) n n 4 2 2 n x n = lim n →∞ n n + 1 4 1 2 2 | x | = lim n →∞ 1 1 + 1 n 4 1 4 | x | = 1 4 | x | ....
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## This note was uploaded on 04/07/2010 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

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A8F08Sol-138 - Math 138 Assignment 8 Solutions Fall 2008 1...

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