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Unformatted text preview: Portillo, Tom – Homework 13 – Due: Nov 28 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the interval of convergence of the se ries ∞ X n = 1 x n √ n + 2 . 1. interval of cgce = [ 1 , 1] 2. interval of cgce = [ 1 , 1) correct 3. interval of cgce = [ 2 , 2] 4. converges only at x = 0 5. interval of cgce = ( 2 , 2] 6. interval of cgce = ( 1 , 1) Explanation: When a n = x n √ n + 2 , then fl fl fl fl a n +1 a n fl fl fl fl = fl fl fl fl x n +1 √ n + 3 √ n + 2 x n fl fl fl fl =  x  µ √ n + 2 √ n + 3 ¶ =  x  r n + 2 n + 3 . But lim n →∞ n + 2 n + 3 = 1 , so lim n →∞ fl fl fl fl a n +1 a n fl fl fl fl =  x  . By the Ratio Test, therefore, the given series (i) converges when  x  < 1, (ii) diverges when  x  > 1. We have still to check what happens at the endpoints x = ± 1. At x = 1 the series becomes ( * ) ∞ X n = 1 1 √ n + 2 . Applying the Integral Test with f ( x ) = 1 √ x + 2 we see that f is continuous, positive, and de creasing on [1 , ∞ ), but the improper integral I = Z ∞ 1 f ( x ) dx diverges, so the infinite series ( * ) diverges also. On the other hand, at x = 1, the series becomes ( ‡ ) ∞ X n = 1 ( 1) n √ n + 2 . which is an alternating series ∞ X n = 1 ( 1) n a n , a n = f ( n ) with f ( x ) = 1 √ x + 2 the same continuous, positive and decreasing function as before. Since lim x →∞ f ( x ) = lim x →∞ 1 √ x + 2 = 0 , however, the Alternating Series Test ensures that ( ‡ ) converges. Consequently, the interval of convergence = [ 1 , 1) . keywords: power series, interval of conver gence, ratio test, integral test, alternating series test Portillo, Tom – Homework 13 – Due: Nov 28 2006, 3:00 am – Inst: David Benzvi 2 002 (part 1 of 1) 10 points Find the interval of convergence of the se ries ∞ X n = 1 ( 1) n x n 4 n 2 + 5 . 1. interval of cgce = [ 4 , 5] 2. interval of cgce = [ 1 , 1] correct 3. interval of cgce = ( 4 , 5] 4. interval of cgce = [ 1 , 1) 5. interval of cgce = ( 1 , 1] 6. converges only at x = 0 Explanation: When a n = ( 1) n x n 4 n 2 + 5 , then fl fl fl fl a n +1 a n fl fl fl fl = fl fl fl fl x n +1 4( n + 1) 2 + 5 4 n 2 + 5 x n fl fl fl fl =  x  µ 4 n 2 + 5 4( n + 1) 2 + 5 ¶ . But 4( n + 1) 2 + 5 = 4 n 2 + 8 n + 9 , while lim n →∞ 4 n 2 + 5 4 n 2 + 8 n + 9 = 1 ....
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This note was uploaded on 02/18/2010 for the course MATH 408L taught by Professor Benzvi during the Spring '06 term at North Texas.
 Spring '06
 Benzvi

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