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Unformatted text preview: Portillo, Tom – Homework 12 – Due: Nov 21 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 20 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 Which one of the following series is conver gent? 1. ∞ X n = 1 3 5 + √ n 2. ∞ X n = 1 ( 1) n 1 1 + √ n correct 3. ∞ X n = 1 ( 1) n 1 1 + √ n 5 + √ n 4. ∞ X n = 1 ( 1) 2 n 5 3 + √ n 5. ∞ X n = 1 ( 1) 3 5 3 + √ n Explanation: Since ∞ X n = 1 ( 1) 3 5 3 + √ n = ∞ X n = 1 5 3 + √ n , use of the Limit Comparison and pseries Tests with p = 1 2 shows that this series is divergent. Similarly, since ∞ X n = 1 ( 1) 2 n 5 3 + √ n = ∞ X n = 1 5 3 + √ n , the same argument shows that this series as well as ∞ X n = 1 3 5 + √ n is divergent. On the other hand, by the Divergence Test, the series ∞ X n = 1 ( 1) n 1 1 + √ n 5 + √ n is divergent because lim n →∞ ( 1) n 1 1 + √ n 5 + √ n 6 = 0 . This leaves only the series ∞ X n = 1 ( 1) n 1 1 + √ n . To see that this series is convergent, set b n = 1 1 + √ n . Then (i) b n +1 ≤ b n , (ii) lim n →∞ b n = 0 . Consequently, by the Alternating Series Test, the series ∞ X n = 1 ( 1) n 1 1 + √ n is convergent. keywords: 002 (part 1 of 1) 10 points Determine whether the series ∞ X n = 1 ( 1) n +1 e 1 /n 5 n is absolutely convergent, conditionally con vergent or divergent. 1. conditionally convergent correct 2. divergent 3. absolutely convergent Explanation: Since ∞ X n = 1 ( 1) n +1 e 1 /n 5 n = 1 5 ∞ X n = 1 ( 1) n e 1 /n n , Portillo, Tom – Homework 12 – Due: Nov 21 2006, 3:00 am – Inst: David Benzvi 2 we have to decide if the series ∞ X n = 1 ( 1) n e 1 /n n is absolutely convergent, conditionally con vergent or divergent. First we check for absolute convergence. Now, since e 1 /n ≥ 1 for all n ≥ 1, e 1 /n 5 n ≥ 1 5 n > . But by the pseries test with p = 1, the series ∞ X n = 1 1 5 n diverges, and so by the Comparison Test, the series ∞ X n = 1 e 1 /n 5 n too diverges; in other words, the given series is not absolutely convergent. To check for conditional convergence, con sider the series ∞ X n = 1 ( 1) n f ( n ) where f ( x ) = e 1 /x x . Then f ( x ) > 0 on (0 , ∞ ). On the other hand, f ( x ) = 1 x 2 e 1 /x x e 1 /x x 2 = e 1 /x ‡ 1 x + x · x 2 = e 1 /x n 1 + x 2 x 3 o . Thus f ( x ) < 0 on (0 , ∞ ), so f ( n ) > f ( n +1) for all n . Finally, since lim x →∞ e 1 /x = 1 , we see that f ( n ) → 0 as n → ∞ . By the Alternating Series Test, therefore, the series ∞ X n = 1 ( 1) n f ( n ) is convergent....
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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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