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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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 Spring '06
 Benzvi

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