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Unformatted text preview: Kim, Jin Homework 12 Due: Nov 20 2007, 3:00 am Inst: Diane Radin 1 This printout should have 17 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 Determine whether the series 2 5 3 6 + 4 7 5 8 + 6 9 . . . is absolutely convergent, conditionally con vergent or divergent. 1. absolutely convergent 2. divergent correct 3. conditionally convergent Explanation: In summation notation, 2 5 3 6 + 4 7 5 8 + 6 9 . . . = X n = 2 a n , with a n given by a n = ( 1) n n n + 3 . However, lim n n n + 3 = 1 , so that as n , a n oscillates between val ues approaching 1 and 1. In particular, therefore, lim n a n 6 = 0 . Consequently, by the Divergence Test, the series is divergent . keywords: 002 (part 1 of 1) 10 points Which one of the following series is conver gent? 1. X n = 1 ( 1) n 1 4 + n correct 2. X n = 1 ( 1) 2 n 5 6 + n 3. X n = 1 ( 1) 3 5 6 + n 4. X n = 1 ( 1) n 1 4 + n 5 + n 5. X n = 1 6 5 + n Explanation: Since X n = 1 ( 1) 3 5 6 + n = X n = 1 5 6 + 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 6 + n = X n = 1 5 6 + n , the same argument shows that this series as well as X n = 1 6 5 + n is divergent. On the other hand, by the Divergence Test, the series X n = 1 ( 1) n 1 4 + n 5 + n is divergent because lim n ( 1) n 1 4 + n 5 + n 6 = 0 . This leaves only the series X n = 1 ( 1) n 1 4 + n . Kim, Jin Homework 12 Due: Nov 20 2007, 3:00 am Inst: Diane Radin 2 To see that this series is convergent, set b n = 1 4 + 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 4 + n is convergent. keywords: 003 (part 1 of 1) 10 points Determine whether the series X n = 1 ( 1) n +1 e 1 /n 2 n is absolutely convergent, conditionally con vergent or divergent. 1. conditionally convergent correct 2. absolutely convergent 3. divergent Explanation: Since X n = 1 ( 1) n +1 e 1 /n 2 n = 1 2 X n = 1 ( 1) n e 1 /n n , 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 2 n 1 2 n > . But by the pseries test with p = 1, the series X n = 1 1 2 n diverges, and so by the Comparison Test, the series X n = 1 e 1 /n 2 n too diverges; in other words, the given series is not absolutely convergent....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

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