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Unformatted text preview: Valenica, Daniel – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Cheng 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 9 11 10 12 + 11 13 12 14 + 13 15 . . . is absolutely convergent, conditionally con vergent or divergent. 1. absolutely convergent 2. conditionally convergent 3. divergent correct Explanation: In summation notation, 9 11 10 12 + 11 13 12 14 + 13 15 . . . = ∞ X n = 9 a n , with a n given by a n = ( 1) n n n + 2 . However, lim n →∞ n n + 2 = 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 2 + √ n 5 + √ n 2. ∞ X n = 1 ( 1) 2 n 5 3 + √ n 3. ∞ X n = 1 ( 1) n 1 2 + √ n correct 4. ∞ X n = 1 ( 1) 3 5 3 + √ n 5. ∞ X n = 1 3 5 + √ 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 2 + √ n 5 + √ n is divergent because lim n →∞ ( 1) n 1 2 + √ n 5 + √ n 6 = 0 . This leaves only the series ∞ X n = 1 ( 1) n 1 2 + √ n . Valenica, Daniel – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Cheng 2 To see that this series is convergent, set b n = 1 2 + √ 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 2 + √ n is convergent. keywords: 003 (part 1 of 1) 10 points Determine whether the series ∞ X n = 1 ( 1) n +1 e 1 /n 7 n is absolutely convergent, conditionally con vergent or divergent. 1. absolutely convergent 2. conditionally convergent correct 3. divergent Explanation: Since ∞ X n = 1 ( 1) n +1 e 1 /n 7 n = 1 7 ∞ 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 7 n ≥ 1 7 n > . But by the pseries test with p = 1, the series ∞ X n = 1 1 7 n diverges, and so by the Comparison Test, the series ∞ X n = 1 e 1 /n 7 n too diverges; in other words, the given series is not absolutely convergent....
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This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas at Austin.
 Fall '07
 CHENG

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