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Homework 12 - Su Yung Homework 12 Due 3:00 am Inst Shinko...

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Su, Yung – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 17 questions. Multiple-choice 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 7 10 - 8 11 + 9 12 - 10 13 + 11 14 - . . . is absolutely convergent, conditionally con- vergent or divergent. 1. absolutely convergent 2. conditionally convergent 3. divergent correct Explanation: In summation notation, 7 10 - 8 11 + 9 12 - 10 13 + 11 14 - . . . = X n = 7 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 6 + n correct 2. X n = 1 ( - 1) 3 2 5 + n 3. X n = 1 ( - 1) 2 n 2 5 + n 4. X n = 1 ( - 1) n - 1 6 + n 2 + n 5. X n = 1 5 2 + n Explanation: Since X n = 1 ( - 1) 3 2 5 + n = - X n = 1 2 5 + n , use of the Limit Comparison and p -series Tests with p = 1 2 shows that this series is divergent. Similarly, since X n = 1 ( - 1) 2 n 2 5 + n = X n = 1 2 5 + n , the same argument shows that this series as well as X n = 1 5 2 + n is divergent. On the other hand, by the Divergence Test, the series X n = 1 ( - 1) n - 1 6 + n 2 + n is divergent because lim n → ∞ ( - 1) n - 1 6 + n 2 + n 6 = 0 . This leaves only the series X n = 1 ( - 1) n - 1 6 + n .
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Su, Yung – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Shinko Harper 2 To see that this series is convergent, set b n = 1 6 + 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 6 + n is convergent. keywords: 003 (part 1 of 1) 10 points Determine whether the series X n = 1 ( - 1) n - 1 e 1 /n 6 n is absolutely convergent, conditionally con- vergent or divergent. 1. divergent 2. absolutely convergent 3. conditionally convergent correct Explanation: Since X n = 1 ( - 1) n - 1 e 1 /n 6 n = - 1 6 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 6 n 1 6 n > 0 . But by the p -series test with p = 1, the series X n = 1 1 6 n diverges, and so by the Comparison Test, the series X n = 1 e 1 /n 6 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 0 ( 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 0 ( 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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