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Unformatted text preview: hyun (hh7953) HW13 gogolev (57440) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the series 2 3 2 4 + 2 5 2 6 + 2 7 . . . is conditionally convergent, absolutely con vergent or divergent. 1. series is conditionally convergent cor rect 2. series is divergent 3. series is absolutely convergent Explanation: In summation notation, 2 3 2 4 + 2 5 2 6 + 2 7 . . . = summationdisplay n =1 ( 1) n 1 f ( n ) , with f ( x ) = 2 x + 2 . Now the improper integral integraldisplay 1 f ( x ) dx = integraldisplay 1 2 x + 2 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 2 n + 2 > 2 n + 1 + 2 = f ( n + 1) , while lim n 2 n + 2 = 0 . Consequently, by the Alternating Series Test, the given series is conditionally convergent . keywords: alternating series, Alternating se ries test, conditionally convergent, absolutely convergent, divergent 002 10.0 points Which one of the following series is conver gent? 1. summationdisplay n =1 5 6 + n 2. summationdisplay n =1 ( 1) n 1 4 + n correct 3. summationdisplay n =1 ( 1) 2 n 6 5 + n 4. summationdisplay n =1 ( 1) 3 6 5 + n 5. summationdisplay n =1 ( 1) n 1 4 + n 6 + n Explanation: Since summationdisplay n =1 ( 1) 3 6 5 + n = summationdisplay n =1 6 5 + n , use of the Limit Comparison and pseries Tests with p = 1 2 shows that this series is divergent. Similarly, since summationdisplay n =1 ( 1) 2 n 6 5 + n = summationdisplay n =1 6 5 + n , the same argument shows that this series as well as summationdisplay n =1 5 6 + n is divergent. On the other hand, by the Divergence Test, the series summationdisplay n =1 ( 1) n 1 4 + n 6 + n hyun (hh7953) HW13 gogolev (57440) 2 is divergent because lim n ( 1) n 1 4 + n 6 + n negationslash = 0 . This leaves only the series summationdisplay n =1 ( 1) n 1 4 + n . 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 summationdisplay n =1 ( 1) n 1 4 + n is convergent. 003 10.0 points Determine whether the series summationdisplay n =1 ( 1) n 1 e 1 /n 4 n is absolutely convergent, conditionally con vergent or divergent. 1. conditionally convergent correct 2. absolutely convergent 3. divergent Explanation: Since summationdisplay n =1 ( 1) n 1 e 1 /n 4 n = 1 4 summationdisplay n =1 ( 1) n e 1 /n n , we have to decide if the series summationdisplay n =1 ( 1) n e 1 /n n is absolutely convergent, conditionally con vergent, or divergent....
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This note was uploaded on 01/19/2010 for the course M 57440 taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus

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