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Unformatted text preview: rhodes (ajr2283) HW05 Gilbert (56380) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the series 3 2 3 3 + 3 4 3 5 + 3 6 . . . is conditionally convergent, absolutely con vergent or divergent. 1. series is conditionally convergent cor rect 2. series is absolutely convergent 3. series is divergent Explanation: In summation notation, 3 2 3 3 + 3 4 3 5 + 3 6 . . . = summationdisplay n =1 ( 1) n 1 f ( n ) , with f ( x ) = 3 x + 1 . Now the improper integral integraldisplay 1 f ( x ) dx = integraldisplay 1 3 x + 1 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 3 n + 1 > 3 n + 1 + 1 = f ( n + 1) , while lim n 3 n + 1 = 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 ( 1) 3 6 3 + n 2. summationdisplay n =1 ( 1) n 1 5 + n 6 + n 3. summationdisplay n =1 ( 1) n 1 5 + n correct 4. summationdisplay n =1 ( 1) 2 n 6 3 + n 5. summationdisplay n =1 3 6 + n Explanation: Since summationdisplay n =1 ( 1) 3 6 3 + n = summationdisplay n =1 6 3 + 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 3 + n = summationdisplay n =1 6 3 + n , the same argument shows that this series as well as summationdisplay n =1 3 6 + n is divergent. On the other hand, by the Divergence Test, the series summationdisplay n =1 ( 1) n 1 5 + n 6 + n rhodes (ajr2283) HW05 Gilbert (56380) 2 is divergent because lim n ( 1) n 1 5 + n 6 + n negationslash = 0 . This leaves only the series summationdisplay n =1 ( 1) n 1 5 + n . To see that this series is convergent, set b n = 1 5 + 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 5 + n is convergent. 003 10.0 points Determine whether the series summationdisplay n =1 ( 1) n 1 e 1 /n 6 n is absolutely convergent, conditionally con vergent or divergent. 1. absolutely convergent 2. conditionally convergent correct 3. divergent Explanation: Since summationdisplay n =1 ( 1) n 1 e 1 /n 6 n = 1 6 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 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.
 Spring '07
 TextbookAnswers
 Multivariable Calculus

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