cal14 - choi (kc24547) HW14 BERG (56525) 1 This print-out...

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Unformatted text preview: choi (kc24547) HW14 BERG (56525) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the series summationdisplay n =2 ( 1) n n 8 ln n is conditionally convergent, absolutely con- vergent, or divergent. 1. series is absolutely convergent 2. series is conditionally convergent 3. series is divergent correct Explanation: By the Divergence Test, a series summationdisplay n = N ( 1) n a n will be divergent for each fixed choice of N if lim n a n negationslash = 0 since it is only the behaviour of a n as n thats important. Now, for the given series, N = 2 and a n = n 8 ln n . But by LHospitals Rule, lim x x ln x = lim x 1 1 /x = . Consequently, by the Divergence Test, the given series is divergent . 002 10.0 points Determine whether the series summationdisplay n =0 3 n + 1 cos n is conditionally convergent, absolutely con- vergent or divergent. 1. divergent 2. absolutely convergent 3. conditionally convergent correct Explanation: Since cos n = ( 1) n , the given series can be rewritten as the alternating series summationdisplay n =0 ( 1) n 3 n + 1 = summationdisplay n =0 ( 1) n f ( n ) with f ( x ) = 3 x + 1 . Now f ( n ) = 3 n + 1 > 3 n + 2 = f ( n + 1) for all n , while lim n f ( n ) = lim n 3 n + 1 = 0 . Consequently, by the Alternating Series Test, the given series converges. On the other hand, by the Limit Comparison Test and the p-series test with p = 1 / 2, we see that the series summationdisplay n =0 f ( n ) is divergent. Consequently, the given series is conditionally convergent . 003 10.0 points To apply the root test to an infinite series n a n the value of = lim n ( a n ) 1 /n choi (kc24547) HW14 BERG (56525) 2 has to be determined. Compute the value of for the series summationdisplay n =1 3 n + 2 n parenleftbigg 3 2 parenrightbigg n . 1. = 4 3 2. = 3 2 correct 3. = 2 3 4. = 3 5. = 9 2 Explanation: After division, 3 n + 2 n = 3 parenleftbigg 1 + 2 3 n parenrightbigg , so ( a n ) 1 /n = parenleftbigg 3 braceleftbigg 1 + 2 3 n bracerightbiggparenrightbigg 1 /n 3 2 . But lim n 3 1 /n parenleftbigg 1 + 2 3 n parenrightbigg 1 /n = 1 as n . Consequently, = 3 2 . 004 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value = lim n a n +1 a n has to be determined. Compute for the series summationdisplay n =1 2 n 6 n 2 + 7 . 1. = 0 2. = 2 7 3. = 2 13 4. = 1 3 5. = 2 correct Explanation: By algebra, a n +1 a n = 2 n +1 2 n bracketleftbigg 6 n 2 + 7 6( n + 1) 2 + 7 bracketrightbigg ....
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This note was uploaded on 04/04/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas at Austin.

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cal14 - choi (kc24547) HW14 BERG (56525) 1 This print-out...

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