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Unformatted text preview: Version 041 L EXAM 3 Henry (54974) 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 sequence { a n } con verges or diverges when a n = ln(4 n 5 ) ln(3 n 2 ) , and if it converges, find the limit. 1. converges with limit = ln 4 ln 3 2. diverges 3. converges with limit = 4 3 4. converges with limit = 5 2 correct 5. converges with limit = 0 Explanation: By properties of logs, ln(4 n 5 ) = ln 4 + 5 ln n , ln(3 n 2 ) = ln 3 + 2 ln n . Thus a n = ln4 + 5 ln n ln3 + 2 ln n = 5 + ln 4 ln n 2 + ln 3 ln n . On the other hand, lim n ln 4 ln n = lim n ln 3 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 5 2 . 002 10.0 points Determine if the sequence { a n } converges when a n = 5 + parenleftbigg 8 7 parenrightbigg n , and if it does, find its limit. 1. limit = 5 1 2. limit = 5 correct 3. the sequence diverges 4. limit = 5 5. limit = 5 + 1 Explanation: It is known that x n as n whenever 1 < x < 1. Now 8 < 7 , so parenleftbigg 8 7 parenrightbigg n as n . Consequently, the given sequence converges and has limit = 5 . 003 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = 7 n + ( 1) n 4 n + 2 . 1. sequence does not converge 2. converges with limit = 7 6 Version 041 L EXAM 3 Henry (54974) 2 3. converges with limit = 3 2 4. converges with limit = 7 4 correct 5. converges with limit = 2 Explanation: After division by n we see that a n = 7 + ( 1) n n 4 + 2 n . But ( 1) n n , 2 n as n , so a n 7 4 as n . Conse quently, the sequence converges and has limit = 7 4 . 004 10.0 points If the n th partial sum of an infinite series is S n = 3 n 2 2 n 2 + 3 , what is the sum of the series? 1. sum = 2 3 2. series diverges 3. sum = 1 4. sum = 3 correct 5. sum = 2 Explanation: By definition sum = lim n S n = lim n parenleftBig 3 n 2 2 n 2 + 3 parenrightBig . Thus sum = 3 . 005 10.0 points Find the values of x for which the series summationdisplay n = 1 x n 4 n converges, and then find the sum of the series for those values of x . 1. converges: 4 < x < 4 , sum = x 4 x correct 2. converges: 4 < x < 4 , sum = 4 4 x 3. converges: 4 x 4 , sum = 4 4 x 4. converges: 4 x < 4 , sum = 4 4 x 5. converges: 4 x 4 , sum = x 4 x 6. converges: 4 < x 4 , sum = x 4 x Explanation: The given series is a geometric series summationdisplay n = 1 a r n 1 in which a = x 4 , r = x 4 . But such a geometric series (i) converges when  r  < 1, with sum a 1 r , and (ii) diverges when  r  1....
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This note was uploaded on 02/06/2012 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus

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