solution Exam 3 _pdf

# solution Exam 3 _pdf - Version 040 Exam 3 Odell (58340) 1...

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Unformatted text preview: Version 040 Exam 3 Odell (58340) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if the sequence { a n } converges when a n = 1 n ln parenleftbigg 4 3 n + 6 parenrightbigg , and if it does, find its limit. 1. the sequence diverges 2. limit = ln 4 9 3. limit = ln 3 4. limit = ln 4 3 5. limit = 0 correct Explanation: After division by n we see that 4 3 n + 6 = 4 n 3 + 6 n , so by properties of logs, a n = 1 n ln 4 n 1 n ln parenleftbigg 3 + 6 n parenrightbigg . But by known limits (or use LHospital), 1 n ln 4 n , 1 n ln parenleftbigg 3 + 6 n parenrightbigg as n . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine if the sequence { a n } converges when a n = n 2 n ( n 5) 2 n , and if it does, find its limit 1. sequence diverges 2. limit = e 5 2 3. limit = e 10 correct 4. limit = 1 5. limit = e 5 2 6. limit = e 10 Explanation: By the Laws of Exponents, a n = parenleftbigg n 5 n parenrightbigg 2 n = parenleftbigg 1 5 n parenrightbigg 2 n = bracketleftBigparenleftBig 1 5 n parenrightBig n bracketrightBig 2 . But parenleftBig 1 + x n parenrightBig n e x as n . Consequently, { a n } converges and has limit = ( e 5 ) 2 = e 10 . 003 10.0 points If the n th partial sum S n of an infinite series summationdisplay n =1 a n is given by S n = 5 n 6 n , find a n for n > 1. 1. a n = 5 parenleftbigg n 6 6 n 1 parenrightbigg 2. a n = 5 parenleftbigg 7 n 6 6 n parenrightbigg Version 040 Exam 3 Odell (58340) 2 3. a n = 5 parenleftbigg 5 n 6 6 n parenrightbigg 4. a n = 5 n 6 6 n correct 5. a n = 7 n 6 6 n 6. a n = n 6 6 n 1 Explanation: By definition, the n th partial sum of summationdisplay n = 1 a n is given by S n = a 1 + a 2 + + a n . In particular, a n = braceleftbigg S n S n 1 , n > 1, S n , n = 1. Thus a n = S n S n 1 = n 1 6 n 1 n 6 n = 6 ( n 1) 6 n n 6 n when n > 1. Consequently, a n = 5 n 6 6 n for n > 1. 004 10.0 points Let g be a continuous, positive, decreasing function on [3 , ). Compare the values of the series A = 19 summationdisplay n = 4 g ( n ) and the integral B = integraldisplay 19 3 g ( x ) dx . 1. A < B correct 2. A = B 3. A > B Explanation: In the figure 3 4 5 6 7 . . . a 4 a 5 a 6 a 7 the bold line is the graph of g on [3 , ) and the areas of the rectangles the terms in the series summationdisplay n = 4 a n , a n = g ( n ) . Clearly from this figure we see that a 4 = g (4) < integraldisplay 4 3 g ( x ) dx, a 5 = g (5) < integraldisplay 5 4 g ( x ) dx , while a 6 = g (6) < integraldisplay 6 5 g ( x ) dx, a 7 = g (7) < integraldisplay 7 6 g ( x ) dx , and so on. Consequently, A < B . Version 040 Exam 3 Odell (58340) 3 keywords: Szyszko 005 10.0 points Which, if any, of the following statements are...
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## This note was uploaded on 09/23/2009 for the course M 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

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solution Exam 3 _pdf - Version 040 Exam 3 Odell (58340) 1...

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