CalcExam3Solutions

CalcExam3Solutions - Version 010 – Exam 3 – Odell...

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Unformatted text preview: Version 010 – 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 2 n + 5 parenrightbigg , and if it does, find its limit. 1. limit = 0 correct 2. limit = − ln 2 3. limit = ln 2 4. the sequence diverges 5. limit = ln 4 7 Explanation: After division by n we see that 4 2 n + 5 = 4 n 2 + 5 n , so by properties of logs, a n = 1 n ln 4 n − 1 n ln parenleftbigg 2 + 5 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 4 n , 1 n ln parenleftbigg 2 + 5 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 5 n ( n − 2) 5 n , and if it does, find its limit 1. limit = e 2 5 2. sequence diverges 3. limit = e − 2 5 4. limit = e − 10 5. limit = e 10 correct 6. limit = 1 Explanation: By the Laws of Exponents, a n = parenleftbigg n − 2 n parenrightbigg − 5 n = parenleftbigg 1 − 2 n parenrightbigg − 5 n = bracketleftBigparenleftBig 1 − 2 n parenrightBig n bracketrightBig − 5 . But parenleftBig 1 + x n parenrightBig n −→ e x as n → ∞ . Consequently, { a n } converges and has limit = ( e − 2 ) − 5 = 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 = 6 − n 3 n , find a n for n > 1. 1. a n = 4 n − 3 3 n 2. a n = 2 n − 3 3 n correct Version 010 – Exam 3 – Odell – (58340) 2 3. a n = 6 parenleftbigg 2 n − 3 3 n parenrightbigg 4. a n = n − 3 3 n − 1 5. a n = 6 parenleftbigg 4 n − 3 3 n parenrightbigg 6. a n = 6 parenleftbigg n − 3 3 n − 1 parenrightbigg 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 3 n − 1 − n 3 n = 3 ( n − 1) 3 n − n 3 n when n > 1. Consequently, a n = 2 n − 3 3 n for n > 1. 004 10.0 points Let f be a continuous, positive, decreasing function on [1 , ∞ ). Compare the values of the integral A = integraldisplay 14 1 f ( x ) dx and the series B = 14 summationdisplay n = 2 f ( n ) . 1. A = B 2. A < B 3. A > B correct Explanation: In the figure 1 2 3 4 5 . . . a 2 a 3 a 4 a 5 the bold line is the graph of f on [1 , ∞ ) and the areas of the rectangles the terms in the series ∞ summationdisplay n = 2 a n , a n = f ( n ) . Clearly from this figure we see that a 2 = f (2) < integraldisplay 2 1 f ( x ) dx, a 3 = f (3) < integraldisplay 3 2 f ( x ) dx , while a 4 = f (4) < integraldisplay 4 3 f ( x ) dx, a 5 = f (5) < integraldisplay 5 4 f ( x ) dx , Version 010 – Exam 3 – Odell – (58340) 3 and so on. Consequently, A > B ....
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This note was uploaded on 04/15/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.

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CalcExam3Solutions - Version 010 – Exam 3 – Odell...

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