# EXAM308 - Version 063 – EXAM 3 – Zheng –(58355 1 This...

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Unformatted text preview: Version 063 – EXAM 3 – Zheng – (58355) 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 + 2 parenrightbigg , and if it does, find its limit. 1. limit = 0 correct 2. limit = ln 4 3 3. the sequence diverges 4. limit = ln 4 5 5. limit = − ln 3 Explanation: After division by n we see that 4 3 n + 2 = 4 n 3 + 2 n , so by properties of logs, a n = 1 n ln 4 n − 1 n ln parenleftbigg 3 + 2 n parenrightbigg . But by known limits (or use L’Hospital), 1 n ln 4 n , 1 n ln parenleftbigg 3 + 2 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 4 n ( n − 7) 4 n , and if it does, find its limit 1. limit = e 28 correct 2. limit = e − 28 3. limit = e 7 4 4. limit = 1 5. sequence diverges 6. limit = e − 7 4 Explanation: By the Laws of Exponents, a n = parenleftbigg n − 7 n parenrightbigg − 4 n = parenleftbigg 1 − 7 n parenrightbigg − 4 n = bracketleftBigparenleftBig 1 − 7 n parenrightBig n bracketrightBig − 4 . But parenleftBig 1 + x n parenrightBig n −→ e x as n → ∞ . Consequently, { a n } converges and has limit = ( e − 7 ) − 4 = e 28 . 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 4 n , find a n for n > 1. 1. a n = 6 parenleftbigg 5 n − 4 4 n parenrightbigg 2. a n = 3 n − 4 4 n correct Version 063 – EXAM 3 – Zheng – (58355) 2 3. a n = 5 n − 4 4 n 4. a n = n − 4 4 n − 1 5. a n = 6 parenleftbigg n − 4 4 n − 1 parenrightbigg 6. a n = 6 parenleftbigg 3 n − 4 4 n 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 4 n − 1 − n 4 n = 4 ( n − 1) 4 n − n 4 n when n > 1. Consequently, a n = 3 n − 4 4 n for n > 1. 004 10.0 points Let g be a continuous, positive, decreasing function on [1 , ∞ ). Compare the values of the integral A = integraldisplay 15 1 g ( x ) dx and the series B = 15 summationdisplay n = 2 g ( n ) . 1. A > B correct 2. A < B 3. A = B Explanation: In the figure 1 2 3 4 5 . . . a 2 a 3 a 4 a 5 the bold line is the graph of g on [1 , ∞ ) and the areas of the rectangles the terms in the series ∞ summationdisplay n = 2 a n , a n = g ( n ) . Clearly from this figure we see that a 2 = g (2) < integraldisplay 2 1 g ( x ) dx, a 3 = g (3) < integraldisplay 3 2 g ( x ) dx , while a 4 = g (4) < integraldisplay 4 3 g ( x ) dx, a 5 = g (5) < integraldisplay 5 4 g ( x ) dx , and so on. Consequently, A > B . Version 063 – EXAM 3 – Zheng – (58355) 3 keywords: Szyszko 005 10.0 points Which, if any, of the following statements are...
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## This note was uploaded on 09/14/2009 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.

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EXAM308 - Version 063 – EXAM 3 – Zheng –(58355 1 This...

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