This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 063 – EXAM 3 – Zheng – (58355) 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 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...
View
Full
Document
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.
 Spring '08
 Cepparo

Click to edit the document details