CalcExam3Solutions - Version 010 Exam 3 Odell (58340) 1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 LHospital), 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 ....
View Full Document

Page1 / 11

CalcExam3Solutions - Version 010 Exam 3 Odell (58340) 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online