This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: moore (jwm2685) HW04 gilbert (55485) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points To apply the root test to an infinite series n a n the value of = lim n ( a n ) 1 /n has to be determined. Compute the value of for the series summationdisplay n =1 3 n + 5 n parenleftbigg 7 4 parenrightbigg n . 1. = 20 7 2. = 35 4 3. = 4 7 4. = 21 4 5. = 7 4 correct Explanation: After division, 3 n + 5 n = 3 parenleftbigg 1 + 5 3 n parenrightbigg , so ( a n ) 1 /n = parenleftbigg 3 braceleftbigg 1 + 5 3 n bracerightbiggparenrightbigg 1 /n 7 4 . But lim n 3 1 /n parenleftbigg 1 + 5 3 n parenrightbigg 1 /n = 1 as n . Consequently, = 7 4 . 002 10.0 points To apply the ratio test to the infinite series summationdisplay n a n , the value = lim n a n +1 a n has to be determined, Compute for the series summationdisplay n =1 1 7 n + 5 sin parenleftbigg 1 n parenrightbigg . 1. = 1 correct 2. = 0 3. = 1 5 4. = 1 7 5. = 1 12 Explanation: By algebra, sin 1 n +1 sin 1 n = sin 1 n +1 1 n +1 1 n sin 1 n n n + 1 . But lim x sin x x = 1 , so lim n sin 1 n +1 1 n +1 = 1 , while lim n 1 n sin 1 n = 1 also. Thus = lim n n n + 1 bracketleftbigg 7 n + 5 7( n + 1) + 5 bracketrightbigg = lim n n n + 1 parenleftbigg 7 n + 5 7 n + 7 + 5 parenrightbigg . moore (jwm2685) HW04 gilbert (55485) 2 Consequently, for the given series, = 1 . 003 10.0 points Decide whether the series summationdisplay n =1 3 n parenleftBig n 2 n parenrightBig n 2 converges or diverges. 1. converges correct 2. diverges Explanation: The given series has the form summationdisplay n =1 a n , a n = 3 n parenleftBig n 2 n parenrightBig n 2 . But then  a n  1 /n = 3 parenleftBig n 1 n parenrightBig n , in which case lim n 1  a n  1 /n = 3 e 2 < 1 , since lim n parenleftBig n 2 n parenrightBig n = lim n parenleftBig 1 2 n parenrightBig n = 1 e 2 < 1 2 2 . Consequently, the Root Test ensures that the given series converges . 004 10.0 points The terms of a series are specified recur sively by the equations a 1 = 5 , a n +1 = parenleftBig 5 n + 1 4 n + 10 parenrightBig a n . Determine whether the series summationdisplay n =1 a n converges or diverges. 1. converges 2. neither converges nor diverges 3. diverges correct Explanation: From the recursive formula we see that lim n a n +1 a n = lim n 5 n + 1 4 n + 10 = 5 4 . Consequently, by the Ratio Test the series diverges. 005 10.0 points Decide whether the series summationdisplay n =1 (2 n )!...
View
Full
Document
 Spring '07
 TextbookAnswers

Click to edit the document details