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Spring 2006 - Zabrocki's Class - Exam 2 (Version B)

Spring 2006 - Zabrocki's Class - Exam 2 (Version B) - score...

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Math 20D - Spring 2006 - Midterm 2- Prof. Mike Zabrocki score 1 24 2 25 3 25 4 26 BONUS 10 total 100 Name: Time of section: 1. Determine if the following sums diverge, converge absolutely or converge conditionally. State which test you use to justify your answer and the conclusion which follows from that test. (a) n =1 (1 - cos(1 /n )) (b) n =2 ( - 1) n n - 2 n 1 - cos(1 /n ) = 1 - 1 + (1 /n ) 2 2! - (1 /n ) 4 4! + · · · This is an alternating series and 1 2 n - n Therefore, lim n →∞ 1 - cos(1 /n ) 1 /n 2 = 1 / 2 is positive and decreasing and lim n →∞ 1 2 n - n = 0 since 1 n 2 converges then by L.C.T. Hence by A.S.T. converges conditionally. then n =1 (1 - cos(1 /n )) converges Since lim n →∞ 1 2 n - n 1 / 2 n = 1 then by L.C.T. absolutely. n =2 ( - 1) n n - 2 n converges absolutely since 1 / 2 n converges. (c) n =1 cos( πn ) 1 - cos( n ) (d) n =1 1 n (1 - e 1 /n ) The series is alternating but 1 n (1 - e 1 /n ) = 1 n (1 - 1 - 1 n - ( 1 n ) 2 2! - · · · ) lim n →∞ 1 1 - cos ( n ) = 0 lim n →∞ 1 n (1 - e 1 /n ) 1 n 3 / 2 = - 1 so the series diverges by divergence test therefore n =1 1 n (1 - e 1 /n ) converges absolutely since 1 n 3 / 2 converges (e)
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