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Unformatted text preview: Math 104, Midterm Examination 2 Solution Instructor: Guoliang Wu 1. (10 points) Find the following limit and justify your answer. lim n 1 n 2 [(2 n )!] 1 /n Solution: Let s n = (2 n )! n 2 n , then we need to find the limit lim n ( s n ) 1 /n . Since s n +1 s n = (2 n + 2)! ( n + 1) 2 n +2 n 2 n (2 n )! = (2 n + 2)(2 n + 1) ( n + 1) 2 n n + 1 2 n = 4 + 2 /n 1 + 1 /n 1 ( 1 + 1 n ) n 2 4 /e 2 , n , by limits laws. Then we apply the theorem in textbook to con clude that lim n  s n  1 /n = lim n s n +1 s n = 4 e 2 . 2. (10 points) Is the following series convergent? Justify your answer. X n =1 ( 1) n cos( n ) Solution: The series diverges. Denote by a n = ( 1) n cos n . Consider the subsequences ( a 2 m ) and ( a 2 m +1 ) . lim m a 2 m = lim m cos 2 m = 1 , lim m a 2 m +1 = lim m  cos 2 m + 1 = 1 ....
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 Winter '08
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 Math

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