math1005test3sol

math1005test3sol - 40 marks-50 minutes MATH1005A,...

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40 marks-50 minutes MATH1005A, Summer2009 Test3 Problem 1 . Evaluate the limits. 1 . ( 3marks ) lim n →∞ n 3 + 2 n 2 n 3 + 4 n 2 + 2 2 . ( 4marks ) lim n →∞ ln n ln 2 n Solution: 1.1 lim n →∞ n 3 + 2 n 2 n 3 + 4 n 2 + 2 = lim n →∞ 1 + (2 /n 2 ) 2 + (4 /n ) + (2 /n 3 ) = lim n →∞ 1 + 0 2 + 0 + 0 = 1 / 2 1.2 Lets look at lim x →∞ ln x ln 2 x . Since it is / we can use l’Hospital’s rule. Then lim x →∞ ln x ln 2 x = lim x →∞ 1 /x 2(1 / 2 x ) = 1 Therefore lim n →∞ ln n ln 2 n = 1 . Problem 2 ( 8 marks) . Find the sum of the telescoping series X n =1 2 n ( n + 2) . Solution: First do the partial function decomposition to get: 2 n ( n + 2) = 1 n - 1 n + 2 . 1
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40 marks-50 minutes MATH1005A, Summer2009 Test3 For n 3 we have S n = ( 1 1 - 1 3 ) + ( 1 2 - 1 4 ) + ( 1 3 - 1 5 ) + . . . + ( 1 n - 2 - 1 n ) + ( 1 n - 1 - 1 n + 1 ) + ( 1 n - 1 n + 2 ) = 1 + 1 2 - 1 n + 1 - 1 n + 2 Therefore X n =1 2 n ( n + 2) = lim n →∞ ( 3 2 - 1 n + 1 - 1 n + 2 ) = 3
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This note was uploaded on 01/18/2012 for the course MATH 1005 taught by Professor Any during the Summer '07 term at Carleton CA.

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math1005test3sol - 40 marks-50 minutes MATH1005A,...

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