# Exam 3 - Version 133 – EXAM 3 – sachse – (56620) 1...

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Unformatted text preview: Version 133 – EXAM 3 – sachse – (56620) 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 whether the sequence { a n } con- verges or diverges when a n = 4 + cos 2 n 5 + 2 n . 1. diverges 2. converges with limit = 4 3. converges with limit = 2 3 4. converges with limit = 4 5 5. converges with limit = 0 correct Explanation: Since ≤ cos 2 n ≤ 1 , we see that 4 5 + 2 n ≤ a n ≤ 5 5 + 2 n . But lim n →∞ 4 5 + 2 n = 0 = lim n →∞ 5 5 + 2 n , so the Squeeze Theorem applies and ensures the sequence { a n } converges with limit = 0 . 002 10.0 points Determine if the sequence { a n } converges, and if it does, find its limit when a n = parenleftBig n − 1 n + 1 parenrightBig − 2 n . 1. limit = e 4 correct 2. limit = 1 3. does not converge 4. limit = e 2 5. limit = e − 2 6. limit = e − 4 Explanation: By the Laws of Exponents, a n = parenleftBig n + 1 n − 1 parenrightBig 2 n = parenleftBig 1 + 1 n parenrightBig 2 n parenleftBig 1 − 1 n parenrightBig 2 n . But lim n →∞ parenleftBig 1 + x n parenrightBig n = e x . Consequently, by Properties of Limits the given limit exists and limit = e 4 . 003 10.0 points Find the n th term, a n , of an infinite series ∑ ∞ n = 1 a n when the n th partial sum, S n , of the series is given by S n = 4 n n + 1 . 1. a n = 4 n ( n + 1) correct 2. a n = 2 n 2 3. a n = 5 2 n 2 4. a n = 5 n ( n + 1) Version 133 – EXAM 3 – sachse – (56620) 2 5. a n = 2 n 6. a n = 5 2 n Explanation: Since S n = a 1 + a 2 + ··· + a n , we see that a 1 = S 1 , a n = S n − S n − 1 ( n > 1) . But S n = 4 n n + 1 = 4 − 4 n + 1 . Thus a 1 = 2, while a n = 4 n − 4 n + 1 , ( n > 1) . Consequently, a n = 4 n − 4 n + 1 = 4 n ( n + 1) for all n . 004 10.0 points Determine whether the series ∞ summationdisplay n = 1 n ( n + 1)2 n converges or diverges. 1. series is divergent 2. series is convergent correct Explanation: We use the Limit Comparison Test with a n = n ( n + 1)2 n , b n = 1 2 n . For lim n →∞ a n b n = lim n →∞ n n + 1 = 1 > . Thus the series ∞ summationdisplay n = 1 n ( n + 1)2 n converges if and only if the series ∞ summationdisplay n = 1 1 2 n converges. But this last series is a geometric series with | r | = 1 2 < 1 , hence convergent. Consequently, the given series is series is convergent . 005 10.0 points Determine whether the series ∞ summationdisplay n =0 2 parenleftbigg 1 3 parenrightbigg n is convergent or divergent, and if convergent, find its sum. 1. convergent, sum = − 7 2 2. convergent, sum = 7 2 3. convergent, sum = 3 2 4. convergent, sum = 3 correct 5. divergent Explanation: The given series is an infinite geometric series ∞ summationdisplay n =0 a r n with a = 2 and r = 1 3 . But the sum of such a series is (i) convergent with sum a 1 − r when | r | < 1, Version 133 – EXAM 3 – sachse – (56620) 3 (ii) divergent when | r | ≥ 1....
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## This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.

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Exam 3 - Version 133 – EXAM 3 – sachse – (56620) 1...

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