# exam307 - Bobbitt, James Exam 3 Due: Dec 4 2007, 11:00 pm...

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Bobbitt, James – Exam 3 – Due: Dec 4 2007, 11:00 pm – Inst: Diane Radin 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Determine whether the sequence { a n } con- verges or diverges when a n = ( - 1) n µ 6 n + 5 7 n + 3 , and iF it does, fnd its limit. 1. limit = ± 6 7 2. limit = 0 3. sequence diverges correct 4. limit = 5 3 5. limit = 6 7 Explanation: AFter division, 6 n + 5 7 n + 3 = 6 + 5 n 7 + 3 n . Now 5 n , 3 n 0 as n → ∞ , so lim n →∞ 6 n + 5 7 n + 3 = 6 7 6 = 0 . Thus as n → ∞ , the values oF a n oscillate be- tween values ever closer to ± 6 7 . Consequently, the sequence diverges . keywords: 002 (part 1 oF 1) 10 points Determine iF the sequence { a n } converges when a n = n 5 n ( n - 7) 5 n , and iF it does, fnd its limit 1. sequence diverges 2. limit = e 35 correct 3. limit = e 7 5 4. limit = e - 7 5 5. limit = 1 6. limit = e - 35 Explanation: By the Laws oF Exponents, a n = µ n - 7 n - 5 n = µ 1 - 7 n - 5 n = h‡ 1 - 7 n · n i - 5 . But 1 + x n · n -→ e x as n → ∞ . Consequently, { a n } converges and has limit = ( e - 7 ) - 5 = e 35 . keywords: sequence, e, exponentials, limit 003 (part 1 oF 1) 10 points Determine whether the series 2 + 3 + 9 2 + 27 4 + ··· is convergent or divergent, and iF convergent, fnd its sum. 1. convergent with sum = 1 4

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Bobbitt, James – Exam 3 – Due: Dec 4 2007, 11:00 pm – Inst: Diane Radin 2 2. divergent correct 3. convergent with sum = 9 4. convergent with sum = 1 9 5. convergent with sum = 4 Explanation: The series 2 + 3 + 9 2 + 27 4 + ··· = X n =1 a r n - 1 is an infnite geometric series in which a = 2 and r = 3 2 . But such a series is (i) convergent with sum a 1 - r when | r | < 1, (ii) divergent when | r | ≥ 1 . Thus the given series is divergent . keywords: 004 (part 1 oF 1) 10 points Determine whether the series X n = 0 2 (cos ) µ 1 3 n is convergent or divergent, and iF convergent, fnd its sum. 1. convergent with sum - 2 3 2. divergent 3. convergent with sum - 3 2 4. convergent with sum 3 2 correct 5. convergent with sum 3 6. convergent with sum - 3 Explanation: Since cos = ( - 1) n , the given series can be rewritten as an infnite geometric series X n =0 2 µ - 1 3 n = X n = 0 a r n in which a = 2 , r = - 1 3 . But the series n =0 ar n is (i) convergent with sum a 1 - r when | r | < 1, and (ii) divergent when | r | ≥ 1. Consequently, the given series is convergent with sum 3 2 . keywords: geometric series, convergent 005 (part 1 oF 1) 10 points Determine whether the infnite series X n =1 2( n + 1) 2 n ( n + 2) converges or diverges, and iF converges, fnd its sum.
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## This note was uploaded on 09/14/2009 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.

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exam307 - Bobbitt, James Exam 3 Due: Dec 4 2007, 11:00 pm...

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