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Unformatted text preview: Su, Yung – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Shinko Harper 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find 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 + 8 7 n + 5 ¶ , and if it does, find its limit. 1. limit = 6 7 2. sequence diverges correct 3. limit = 8 5 4. limit = ± 6 7 5. limit = 0 Explanation: After division, 6 n + 8 7 n + 5 = 6 + 8 n 7 + 5 n . Now 8 n , 5 n → 0 as n → ∞ , so lim n →∞ 6 n + 8 7 n + 5 = 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 3 n ( n 8) 3 n , and if it does, find its limit 1. limit = e 8 3 2. limit = e 24 3. limit = 1 4. sequence diverges 5. limit = e 24 correct 6. limit = e 8 3 Explanation: By the Laws of Exponents, a n = µ n 8 n ¶ 3 n = µ 1 8 n ¶ 3 n = h‡ 1 8 n · n i 3 . But ‡ 1 + x n · n→ e x as n → ∞ . Consequently, { a n } converges and has limit = ( e 8 ) 3 = e 24 . keywords: sequence, e, exponentials, limit 003 (part 1 of 1) 10 points Determine whether the series 4 + 3 + 9 4 + 27 16 + ··· is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 9 Su, Yung – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Shinko Harper 2 2. divergent 3. convergent with sum = 1 16 4. convergent with sum = 16 correct 5. convergent with sum = 1 9 Explanation: The series 4 + 3 + 9 4 + 27 16 + ··· = ∞ X n =1 a r n 1 is an infinite geometric series in which a = 4 and r = 3 4 . 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 convergent with sum = 16 . keywords: 004 (part 1 of 1) 10 points Determine whether the series ∞ X n = 0 4 (cos nπ ) µ 1 2 ¶ n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum 3 8 2. convergent with sum 8 3. convergent with sum 8 3 4. divergent 5. convergent with sum 8 6. convergent with sum 8 3 correct Explanation: Since cos nπ = ( 1) n , the given series can be rewritten as an infinite geometric series ∞ X n =0 4 µ 1 2 ¶ n = ∞ X n = 0 a r n in which a = 4 , r = 1 2 . 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 8 3 . keywords: geometric series, convergent 005 (part 1 of 1) 10 points Determine whether the infinite series ∞ X n =1 4( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum....
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 Spring '09
 Harper
 Calculus, Mathematical Series, Shinko Harper

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