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Integral Calculus Exam 3

# Integral Calculus Exam 3 - Panjwani Sameer Exam 3 Due Dec 5...

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Panjwani, Sameer – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: James Rath 1 This print-out should have 18 questions. Multiple-choice 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 2 n + 4 n + 3 , and if it does, find its limit. 1. limit = 0 2. sequence diverges correct 3. limit = 2 4. limit = ± 2 5. limit = 4 3 Explanation: After division, 2 n + 4 n + 3 = 2 + 4 n 1 + 3 n . Now 4 n , 3 n 0 as n → ∞ , so lim n → ∞ 2 n + 4 n + 3 = 2 6 = 0 . Thus as n → ∞ , the values of a n oscillate be- tween values ever closer to ± 2. Consequently, the sequence diverges . keywords: 002 (part 1 of 1) 10 points Determine if the sequence { a n } converges when a n = n 2 n ( n - 6) 2 n , and if it does, find its limit 1. limit = e 12 correct 2. sequence diverges 3. limit = e 3 4. limit = e - 3 5. limit = e - 12 6. limit = 1 Explanation: By the Laws of Exponents, a n = n - 6 n - 2 n = 1 - 6 n - 2 n = h ‡ 1 - 6 n · n i - 2 . But 1 + x n · n -→ e x as n . Consequently, { a n } converges and has limit = ( e - 6 ) - 2 = e 12 . keywords: sequence, e, exponentials, limit 003 (part 1 of 1) 10 points Determine whether the series 3 + 2 + 4 3 + 8 9 + · · · is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 1 9

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Panjwani, Sameer – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: James Rath 2 2. divergent 3. convergent with sum = 4 4. convergent with sum = 1 4 5. convergent with sum = 9 correct Explanation: The series 3 + 2 + 4 3 + 8 9 + · · · = X n =1 a r n - 1 is an infinite geometric series in which a = 3 and r = 2 3 . 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 = 9 . keywords: 004 (part 1 of 1) 10 points Determine whether the series X n = 0 3 (cos ) 3 4 n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum - 12 2. convergent with sum 12 3. convergent with sum - 7 12 4. convergent with sum - 12 7 5. convergent with sum 12 7 correct 6. divergent Explanation: Since cos = ( - 1) n , the given series can be rewritten as an infinite geometric series X n =0 3 - 3 4 n = X n = 0 a r n in which a = 3 , r = - 3 4 . 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 12 7 . keywords: geometric series, convergent 005 (part 1 of 1) 10 points Determine whether the infinite series X n =1 ( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum. 1. diverges correct 2. converges with sum 1 2 3. converges with sum 1 4
Panjwani, Sameer – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: James Rath 3 4. converges with sum 1 8 5. converges with sum 1 6 Explanation: The infinite series X n =1 a n diverges when lim n →∞ a n 6 = 0 .

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Integral Calculus Exam 3 - Panjwani Sameer Exam 3 Due Dec 5...

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