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solution_pdf - schuem Exam3 Practice Problems Schultz(58385...

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schuem – Exam3 Practice Problems – Schultz – (58385) 1 This print-out should have 23 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 = ( 1) n parenleftbigg 2 n + 1 5 n + 1 parenrightbigg , and if it does, find its limit. 1. limit = 2 5 2. limit = ± 2 5 3. limit = 1 4. sequence diverges correct 5. limit = 0 Explanation: After division, 2 n + 1 5 n + 1 = 2 + 1 n 5 + 1 n . Now 1 n , 1 n 0 as n → ∞ , so lim n → ∞ 2 n + 1 5 n + 1 = 2 5 negationslash = 0 . Thus as n → ∞ , the values of a n oscillate be- tween values ever closer to ± 2 5 . Consequently, the sequence diverges . 002 10.0 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 4 2. convergent with sum = 9 correct 3. convergent with sum = 4 4. convergent with sum = 1 9 5. divergent Explanation: The series 3 + 2 + 4 3 + 8 9 + · · · = summationdisplay 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 . 003 10.0 points Determine whether the infinite series summationdisplay n =1 ( n + 1) 2 n ( n + 2) converges or diverges, and if converges, find its sum. 1. converges with sum = 1 2 2. converges with sum = 1 8 3. converges with sum = 1 4
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schuem – Exam3 Practice Problems – Schultz – (58385) 2 4. diverges correct 5. converges with sum = 1 Explanation: By the Divergence Test, an infinite series n a n diverges when lim n →∞ a n negationslash = 0 . Now, for the given series, a n = ( n + 1) 2 n ( n + 2) = n 2 + 2 n + 1 n 2 + 2 n . But then, lim n → ∞ a n = 1 negationslash = 0 . Consequently, the Divergence Test says that the given series diverges . keywords: infinite series, Divergence Test, ra- tional function 004 10.0 points To apply the root test to an infinite series n a n the value of ρ = lim n → ∞ | a n | 1 /n has to be determined. Compute the value of ρ for the series summationdisplay n =1 parenleftbigg 3 n + 5 4 n parenrightbigg 2 n . 1. ρ = 3 4 2. ρ = 25 9 3. ρ = 9 16 correct 4. ρ = 25 16 5. ρ = 5 4 Explanation: After division, 3 n + 5 4 n = 3 + 5 /n 4 , so | a n | 1 /n = parenleftBig 3 + 5 /n 4 parenrightBig 2 . On the other hand, lim n → ∞ 3 + 5 /n 4 = 3 4 . Consequently, ρ = 9 16 . keywords: /* If you use any of these, fix the comment symbols. 005 10.0 points Determine whether the series summationdisplay k =1 ( 1) k 1 cos parenleftBig 1 3 k parenrightBig is absolutely convergent, conditionally con- vergent or divergent. 1. absolutely convergent 2. conditionally convergent 3. series is divergent correct Explanation: The given series can be written as summationdisplay k =1 ( 1) k 1 a k
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schuem – Exam3 Practice Problems – Schultz – (58385) 3 with a k = cos parenleftBig 1 3 k parenrightBig > 0 , and so is an alternating series. But lim k → ∞ cos parenleftBig 1 3 k parenrightBig = 1 , in which case lim k → ∞ ( 1) k 1 a k does not exist. Thus by the Divergence test, the given
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