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ReviewCalc3

# ReviewCalc3 - nav277 Review plus(exam 3 Odell(58340 This...

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nav277 – Review plus (exam 3) – Odell – (58340) 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 5 n + 6 6 n + 2 parenrightbigg , and if it does, find its limit. 1. limit = 5 6 2. limit = ± 5 6 3. limit = 0 4. limit = 3 5. sequence diverges correct Explanation: After division, 5 n + 6 6 n + 2 = 5 + 6 n 6 + 2 n . Now 6 n , 2 n 0 as n → ∞ , so lim n → ∞ 5 n + 6 6 n + 2 = 5 6 negationslash = 0 . Thus as n → ∞ , the values of a n oscillate be- tween values ever closer to ± 5 6 . Consequently, the sequence diverges . 002 10.0 points Determine whether the series 1 + 3 + 9 + 27 + · · · is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 9 2 2. divergent correct 3. convergent with sum = 1 2 4. convergent with sum = 2 5. convergent with sum = 2 9 Explanation: The series 1 + 3 + 9 + 27 + · · · = summationdisplay n =1 a r n 1 is an infinite geometric series in which a = 1 and r = 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 divergent . 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. diverges correct 2. converges with sum = 1 3. converges with sum = 1 2

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nav277 – Review plus (exam 3) – Odell – (58340) 2 4. converges with sum = 1 8 5. converges with sum = 1 4 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 6 n + 3 2 n parenrightbigg 2 n . 1. ρ = 3 2. ρ = 3 2 3. ρ = 9 correct 4. ρ = 1 4 5. ρ = 9 4 Explanation: After division, 6 n + 3 2 n = 6 + 3 /n 2 , so | a n | 1 /n = parenleftBig 6 + 3 /n 2 parenrightBig 2 . On the other hand, lim n → ∞ 6 + 3 /n 2 = 3 . Consequently, ρ = 9 . 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 5 k parenrightBig is absolutely convergent, conditionally con- vergent or divergent. 1. series is divergent correct 2. absolutely convergent 3. conditionally convergent Explanation: The given series can be written as summationdisplay k =1 ( 1) k 1 a k
nav277 – Review plus (exam 3) – Odell – (58340) 3 with a k = cos parenleftBig 1 5 k parenrightBig > 0 , and so is an alternating series. But lim k → ∞ cos parenleftBig 1 5 k parenrightBig = 1 , in which case lim k → ∞ ( 1) k 1 a k does not exist. Thus by the Divergence test, the given series is divergent .

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