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**Unformatted text preview: **truong (at28889) – HW13 – zupan – (56530) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the series ∞ summationdisplay k = 1 ( − 1) k − 1 cos 2 ( k ) 5 k is absolutely convergent, conditionally con-vergent or divergent 1. divergent 2. conditionally convergent 3. absolutely convergent correct Explanation: To check for absolute convergence we have to decide if the series ∞ summationdisplay k =1 cos 2 ( k ) 5 k is convergent. For this we can use the Com-parison Test with a k = cos 2 ( k ) 5 k , b k = 1 5 k . For then ≤ a k ≤ b k , since 0 ≤ cos 2 ( k ) ≤ 1. Thus the series ∞ summationdisplay k =1 cos 2 ( k ) 5 k converges if the series ∞ summationdisplay k =1 1 5 k converges. But this last series is a geometric series with r = 1 5 < 1 , hence convergent. Consequently, the given series is absolutely convergent . 002 10.0 points Determine whether the series 5 8 − 6 9 + 7 10 − 8 11 + 9 12 − . . . is absolutely convergent, conditionally con-vergent or divergent. 1. absolutely convergent 2. conditionally convergent 3. divergent correct Explanation: In summation notation, 5 8 − 6 9 + 7 10 − 8 11 + 9 12 − . . . = ∞ summationdisplay n = 5 a n , with a n given by a n = ( − 1) n n n + 3 . However, lim n →∞ n n + 3 = 1 , so that as n → ∞ , a n oscillates between val-ues approaching 1 and − 1. In particular, therefore, lim n →∞ a n negationslash = 0 . Consequently, by the Divergence Test, the series is divergent . 003 10.0 points Determine whether the series ∞ summationdisplay n =1 ( − 1) n +1 e 1 /n 6 n truong (at28889) – HW13 – zupan – (56530) 2 is absolutely convergent, conditionally con-vergent or divergent. 1. absolutely convergent 2. conditionally convergent correct 3. divergent Explanation: Since ∞ summationdisplay n =1 ( − 1) n +1 e 1 /n 6 n = − 1 6 ∞ summationdisplay n = 1 ( − 1) n e 1 /n n , we have to decide if the series ∞ summationdisplay n =1 ( − 1) n e 1 /n n is absolutely convergent, conditionally con-vergent, or divergent. First we check for absolute convergence. Now, since e 1 /n ≥ 1 for all n ≥ 1, e 1 /n 6 n ≥ 1 6 n > . But by the p-series test with p = 1, the series ∞ summationdisplay n = 1 1 6 n diverges, and so by the Comparison Test, the series ∞ summationdisplay n = 1 e 1 /n 6 n too diverges; in other words, the given series is not absolutely convergent. To check for conditional convergence, con-sider the series ∞ summationdisplay n = 1 ( − 1) n f ( n ) where f ( x ) = e 1 /x 6 x . Then f ( x ) > 0 on (0 , ∞ ). On the other hand, f ′ ( x ) = − 1 6 x 3 e 1 /x − e 1 /x 6 x 2 = − e 1 /x parenleftBig 1 + x 6 x 3 parenrightBig ....

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