int cal 13

# int cal 13 - pokharel (yp624) – HW13 – Radin –...

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Unformatted text preview: pokharel (yp624) – HW13 – Radin – (57410) 1 This print-out should have 20 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 2 3 − 2 4 + 2 5 − 2 6 + 2 7 − . . . is conditionally convergent, absolutely con- vergent or divergent. 1. series is divergent 2. series is absolutely convergent 3. series is conditionally convergent cor- rect Explanation: In summation notation, 2 3 − 2 4 + 2 5 − 2 6 + 2 7 − . . . = ∞ summationdisplay n =1 ( − 1) n − 1 f ( n ) , with f ( x ) = 2 x + 2 . Now the improper integral integraldisplay ∞ 1 f ( x ) dx = integraldisplay ∞ 1 2 x + 2 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 2 n + 2 > 2 n + 1 + 2 = f ( n + 1) , while lim n →∞ 2 n + 2 = 0 . Consequently, by the Alternating Series Test, the given series is conditionally convergent . keywords: alternating series, Alternating se- ries test, conditionally convergent, absolutely convergent, divergent 002 10.0 points Which one of the following series is conver- gent? 1. ∞ summationdisplay n =1 4 5 + √ n 2. ∞ summationdisplay n =1 ( − 1) 2 n 5 4 + √ n 3. ∞ summationdisplay n =1 ( − 1) n − 1 3 + √ n correct 4. ∞ summationdisplay n =1 ( − 1) 3 5 4 + √ n 5. ∞ summationdisplay n =1 ( − 1) n − 1 3 + √ n 5 + √ n Explanation: Since ∞ summationdisplay n =1 ( − 1) 3 5 4 + √ n = − ∞ summationdisplay n =1 5 4 + √ n , use of the Limit Comparison and p-series Tests with p = 1 2 shows that this series is divergent. Similarly, since ∞ summationdisplay n =1 ( − 1) 2 n 5 4 + √ n = ∞ summationdisplay n =1 5 4 + √ n , the same argument shows that this series as well as ∞ summationdisplay n =1 4 5 + √ n is divergent. On the other hand, by the Divergence Test, the series ∞ summationdisplay n =1 ( − 1) n − 1 3 + √ n 5 + √ n pokharel (yp624) – HW13 – Radin – (57410) 2 is divergent because lim n →∞ ( − 1) n − 1 3 + √ n 5 + √ n negationslash = 0 . This leaves only the series ∞ summationdisplay n =1 ( − 1) n − 1 3 + √ n . To see that this series is convergent, set b n = 1 3 + √ n . Then (i) b n +1 ≤ b n , (ii) lim n →∞ b n = 0 . Consequently, by the Alternating Series Test, the series ∞ summationdisplay n =1 ( − 1) n − 1 3 + √ n is convergent. 003 10.0 points Determine whether the series ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 7 n is absolutely convergent, conditionally con- vergent or divergent. 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: Since ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 7 n = − 1 7 ∞ 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....
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## This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas.

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int cal 13 - pokharel (yp624) – HW13 – Radin –...

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