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Unformatted text preview: moseley (cmm3869) – HW05 – Gilbert – (56380) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the series 3 2 − 3 3 + 3 4 − 3 5 + 3 6 − . . . is conditionally convergent, absolutely con vergent or divergent. 1. series is conditionally convergent cor rect 2. series is absolutely convergent 3. series is divergent Explanation: In summation notation, 3 2 − 3 3 + 3 4 − 3 5 + 3 6 − . . . = ∞ summationdisplay n =1 ( − 1) n − 1 f ( n ) , with f ( x ) = 3 x + 1 . Now the improper integral integraldisplay ∞ 1 f ( x ) dx = integraldisplay ∞ 1 3 x + 1 dx is divergent, so by the Integral Test, the given series is not absolutely convergent. On the other hand, f ( n ) = 3 n + 1 > 3 n + 1 + 1 = f ( n + 1) , while lim n →∞ 3 n + 1 = 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 ( − 1) 3 4 3 + √ n 2. ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n 4 + √ n 3. ∞ summationdisplay n =1 3 4 + √ n 4. ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n correct 5. ∞ summationdisplay n =1 ( − 1) 2 n 4 3 + √ n Explanation: Since ∞ summationdisplay n =1 ( − 1) 3 4 3 + √ n = − ∞ summationdisplay n =1 4 3 + √ n , use of the Limit Comparison and pseries Tests with p = 1 2 shows that this series is divergent. Similarly, since ∞ summationdisplay n =1 ( − 1) 2 n 4 3 + √ n = ∞ summationdisplay n =1 4 3 + √ n , the same argument shows that this series as well as ∞ summationdisplay n =1 3 4 + √ n is divergent. On the other hand, by the Divergence Test, the series ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n 4 + √ n moseley (cmm3869) – HW05 – Gilbert – (56380) 2 is divergent because lim n →∞ ( − 1) n − 1 1 + √ n 4 + √ n negationslash = 0 . This leaves only the series ∞ summationdisplay n =1 ( − 1) n − 1 1 + √ n . To see that this series is convergent, set b n = 1 1 + √ 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 1 + √ n is convergent. 003 10.0 points Determine whether the series ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 3 n is absolutely convergent, conditionally con vergent or divergent. 1. absolutely convergent 2. divergent 3. conditionally convergent correct Explanation: Since ∞ summationdisplay n =1 ( − 1) n − 1 e 1 /n 3 n = − 1 3 ∞ 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 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.
 Spring '07
 Sadler
 Multivariable Calculus

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