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**Unformatted text preview: **Version 023 – EXAM 3 – lawn – (55930) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find a power series representation centered at the origin for the function f ( y ) = 1 (7 − y ) 2 . 1. f ( y ) = ∞ summationdisplay n = 0 1 7 n +1 y n 2. f ( y ) = ∞ summationdisplay n = 1 n 7 n y n − 1 3. f ( y ) = ∞ summationdisplay n = 1 1 7 n +1 y n 4. f ( y ) = ∞ summationdisplay n = 0 n + 1 7 n y n 5. f ( y ) = ∞ summationdisplay n = 1 n 7 n +1 y n − 1 correct 6. f ( y ) = ∞ summationdisplay n = 0 ( n + 1) y n Explanation: By the known result for geometric series, 1 7 − y = 1 7 parenleftBig 1 − y 7 parenrightBig = 1 7 ∞ summationdisplay n = 0 parenleftBig y 7 parenrightBig n = ∞ summationdisplay n = 0 1 7 n +1 y n . This series converges on ( − 7 , 7). On the other hand, 1 (7 − y ) 2 = d dy parenleftBig 1 7 − y parenrightBig , and so on ( − 7 , 7), 1 (7 − y ) 2 = d dy parenleftBigg ∞ summationdisplay n = 0 y n 7 n +1 parenrightBigg = ∞ summationdisplay n = 1 n 7 n +1 y n − 1 . Consequently, f ( y ) = ∞ summationdisplay n = 1 n 7 n +1 y n − 1 . 002 10.0 points Which, if any, of the following series converge? ( A ) ∞ summationdisplay n = 1 1 + sin( n ) 4 n ( B ) ∞ summationdisplay n = 4 3 √ n n − 3 1. neither A nor B 2. A and B 3. A but not B correct 4. B but not A Explanation: ( A ) Since 0 < 1 + sin( n ) < 2, we see that < 1 + sin( n ) 4 n < 2 4 n . On the other hand, the series ∞ summationdisplay n = 1 1 4 n is a convergent geometric series. Conse- quently, by the Comparison Test the series converges . ( B ) Since lim n →∞ √ n parenleftBig 3 √ n n − 3 parenrightBig = 3 , Version 023 – EXAM 3 – lawn – (55930) 2 the Limit Comparison Test applies. But the series ∞ summationdisplay n = 1 1 √ n diverges by the p-series test with p = 1 / 2. Consequently, the series diverges . keywords: 003 10.0 points Determine whether the series ∞ summationdisplay m = 1 ( m √ 6 − 1) m converges or diverges. 1. diverges 2. converges correct Explanation: The given series can be written in the form ∞ summationdisplay m = 1 a m where a m = ( m √ 6 − 1) m . But then m radicalbig | a m | = 6 1 /m − 1 , in which case lim m →∞ m radicalbig | a m | = 1 − 1 = 0 . Consequently, by the Root Test, the given series converges . 004 10.0 points Determine whether the series ∞ summationdisplay n = 1 n 3 4 n 3 + 3 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 4 2. convergent with sum = 1 4 3. convergent with sum = 7 4. divergent correct 5. convergent with sum = 1 7 Explanation: The infinite series ∞ summationdisplay n =1 a n is divergent when lim n →∞ a n exists but lim n →∞ a n negationslash = 0 ....

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- Fall '11
- Gramlich
- Accounting, Mathematical Series, lim