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Unformatted text preview: Version 099 EXAM 3 Neitzke (56585) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points If the n th partial sum of an infinite series is S n = 2 n 2 3 5 n 2 + 2 , what is the sum of the series? 1. sum = 3 10 2. sum = 2 5 correct 3. sum = 1 4 4. sum = 9 20 5. sum = 7 20 Explanation: By definition sum = lim n S n = lim n parenleftBig 2 n 2 3 5 n 2 + 2 parenrightBig . Thus sum = 2 5 . 002 10.0 points First find a n so that summationdisplay n =1 a n = 6 + 3 2 + 2 3 + 3 4 + 6 5 5 + . . . and then determine whether the series con verges or diverges. 1. a n = 6 n 3 / 2 , series diverges 2. a n = 6 n 1 / 2 , series diverges 3. a n = 3 2 n 3 / 2 , series diverges 4. a n = 3 2 n 3 / 2 , series converges 5. a n = 6 n 1 / 2 , series converges 6. a n = 6 n 3 / 2 , series converges correct Explanation: By inspection a n = 6 n 3 / 2 . To test for convergence we use the Integral test with f ( x ) = 6 x 3 / 2 . This is a positive, continuous, decreasing function on [1 , ). Furthermore, integraldisplay 1 f ( x ) dx = lim n integraldisplay n 1 6 x 3 / 2 dx = lim n bracketleftbigg 12 x 1 / 2 bracketrightbigg n 1 = 12 , so the improper integral integraldisplay 1 f ( x ) dx is convergent, which by the Integral test means that the series converges . 003 10.0 points Which of the following series ( A ) summationdisplay n = 1 4 n 6 n 2 + 5 ( B ) summationdisplay n =18 parenleftbigg 6 7 parenrightbigg n converge(s)? Version 099 EXAM 3 Neitzke (56585) 2 1. both of them 2. B only correct 3. neither of them 4. A only Explanation: ( A ) Because of the way the n th term is defined as a quotient of polynomials in the series, use of the integral test is suggested. Set f ( x ) = 4 x 6 x 2 + 5 . Then f is continuous, positive and decreasing on [1 , ); thus summationdisplay n = 1 4 n 6 n 2 + 5 converges if and only if the improper integral integraldisplay 1 4 x 6 x 2 + 5 dx converges, which requires us to evaluate the integral I n = integraldisplay n 1 4 x 6 x 2 + 5 dx . Now after substitution (set u = x 2 ), we see that I n = bracketleftbigg 1 3 ln(6 x 2 + 5) bracketrightbigg n 1 = 1 3 ( ln(6 n 2 + 5) ln(6 + 5) ) . But ln(6 n 2 + 5) as n , so the infinite series summationdisplay n = 1 4 n 6 n 2 + 5 diverges. ( B ) The series summationdisplay n =18 parenleftbigg 6 7 parenrightbigg n is a geometric series, but the summmation starts at n = 18 instead of at n = 1, as it usually does. We can still show, however, that this series converges. Indeed, summationdisplay n = 18 parenleftbigg 6 7 parenrightbigg n = parenleftbigg 6 7 parenrightbigg 18 + parenleftbigg 6 7 parenrightbigg 19 + parenleftbigg 6 7 parenrightbigg 20 + . . . ....
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 Fall '09
 GOGOLEV

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