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Unformatted text preview: Version 035 – EXAM 3 – BERG – (56525) 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 = 3 n 2 1 4 n 2 + 1 , what is the sum of the series? 1. sum = 3 4 correct 2. sum = 7 8 3. sum = 15 16 4. sum = 1 5. sum = 13 16 Explanation: By definition sum = lim n →∞ S n = lim n →∞ parenleftBig 3 n 2 1 4 n 2 + 1 parenrightBig . Thus sum = 3 4 . 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 1 / 2 , series converges 2. a n = 3 2 n 3 / 2 , series converges 3. a n = 6 n 3 / 2 , series diverges 4. a n = 3 2 n 3 / 2 , series diverges 5. a n = 6 n 1 / 2 , series diverges 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 2 n 6 n 2 + 2 ( B ) ∞ summationdisplay n =21 parenleftbigg 6 7 parenrightbigg n converge(s)? Version 035 – EXAM 3 – BERG – (56525) 2 1. neither of them 2. both of them 3. A only 4. B only correct 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 ) = 2 x 6 x 2 + 2 . Then f is continuous, positive and decreasing on [1 , ∞ ); thus ∞ summationdisplay n = 1 2 n 6 n 2 + 2 converges if and only if the improper integral integraldisplay ∞ 1 2 x 6 x 2 + 2 dx converges, which requires us to evaluate the integral I n = integraldisplay n 1 2 x 6 x 2 + 2 dx . Now after substitution (set u = x 2 ), we see that I n = bracketleftbigg 1 6 ln(6 x 2 + 2) bracketrightbigg n 1 = 1 6 ( ln(6 n 2 + 2) ln(6 + 2) ) . But ln(6 n 2 + 2)→ ∞ as n → ∞ , so the infinite series ∞ summationdisplay n = 1 2 n 6 n 2 + 2 diverges. ( B ) The series ∞ summationdisplay n =21 parenleftbigg 6 7 parenrightbigg n is a geometric series, but the summmation starts at n = 21 instead of at n = 1, as it usually does. We can still show, however, that this series converges. Indeed, ∞ summationdisplay n = 21 parenleftbigg 6 7 parenrightbigg n = parenleftbigg 6 7 parenrightbigg 21 + parenleftbigg 6 7 parenrightbigg 22 + parenleftbigg 6 7 parenrightbigg 23 + . . . ....
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This note was uploaded on 04/04/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.
 Spring '10
 ZHENG
 Calculus

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