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# solution0_pdf - dang(rtd356 – Wk 3 Check Knowledge –...

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Unformatted text preview: dang (rtd356) – Wk 3 Check Knowledge – Cepparo – (57045) 1 This print-out should have 8 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 ∞ summationdisplay n = 1 n ( n + 1)4 n converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: We use the Limit Comparison Test with a n = n ( n + 1)4 n , b n = 1 4 n . For lim n →∞ a n b n = lim n →∞ n n + 1 = 1 > . Thus the series ∞ summationdisplay n = 1 n ( n + 1)4 n converges if and only if the series ∞ summationdisplay n = 1 1 4 n converges. But this last series is a geometric series with | r | = 1 4 < 1 , hence convergent. Consequently, the given series is series is convergent . 002 10.0 points Which of the following series ( A ) ∞ summationdisplay n = 1 4 n 6 n 2 + 4 ( B ) ∞ summationdisplay n = 1 parenleftbigg 1 2 parenrightbigg n ( C ) ∞ summationdisplay n = 21 parenleftbigg 6 7 parenrightbigg n converge(s)? 1. A, B, and C 2. B and C only correct 3. C only 4. B only 5. A and B 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 + 4 . Then f is continuous, positive and decreasing on [1 , ∞ ); thus ∞ summationdisplay n = 1 4 n 6 n 2 + 4 converges if and only if the improper integral integraldisplay ∞ 1 4 x 6 x 2 + 4 dx converges, which requires us to evaluate the integral I n = integraldisplay n 1 4 x 6 x 2 + 4 dx . dang (rtd356) – Wk 3 Check Knowledge – Cepparo – (57045) 2 Now after substitution (set u = x...
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## This note was uploaded on 06/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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solution0_pdf - dang(rtd356 – Wk 3 Check Knowledge –...

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