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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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solution0_pdf - dang (rtd356) Wk 3 Check Knowledge Cepparo...

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