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Unformatted text preview: Su, Yung – Homework 13 – Due: Nov 27 2007, 3:00 am – Inst: Shinko Harper 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points For the series ∞ X n = 1 ( 1) n n + 6 x n , (i) determine its radius of convergence, R . 1. R = (∞ , ∞ ) 2. R = 1 correct 3. R = 0 4. R = 1 6 5. R = 6 Explanation: The given series has the form ∞ X n = 1 a n x n with a n = ( 1) n n + 6 . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x , while 0 < R < ∞ , (iii) if it converges when  x  < R , and (iv) diverges when  x  > R . But lim n →∞ fl fl fl a n +1 a n fl fl fl = lim n →∞ n + 6 n + 7 = 1 . By the Ratio Test, therefore, the given series converges when  x  < 1 and diverges when  x  > 1. Consequently, R = 1 . 002 (part 2 of 2) 10 points (ii) Determine the interval of convergence of the series. 1. interval convergence = [ 6 , 6) 2. converges only at x = 0 3. interval convergence = ( 1 , 1] correct 4. interval convergence = ( 1 , 1) 5. interval convergence = ( 6 , 6) 6. interval convergence = [ 1 , 1) 7. interval convergence = ( 6 , 6] Explanation: Since R = 1, the given series (i) converges when  x  < 1, and (ii) diverges when  x  > 1. On the other hand, at the point x = 1 and x = 1, the series reduces to ∞ X n = 1 ( 1) n n + 6 , ∞ X n = 1 1 n + 6 respectively. But by the Alternating Series Test, the first of these series converges, while by the pseries Test with p = 1, the second of these series diverges. Consequently, interval convergence = ( 1 , 1] . keywords: 003 (part 1 of 1) 10 points Su, Yung – Homework 13 – Due: Nov 27 2007, 3:00 am – Inst: Shinko Harper 2 Determine the radius of convergence, R , of the series ∞ X n = 1 x n ( n + 5)! . 1. R = 5 2. R = 0 3. R = ∞ correct 4. R = 1 5. R = 1 5 Explanation: The given series has the form ∞ X n = 1 a n x n with a n = 1 ( n + 5)! . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x , while 0 < R < ∞ , (iii) if it converges when  x  < R , and (iv) diverges when  x  > R . But lim n →∞ fl fl fl a n +1 a n fl fl fl = lim n →∞ 1 n + 6 = 0 . By the Ratio Test, therefore, the given series converges for all x . Consequently, R = ∞ . keywords: power series, Ratio test, radius convergence 004 (part 1 of 1) 10 points Find the interval of convergence of the power series ∞ X n = 1 ‡ 2 n + 7 8 n · n x n . 1. interval of convergence = ‡ 4 , 4 · correct 2. interval of convergence = h 8 , 8 · 3. interval of convergence = h 1 4 , 1 4 · 4. interval of convergence = ‡ 1 4 , 1 4 · 5. interval of convergence = h 4 , 4 · Explanation: We first apply the root test to the infinite series ∞ X n = 1 ‡ 2 n + 7 8 n · n  x  n...
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This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas.
 Spring '09
 Harper
 Calculus

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