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Unformatted text preview: Kim, Jin Homework 13 Due: Nov 27 2007, 3:00 am Inst: Diane Radin 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 + 2 x n , (i) determine its radius of convergence, R . 1. R = 2 2. R = 0 3. R = 1 2 4. R = 1 correct 5. R = ( , ) Explanation: The given series has the form X n = 1 a n x n with a n = ( 1) n n + 2 . 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 + 2 n + 3 = 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 = [ 2 , 2) 2. interval convergence = ( 2 , 2) 3. interval convergence = ( 1 , 1) 4. interval convergence = ( 1 , 1] correct 5. interval convergence = [ 1 , 1) 6. converges only at x = 0 7. interval convergence = ( 2 , 2] 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 + 2 , X n = 1 1 n + 2 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 Kim, Jin Homework 13 Due: Nov 27 2007, 3:00 am Inst: Diane Radin 2 Determine the radius of convergence, R , of the series X n = 1 x n ( n + 5)! . 1. R = 1 2. R = correct 3. R = 5 4. R = 1 5 5. R = 0 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 + 5 7 n n x n . 1. interval of convergence =  7 2 , 7 2 correct 2. interval of convergence = h 7 2 , 7 2 3. interval of convergence =  2 7 , 2 7 4. interval of convergence = h 7 , 7 5. interval of convergence = h 2 7 , 2 7 Explanation: We first apply the root test to the infinite series X n = 1...
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 Spring '09
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