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Unformatted text preview: Valenica, Daniel – Homework 13 – Due: Nov 27 2007, 3:00 am – Inst: Cheng 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 + 4 x n , (i) determine its radius of convergence, R . 1. R = 1 correct 2. R = 4 3. R = 0 4. R = 1 4 5. R = (∞ , ∞ ) Explanation: The given series has the form ∞ X n = 1 a n x n with a n = ( 1) n n + 4 . 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 + 4 n + 5 = 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 = [ 1 , 1) 2. interval convergence = ( 1 , 1] correct 3. interval convergence = ( 4 , 4) 4. interval convergence = [ 4 , 4) 5. converges only at x = 0 6. interval convergence = ( 1 , 1) 7. interval convergence = ( 4 , 4] 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 + 4 , ∞ X n = 1 1 n + 4 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 Valenica, Daniel – Homework 13 – Due: Nov 27 2007, 3:00 am – Inst: Cheng 2 Determine the radius of convergence, R , of the series ∞ X n = 1 x n ( n + 4)! . 1. R = 1 2. R = 1 4 3. R = ∞ correct 4. R = 4 5. R = 0 Explanation: The given series has the form ∞ X n = 1 a n x n with a n = 1 ( n + 4)! . 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 + 5 = 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 ‡ 6 n + 7 3 n · n x n . 1. interval of convergence = h 1 2 , 1 2 · 2. interval of convergence = h 3 , 3 · 3. interval of convergence = h 2 , 2 · 4. interval of convergence = ‡ 2 , 2 · 5. interval of convergence = ‡ 1 2 , 1 2 · correct Explanation: We first apply the root test to the infinite series ∞ X n = 1 ‡ 6 n + 7 3 n · n  x  n ....
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 Fall '07
 CHENG
 Mathematical Series, Convergence, lim, Valenica

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