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Unformatted text preview: rhodes (ajr2283) – HW06 – Gilbert – (56380) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points For the series ∞ summationdisplay n = 1 ( 1) n n + 2 x n , (i) determine its radius of convergence, R . 1. R = (∞ , ∞ ) 2. R = 2 3. R = 1 correct 4. R = 0 5. R = 1 2 Explanation: The given series has the form ∞ summationdisplay n = 1 a n with a n = ( 1) n x 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 →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n →∞  x  n + 2 n + 3 =  x  . 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) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = ( 1 , 1) 2. converges only at x = 0 3. interval convergence = ( 1 , 1] correct 4. interval convergence = ( 2 , 2) 5. interval convergence = [ 1 , 1) 6. interval convergence = [ 2 , 2) 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 ∞ summationdisplay n = 1 ( 1) n n + 2 , ∞ summationdisplay n = 1 1 n + 2 respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set a n = 1 n + 2 , b n = 1 n , then lim n →∞ a n b n = lim n →∞ n n + 2 = 1 . By the pseries Test with p = 1, however, the series ∑ n b n diverges. Thus by the Limit Comparison test, the series ∑ n a n also di verges. Consequently, the given series has interval convergence = ( 1 , 1] . rhodes (ajr2283) – HW06 – Gilbert – (56380) 2 keywords: 003 (part 1 of 2) 10.0 points For the series ∞ summationdisplay n =1 ( 1) n n 3 n ( x + 1) n , (i) determine its radius of convergence, R . 1. R = 0 2. R = 1 3. R = 3 correct 4. R = ∞ 5. R = 1 3 Explanation: The given series has the form summationdisplay n = 1 a n ( x + 1) n with a n = ( 1) n n 3 n . Now for this series (i) R = 0 if it converges only at x = 1, (ii) R = ∞ if it converges for all x , while if R > 0, (iii) it coverges when  x + 1  < R , and (iv) diverges when  x + 1  > R . But lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n →∞ n 3( n + 1) = 1 3 ....
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 Spring '07
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 Multivariable Calculus, Mathematical Series, Rhodes, interval convergence

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