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Unformatted text preview: momin (rrm497) – Homework 13 – cheng – (58520) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the interval of convergence of the power series ∞ summationdisplay n = 1 4 x n √ n . 1. interval of cgce = ( − 4 , 4) 2. interval of cgce = [ − 4 , 4) 3. interval of cgce = ( − 4 , 4] 4. interval of cgce = [ − 1 , 1) correct 5. interval of cgce = [ − 4 , , 4] 6. interval of cgce = [ − 1 , 1] 7. interval of cgce = ( − 1 , 1) 8. interval of cgce = ( − 1 , 1] Explanation: When a n = 4 x n √ n , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle x n +1 √ n + 1 · √ n x n vextendsingle vextendsingle vextendsingle vextendsingle =  x  √ n √ n + 1 =  x  parenleftbiggradicalbigg n n + 1 parenrightbigg . Thus lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle =  x  . By the Ratio Test, therefore, the given series converges when  x  < 1, and diverges when  x  > 1. We have still to check for convergence at x = ± 1. But when x = 1, the series reduces to ∞ summationdisplay n =1 4 √ n which diverges by the pseries test with p = 1 2 ≤ 1. On the other hand, when x = − 1, the series reduces to ∞ summationdisplay n =1 ( − 1) n 4 √ n which converges by the Alternating Series Test. Thus the interval of convergence = [ − 1 , 1) . keywords: 002 (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 = 0 3. R = 1 correct 4. R = 2 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 ....
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 Fall '08
 RAdin
 Calculus, Mathematical Series, lim

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