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Unformatted text preview: suleimenov (bs26835) – FINAL REVIEW #2 – rusin – (55565) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. More of the final review. 001 10.0 points Find the interval of convergence of the se ries ∞ summationdisplay n =1 x n √ n + 4 . 1. converges only at x = 0 2. interval of cgce = [ − 1 , 1] 3. interval of cgce = [ − 1 , 1) correct 4. interval of cgce = ( − 4 , 4] 5. interval of cgce = ( − 1 , 1) 6. interval of cgce = [ − 4 , 4] Explanation: When a n = x n √ n + 4 , then vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle x n +1 √ n + 5 √ n + 4 x n vextendsingle vextendsingle vextendsingle vextendsingle =  x  parenleftbigg √ n + 4 √ n + 5 parenrightbigg =  x  radicalbigg n + 4 n + 5 . But lim n →∞ n + 4 n + 5 = 1 , so lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle =  x  . By the Ratio Test, therefore, the given series (i) converges when  x  < 1, (ii) diverges when  x  > 1. We have still to check what happens at the endpoints x = ± 1. At x = 1 the series becomes ( ∗ ) ∞ summationdisplay n =1 1 √ n + 4 . Applying the Integral Test with f ( x ) = 1 √ x + 4 we see that f is continuous, positive, and de creasing on [1 , ∞ ), but the improper integral I = integraldisplay ∞ 1 f ( x ) dx diverges, so the infinite series ( ∗ ) diverges also. On the other hand, at x = − 1, the series becomes ( ‡ ) ∞ summationdisplay n =1 ( − 1) n √ n + 4 . which is an alternating series ∞ summationdisplay n =1 ( − 1) n a n , a n = f ( n ) with f ( x ) = 1 √ x + 4 the same continuous, positive and decreasing function as before. Since lim x →∞ f ( x ) = lim x →∞ 1 √ x + 4 = 0 , however, the Alternating Series Test ensures that ( ‡ ) converges. Consequently, the interval of convergence = [ − 1 , 1) . 002 10.0 points suleimenov (bs26835) – FINAL REVIEW #2 – rusin – (55565) 2 Determine the interval of convergence of the series ∞ summationdisplay k = 1 ( − 1) k − 1 1 k 2 ( x − 1) k . 1. interval of convergence = [ − 2 , 0 ) 2. interval of convergence = (0 , 2 ) 3. interval of convergence = [0 , 2 ] cor rect 4. interval of convergence = (0 , 2 ] 5. interval of convergence = [0 , 2 ) 6. interval of convergence = ( − 2 , 0 ] 7. interval of convergence = [ − 2 , 0 ] 8. interval of convergence = ( − 2 , 0 ) Explanation: The given series has the form ∞ summationdisplay k = 1 c k ( x − 1 ) k , c k = ( − 1) k − 1 k 2 ....
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 Spring '11
 RUSIN
 Power Series, Taylor Series, lim

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