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Unformatted text preview: Mathematics 104, Quiz 5, Friday, April 4, 2008 1. Consider the power series summationdisplay n =1 n 3 x n 3 n . For which values of x is this series convergent? As usual, explain your reasoning. vextendsingle vextendsingle vextendsingle vextendsingle ( n + 1) 3 x n +1 3 n +1 3 n n 3 x n vextendsingle vextendsingle vextendsingle vextendsingle = parenleftbigg n + 1 n parenrightbigg 3  x  3  x  3 as n goes to infinity since ( n + 1) /n goes to 1 by LHopitals rule. So by the ratio test this series is absolutely convergent when  x  / 3 < 1 or in other words if x ( 3 , 3). It is divergent if x > 3 or if x < 3 by the ratio test. It remains to check the endpoints. If x = 3 then the series becomes summationdisplay 1 n 3 which diverges because it fails the nth term test since n 3 as n goes to infinity. Similarly when x = 3 we get the series summationdisplay 1 ( 1) n n 3 , which also diverges because it too fails the n th term test....
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This note was uploaded on 04/07/2008 for the course MAT 104 taught by Professor Edwardnelson during the Spring '08 term at Princeton.
 Spring '08
 EdwardNelson
 Calculus, Power Series

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