M408D Quest Homework 3-solutions

# M408D Quest Homework 3-solutions - hernandez(ah29758 M408D...

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Unformatted text preview: hernandez (ah29758) M408D Quest Homework 3 pascaleff (54550) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine whether the series summationdisplay n =1 ( 1) n 1 e 1 /n 4 n is absolutely convergent, conditionally con- vergent or divergent. 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: Since summationdisplay n =1 ( 1) n 1 e 1 /n 4 n = 1 4 summationdisplay n =1 ( 1) n e 1 /n n , we have to decide if the series summationdisplay n =1 ( 1) n e 1 /n n is absolutely convergent, conditionally con- vergent, or divergent. First we check for absolute convergence. Now, since e 1 /n 1 for all n 1, e 1 /n 4 n 1 4 n &amp;gt; . But by the p-series test with p = 1, the series summationdisplay n =1 1 4 n diverges, and so by the Comparison Test, the series summationdisplay n =1 e 1 /n 4 n too diverges; in other words, the given series is not absolutely convergent. To check for conditional convergence, con- sider the series summationdisplay n =1 ( 1) n f ( n ) where f ( x ) = e 1 /x 4 x . Then f ( x ) &amp;gt; 0 on (0 , ). On the other hand, f ( x ) = 1 4 x 3 e 1 /x e 1 /x 4 x 2 = e 1 /x parenleftBig 1 + x 4 x 3 parenrightBig . Thus f ( x ) &amp;lt; 0 on (0 , ), so f ( n ) &amp;gt; f ( n + 1) for all n . Finally, since lim x e 1 /x = 1 , we see that f ( n ) 0 as n . By the Alternating Series Test, therefore, the series summationdisplay n =1 ( 1) n f ( n ) is convergent. Consequently, the given series is conditionally convergent . keywords: 002 10.0 points Which one of the following properties does the series summationdisplay m =1 ( 1) m 1 1 1 + m 2 have? 1. absolutely convergent 2. divergent 3. conditionally convergent correct Explanation: hernandez (ah29758) M408D Quest Homework 3 pascaleff (54550) 2 The given series has the form of an alter- nating series summationdisplay m =1 ( 1) m 1 a m , a m = 1 1 + m 2 . To check for absolute convergence we apply the Limit Comparison Test with a m = 1 1 + m 2 , b m = 1 m . For then lim m a m b m = lim m m 1 + m 2 = 1 &amp;gt; . Thus the series a m converges if and only if the series summationdisplay m =1 1 m converges. But, by the p-series test with p = 1, this last series diverges; in particular, the given series does not converge absolutely. To check if the given series converges con- ditionally, consider first the function f ( x ) = 1 1 + x 2 . Then, by the Chain Rule, f ( x ) = x (1 + x 2 ) 3 / 2 &amp;lt; for all x &amp;gt; 0, while lim x f ( x ) = 0 ....
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## This note was uploaded on 11/15/2011 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas.

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M408D Quest Homework 3-solutions - hernandez(ah29758 M408D...

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