quiz_sl1 - M408D Quiz Solution You must show your work to...

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M408D Quiz Solution You must show your work to get full credit. 1. Determine whether the series X n =1 ( - 1) n n n 3 + 2 is absolutely convergent, condi- tionally convergent or divergent. We always check absolutely convergence rst because absolutely convergence implies con- vergence. Let a n = ( - 1) n n n 3 + 2 , then | a n | = ( - 1) n n n 3 + 2 = n n 3 + 2 n n 3 + 3 n 3 = 1 2 n = 1 2 n 1 / 2 . Since X 1 n 1 / 2 is divergent ( p -series with p = 1 / 2 1 ), X 1 2 n 1 / 2 is also divergent. Therefore by comparison test, the series X | a n | is divergent. The series a n is an alternating series with b n = n n 3 + 2 . We can use the alternating series test to check convergence. lim n →∞ b n = lim n →∞ n n 3 + 2 = lim n →∞ 1 / n p 1 + 2 /n 3 = 0 Let f ( x ) = x 2 x 3 + 2 . Then f 0 ( x ) = 2 x 4 + 2 x - 3 x 4 ( x 3 + 2) 2 = - x 4 + 2 x ( x 3 + 2) 2 < 0 , x 2 Since square root is an increasing function and
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This note was uploaded on 04/02/2010 for the course M 408d taught by Professor Sadler during the Fall '07 term at University of Texas.

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quiz_sl1 - M408D Quiz Solution You must show your work to...

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