SampleMidtermwithSolutions1

SampleMidtermwithSolutions1 - MATH 262: Sample Midterm with...

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Unformatted text preview: MATH 262: Sample Midterm with Solutions Problem 1. (a) Find the sum of the series X n =1 3 n 6 2 n +1 . (b) Find out for which values of x the series X n =1 (- 1) n- 1 ( x- 5) 2 n n 3 n converges absolutely, converges conditionally, diverges. Solution. (a) X n =1 3 n 6 2 n +1 = 3 6 3 + 3 2 6 5 + = 3 6 3 1 + 3 6 2 + 3 2 6 4 + = 3 6 3 1 + 3 6 2 + 3 6 2 2 + ! = 3 6 3 1 1- 3 6 2 = 1 66 . (b) At first we test the series for absolute convergence using the Ratio Test. lim n fl fl fl fl a n +1 a n fl fl fl fl = lim n fl fl fl fl (- 1) n ( x- 5) 2( n +1) ( n + 1) 3 n +1 (- 1) n- 1 ( x- 5) 2 n n 3 n fl fl fl fl = lim n fl fl fl fl ( x- 5) 2( n +1) n 3 n ( x- 5) 2 n ( n + 1) 3 n +1 fl fl fl fl = lim n fl fl fl fl ( x- 5) 2 3 n n + 1 fl fl fl fl = ( x- 5) 2 3 . Hence, the series converges absolutely if and only if ( x- 5) 2 3 < 1, or equivalently, if | x- 5 | < 3. Assume | x- 5 | > 3. Then a = ( x- 5) 2 3 > 1 and the limit lim n (- 1) n- 1 ( x- 5) 2 n n 3 n = lim n (- 1) n- 1 a n n does not exist since a n n as n . Hence, the series cant converge if | x- 5 | > 3. Finally, we check the endpoints of the interval of convergence, that is, when | x- 5 | = 3, equivalently, x = 5 3. We have X n =1 (- 1) n- 1 ( x- 5) 2 n n 3 n = X n =1 (- 1) n- 1 ( 3) 2 n n 3 n = X n =1 (- 1) n- 1 3 n n 3 n = X n =1 (- 1) n- 1 1 n which converges conditionally by the Alternating Series Test.which converges conditionally by the Alternating Series Test....
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SampleMidtermwithSolutions1 - MATH 262: Sample Midterm with...

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