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Unformatted text preview: Practice Prelim 3 Solutions, Math 191, Fall 2005 Problem 1) Decide, giving reasons, whether the following series converges ab solutely, converges conditionally, or diverges? Note: all the series have positive terms, so either converge absolutely, or diverge. There is no conditional convergence. a ) X n =1 (ln n ) 2 n 3 / 2 [10 points ] (ln n ) 2 n 3 / 2 = 1 n 5 4 . ln n n 1 8 2 Since ln n n 1 8 0 by LHopitals rule as n we see that for n large enough (ln n ) 2 n 3 / 2 = 1 n 5 4 . ln n n 1 8 2 1 n 5 4 . Since n =1 1 n 5 4 converges as it is a pseries with p = 5 4 > 1, it follows by the Comparison Test that n =1 (ln n ) 2 n 3 / 2 also converges. b ) X n =2 1 n + sin n [5 points ] n +sin n n = 1 + sin n n 1 as n since  sin n  is bounded. Comparison with the harmonic series n =2 1 n , which diverges, therefore shows by the Limit Comparison Test that n =2 1 n +sin n also diverges....
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This test prep was uploaded on 02/15/2008 for the course MATH 1910 taught by Professor Berman during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 BERMAN
 Math

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