practice_prelim3_fa05_solutions - Practice Prelim 3...

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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 ) 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 L’Hopital’s 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 p -series with p = 5 4 > 1, it follows by the Comparison Test that n =1 (ln n ) 2 n 3 / 2 also converges. b ) 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. c ) n =2 n n + 1 n 3 + 3 n + 1 [5 points ] Let a n be the n th term of the above series, and let b n be 1 n 3 2 . Then a n b n = n 5 2

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