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Unformatted text preview: Math 1B. Sample Answers to Second Midterm 1. (12 points) Determine whether the series X n =1 ( 1) n n + 1 n 2 + 1 is absolutely convergent, conditionally convergent, or divergent. Following Example 2 on pages 711712, we note that the series is alternating, but lim n n + 1 n 2 + 1 = lim n 1 + 1 /n p 1 + 1 /n 2 = 1 , so the Alternating Series Test cannot be applied. Instead, we look at the limit of the n th term of the series: lim n ( 1) n n + 1 n 2 + 1 . This limit does not exist (the terms alternate between being close to 1 and 1), so the series diverges, by the Test for Divergence. 2. (14 points) Determine whether the series X n =2 cos n n 2 1 is absolutely convergent, conditionally convergent, or divergent. Since 1 cos n 1, we have cos n n 2 1 1 n 2 1 . Therefore, we can test for absolute convergence by the Comparison Test, comparing it with the series 1 / ( n 2 1) ( provided the series 1 / ( n 2 1) converges). This latter series has positive terms, so we can use the Limit Comparison Test to compare it with the pseries 1 /n 2 : lim n 1 n 2 1 1 n 2 = lim n 1 1 1 /n 2 = 1 . Since the pseries converges ( p = 2 > 1), the series 1 / ( n 2 1) also converges (by the Limit Comparison Test), and so the original series converges absolutely (by the ordinary Comparison Test)....
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This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Reshetiken
 Math

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