MTH
mth.122.handout.18

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This document was prepared by Ron Bannon ( [email protected] ) using L A T E X 2 ε . Last revised January 10, 2009. 1

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in calculus I) lim n →∞ a n b n = lim n →∞ sin (1 /n ) 1 /n = lim x 0 + sin x x = 1 . The limit comparison test applies with c = 1. Since the p -series diverges, and c = 1 > 0, this test shows that X n =1 sin 1 n diverges . 2. Alternating Series Test : If the alternating series X n =1 ( - 1) n - 1 a n = a 1 - a 2 + a 3 - a 4 + · · · ( a n > 0) satisfies (a) a n +1 a n for all n ; (b) and lim n →∞ a n = 0, then the series converges. Example: Show that the alternating harmonic series is convergent. X n =1 ( - 1) n - 1 n Work: We need to show. (a) a n +1 a n for all n . 1 n + 1 < 1 n 0 < 1 . (b) and lim n →∞ a n = 0. lim n →∞ 1 n = 0 . Since both conditions are satisfied, we can state that the alternating harmonic series converges . However, this test is for alternating series only. 3. Definition : The series X n =1 a n is called absolutely convergent if the series of the absolute values X n =1 | a n | is convergent. 2
4. Definition : The series X n =1 a n is called conditionally convergent if it is convergent, but not absolutely convergent. 5. Theorem : If the series X n =1 a n is absolutely convergent, then it is convergent. Determine if the series is absolutely convergent or conditionally convergent. (a) X n =1 ( - 1) n - 1 n 2 Work: This series is absolutely convergent because the series converges and X n =1 ( - 1) n - 1 n 2 = X n =1 1 n 2 is a convergent p -series.

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