mth.122.handout.19

mth.122.handout.19 - MTH 122 Calculus II Essex County...

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MTH 122 — Calculus II Essex County College — Division of Mathematics and Physics 1 Lecture Notes #19 — Sakai Web Project Material 1 Introduction to Sequences and Series, Part VI 1. The Root Test : (a) If lim n →∞ n p | a n | = L < 1, then the series X n =1 a n is absolutely convergent. (b) If lim n →∞ n p | a n | = L > 1, or lim n →∞ n p | a n | = , then the series X n =1 a n is divergent. (c) If lim n →∞ n p | a n | = 1, the root test fails, and we can draw no conclusion. This test works because lim n →∞ n a n = r tells use that the series is comparable to a geometric series with ration r . Example: Use the root test to determine if the series converges. X n =1 ± 2 n ² n Work: lim n →∞ n s ± 2 n ² n = lim n →∞ 2 n = 0 < 1 So the series converges absolutely. Example: Use the root test to determine if the series converges. X n =1 ± 5 n 2 + 1 3 n 2 ² n Work: lim n →∞ n s ± 5 n 2 + 1 3 n 2 ² n = lim n →∞ 5 n 2 + 1 3 n 2 = 5 3 > 1 So the series diverges . 1 This document was prepared by Ron Bannon ( [email protected] ) using L A T E X2 ε . Last revised January 10, 2009. 1
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2 Summary for Testing Series Before you come to the exam, I strongly suggest that you review the following listed items. I also think it might be appropriate to make a one page review sheet covering these items, and this should be prepared well in advance of any exam/quiz related to this material. Representative problems are provided, but we’ll only be doing a few of these in class. 1. Know the p -series! The p -series X n =1 n - p is convergent if p > 1, and is divergent if p 1. 2. Know the geometric series! The geometric series X n =1 ar n - 1 = a + ar + ar 2 + ··· is convergent if | r | < 1, and its sum is X n =1 ar n - 1 = a + ar + ar 2 + ··· = a 1 - r . If | r | ≥ 1, the geometric series is divergent. 3. Know the test for divergence! If the series X n =1 a n is convergent, then lim n →∞ a n = 0 . 4. Know the integral test! The Integral Test : Suppose f is a continuous, positive, decreasing function on [1 , ) and a n = f ( n ). Then the series X n =1 a n is convergent if and only if the improper integral Z 1 f ( x ) d x is convergent. In other words: (a) If Z 1 f ( x ) d x is convergent, then X n =1 a n is convergent. (b) If Z 1 f ( x ) d x is divergent, then X n =1 a n is divergent. 2
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5. Know the (limit) comparison tests! Comparison Test : Suppose that X n =1 a n and X n =1 b n are series with positive terms. (a) If X n =1 b n is convergent, and 0 < a n b n for all n , then X n =1 a n is also convergent. (b) If
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mth.122.handout.19 - MTH 122 Calculus II Essex County...

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