MA104 week 12 report solutions - MA 104 Week 12 Report Absolute\/Conditional Convergence Power Series Name Lab 1 P Winter 2006 n 2 7 n=1(a Determine

MA104 week 12 report solutions - MA 104 Week 12 Report...

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MA 104 Week 12 Report & Absolute/Conditional Convergence; Power Series Name: Lab: Winter 2006 1. [12 marks ] Consider the series 1 P n =1 3 n +1 2 7 ± n . (a) Determine whether the series converges absolutely, conditionally, or neither (i.e. diverges). Using the Ratio Test: lim n !1 ² ² ² ² a n +1 a n ² ² ² ² = lim n !1 ² ² ² ² ² ² ² ² ² 3 n +2 2 7 ± n +1 3 n +1 2 7 ± n ² ² ² ² ² ² ² ² ² = lim n !1 ² ² ² ² 3 ± 2 7 ±² ² ² ² = 6 7 ) The series 1 P n =1 3 n +1 2 7 ± n converges absolutely since lim n !1 ² ² ² ² a n +1 a n ² ² ² ² = 6 7 < 1 (b) Determine the sum of the series, s = 1 P n =1 3 n +1 2 7 ± n . 1 P n =1 a n = 1 P n =1 3 n +1 2 7 ± n = 1 P n =1 ³ 3 2 )(3 n 1 ´ ± 2 7 ± 1 2 7 ± n 1 = 1 P n =1 18 7 6 7 ± n 1 ) 1 P n =1 a n is a geometric series with a = 18 7 and j r j = 6 7 < 1 ) s = 1 P n =1 3 n +1 2 7 ± n = 1 P n =1 18 7 6 7 ± n 1 = 18 7 1 ³ 6 7 ´ = 18 13 (c) Since 1 P n =1 3 n +1 2 7 ± n is absolutely convergent with sum s , any rearrangement of the terms in the sum of the series will also have sum s: (i) Determine s + , the sum of the positive terms of the series. s + = 108 49 + 3888 2401 + 139968 117649 ±±± = 1 P n =1 108 49 36 49 ± n 1 which is geometric with a = 108 49 and r = 36 49 ) s + = 108 49 1 ³ 36 49 ´ = 108 49 ± 49 13 = 108

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