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MATH 101, Section 212 (CSP) Week 10: Marked Homework Solutions University of British Columbia - MATH 101

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MATH 101, Section 212 (CSP) Week 10: Marked Homework Solutions 2011 Mar 24 1. (a) [5] 1 − 13 + 15 − 1 7 + · · · = P1 n=1(−1)n−1 2n1−1 is an alternating series. The series of its absol ... ute values P1 n=1 (−1)n−1 2n1−1 = P1 n=1 2n1−1 = 1 + 13 + 1 5 + 17 + : : : is divergent by the Integral Test: f(x) = 2x1−1 is continuous for x > 1 2, positive for x > 1 2, and f0(x) = −(2x−2 1)2 < 0 so f(x) is decreasing for x > 12. Therefore f is continuous, positive and decreasing on [1; 1). The improper integral Z11 2x1− 1 dx = tlim !1 Z1t(2x − 1)−1 dx (substitution u = 2x − 1, du = 2 dx) = lim t!1 12 Z1 2t−1 u−1 du = 12 lim t!1 ln juj]2 1t−1 = 12 lim t!1 ln(2t − 1) = 1 is divergent. However, the series itself is convergent: bn = janj = 2n1−1 satisfies bn ≥ bn+1 for all n = 1; 2; 3; : : : (2n1−1 ≥ 2(n+1) 1 −1 iff 2(n + 1) − 1 ≥ 2n − 1 iff 2n + 1 ≥ 2n − 1 iff 1 ≥ −1 which is true; alternatively, show that f(x) = 2x1−1 is decreasing by showing that f0(x) is negative) and limn!1 bn = 0 (limn!1 2n1−1 = limn!1 2−1(1 =n=n) = 0), so by the Alternating Series Test P1 n=1(−1)n−1 2n1−1 is convergent. The series is convergent but the series of its absolute values is divergent, so the series 1 − 1 3 + 15 − 1 7 + : : : is conditionally convergent. (b) [3] P1 j=2 j((−lnj 1))j3 = 2 (ln12)3 − 3 (ln13)3 + 4 (ln14)3 − 5 (ln15)3 + : : : is an alternating series. The series of its absolute values P1 j=2 j((−lnj 1))j3 = P1 j=2 j (lnj 1 )3 is convergent, by the Integral Test (see Week 9: Marked Homework Assignment, question 1(b)). Therefore the series P1 j=2 j((−lnj 1))j3 is absolutely convergent (which implies that it is convergent, so we don’t need the to apply the Alternating Series Test in this case). (c) [3] limk!1 ak = limk!1(−1)k does not exist (fakg is the sequence f1; −1; 1; −1; 1; −1; : : :g which does not get close to any unique finite number for all sufficiently large k) so by the Test for Divergence, the series Pk=0(−1)k is divergent. (d) [3] P1 ‘=0(−1)‘ 100 (2‘+1)! 2‘+1 = 100 − 100 3!3 + 100 5!5 − 100 7!7 + : : : is an alternating series. We use the Ratio Test: lim ‘!1 ja‘+1j ja‘j = ‘lim !1 1002(‘+1)+1 (2(‘+1)+1)! 1002‘+1 (2‘+1)! = lim ‘!1 1002‘+3 1002‘+1 (2‘ + 1)! (2‘ + 3)! = lim ‘!1 1002 (2‘ + 2)(2‘ + 3) = ‘lim !1 1002 ‘2 2 + 2‘ 2 + 3‘ = 0 [Show More]

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