asst7soln

# asst7soln - MATH 138 Assignment 7 Solutions Problem 1 Find...

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Unformatted text preview: MATH 138 Assignment 7 Solutions Problem 1: Find the sums of the following series. (a) ∞ X n =0 π n 2 3 n- 1 Solution: ∞ X n =0 π n 2 3 n- 1 = ∞ X n =0 2 π n (2 3 ) n = ∞ X n =0 2 π 8 n Note: Geometric Series with a = 2, | r | < 1 = 2 1- π 8 = 16 8- π (b) ∞ X n =0 3 + 2 n 3 n +2 Solution: ∞ X n =0 3 + 2 n 3 n +2 = ∞ X n =0 3 3 n +2 + ∞ X n =0 2 n 3 n +2 ∞ X n =0 3 3 n +2 = ∞ X n =0 1 3 1 3 n Note: Geometric Series with a = 1 3 , | r | < 1 = 1 3 1- 1 3 = 1 3 2 3 = 1 2 ∞ X n =0 2 n 3 n +2 = ∞ X n =0 1 9 2 3 n Note: Geometric Series with a = 1 9 , | r | < 1 = 1 9 1- 2 3 = 1 9 1 3 = 1 3 ∞ X n =0 3 + 2 n 3 n +2 = 1 2 + 1 3 = 5 6 Problem 2a: Use the integral test to show that the series ∞ X n =1 ln n n 2 converges. Solution: Let f ( x ) = ln x x 2 . For x ≥ 1, f is continuous and f ( x ) ≥ 0. f ( x ) = 1 x · 1 x 2 + ln x ·- 2 x 3 = 1- 2 ln x x 3 f ( x ) < 0 for x ≥ 2. We can apply the integral test for the series ∞ X n =2 ln n n 2 . Z ∞ 2 ln x x 2 dx Let u = ln x dv = 1 x 2 dx du = 1 x dx v =- 1 x = lim t →∞- ln x x t 2- Z t 2- 1 x 2 dx = lim t →∞- ln t t + ln 2 2- 1 x t 2 = lim t →∞- 2 ln t t + ln 2 2 + 1 2 = ln 2 2 + 1 2 + lim t →∞- 2 t 1 by L’Hopital’s Rule = ln 2 2 + 1 2 Since R ∞ 2 ln x x 2 dx converges, so does ∞ X n =2 ln n n 2 . We can then add the case when n = 1. Thus, ∞ X n =1 ln n n 2 converges. Problem 2b: If s 50 is the partial sum of the above series and s is the full sum of this series, use the remainder estimate for the integral test to show that 9 / 100 ≤ s- s 50 ≤ 10 / 100 . Does this series feel like it is converging fast or slowly? This is a matter of opinion and so there is no wrong answer....
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asst7soln - MATH 138 Assignment 7 Solutions Problem 1 Find...

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