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math1B-Fall_06-mt2-Tataru-soln

# math1B-Fall_06-mt2-Tataru-soln - Mathematics 1B Fall...

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Unformatted text preview: Mathematics 1B. Fall Semester 2006 Professor: Daniel Tataru Midterm 2 Solutions 1 (20) . Determine the interval of convergence of the following series. Do they converge at endpoints ? a ) ∞ X n =1 ( x- 1) 2 n √ n 4 n Solution: Using the ratio test we compute lim n →∞ ( x- 1) 2( n +1) √ n + 1 4 n +1 ( x- 1) 2 n √ n 4 n = lim n →∞ ( x- 1) 2 4 √ n √ n + 1 = ( x- 1) 2 4 The limit is less than 1 if | x- 1 | < 2. Hence the radius of convergence is R = 2. At the endpoints we have x- 1 = ± 2 and the series becomes ∞ X n =1 1 √ n which is a divergent p-series. Thus the interval of convergence is (- 1 , 3). b ) ∞ X n =2 ln n + 1 n- 1 x n Solution: Using the Taylor expansion for ln(1 + x ) we write ln n + 1 n- 1 = ln 1 + 2 n- 1 = 2 n- 1- 2 ( n- 1) 2 + ··· Then for the ratio test we compute lim n →∞ ln n + 2 n x n +1 ln n + 1 n- 1 x n = x lim n →∞ 2 n- 2 n 2 + ··· 2 n- 1- 2 ( n- 1) 2 + ··· = x The limit is less than 1 if | x | < 1. Hence the radius of convergence is R = 1. At the endpoint x =- 1 we obtain the alternating series ∞ X n =2 (- 1) n ln 1 + 2 n- 1 Due to the expansion above we have ln n + 1 n- 1 & 0 as n → ∞ therefore the series converges by the alternating test. At the endpoint x = 1 we obtain the series ∞ X n =2 ln 1 + 2 n- 1 Due to the expansion above this is comparable to the harmonic series ∞ X n =2...
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math1B-Fall_06-mt2-Tataru-soln - Mathematics 1B Fall...

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