142p23s11solution[1]

1 x 3 1 for x 3 2 that is 1 x 5 so series converges 2

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Unformatted text preview: st with a p-series, p > 1. iv) Diverges: Ratio test. v) Diverges: Comparison or limit comparison with the harmonic series. a) Answer: i) b) Answer: v) 3.6 Determine if the following series converge or diverge. Give your reasoning using complete sentences. a) ∞ n=1 b) (−1)n ∞ n=2 n n+2 (−1)n+1 1 ln n a) An alternating series, but the terms do not approach 0 (they approach 1) so divergent. b) An alternating series with terms decreasing to 0 so convergent. 3.7 a) Find the interval of convergence for the power series b) Find the radius of convergence for ∞ n! n x n=1 (2n)! ∞ 1 (x − 3)n n n=1 n2 MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 4 a) 1 (x − 3)n+1 (n+1)2n+1 1 (x − 3)n n2n = n2n (x − 3)n+1 1 n 1 =· |x − 3| → |x − 3| n+1 (x − 3)n (n + 1)2 2 n+1 2 as n → ∞. 1 |x − 3| < 1 for |x − 3| < 2, that is, 1 < x < 5 so series converges 2 if 1 < x < 5 and diverges if x > 5 or x < 1. At x = 5 the series becomes the 1 harmonic series n n which diverges, but at x = 1 it becomes the alternating 1 series n (−1)n n which converges. The interval of convergence is therefore [1, 5). b) (n+1)! n+1 x (2n+2)! n! xn (2n)! = n+1 |x| → 0 &...
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This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

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