Math.112HW9Solns.(2008-9,Spring)

Math.112HW9Solns.(2008-9,Spring) - Solutions of Homework 9...

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Unformatted text preview: Solutions of Homework 9 Question 1. Let g be a decreasing differentiable function on (100 , ∞ ) . Assume that lim x →∞ x 3 g ( x ) = 0 and lim x →∞ g ( x ) = 1 . Show that the series X ( 1- g ( n ) ) is convergent. Solution: As g is decreasing and as lim x →∞ g ( x ) = 1 , it follows that g ( n )- 1 > 0 eventually. Let a n = g ( n )- 1 and b n = 1 n 2 > . Then, lim n →∞ a n b n = lim x →∞ g ( x )- 1 1 x 2 = |{z} / , L Hosp . lim x →∞ g ( x )- 2 x 3 = lim x →∞ x 3 g ( x )- 2 = 0 . As X b n = X 1 n 2 converges, it follows from the Limit Comparison Test that X ( 1- g ( n ) ) =- X a n converges. Question 2. For which values of p is the series X (- 1) n sin 1 n p absolutely con- vergent, conditionally convergent, divergent? Solution: Let a n = (- 1) n sin 1 n p If p ≤ 0 then a n does not go to 0 so that by the n th-term test the series diverges. Assume now that p > . Note that n p → ∞ and so sin 1 n p ≥ 0 eventually. Let b n = 1 n p ....
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This note was uploaded on 10/29/2010 for the course MATH 112 taught by Professor Ahmetguloglu during the Spring '10 term at Bilkent University.

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Math.112HW9Solns.(2008-9,Spring) - Solutions of Homework 9...

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