Explanation Result used: ( 1) Ratio test: Let ∑ k = 1 ∞ a k be an infinite series and let r = lim k → ∞ | a k + 1 a k | . If r < 1 , then the series ∑ k = 1 ∞ a k is converges. (2) Every infinite sum of a nonzero constant diverges. (2) The p -series ∑ k = 1 ∞ 1 k p converges for p > 1 and diverges for p ≤ 1 . Given: The power series ∑ ( x + 2 ) k 2 k ln k . Calculation: Let a k = ( x + 2 ) k 2 k ln k . Then, a k + 1 = ( x + 2 ) k + 1 2 k + 1 ln ( k + 1 ) . Use the Ratio test to compute the value | a k + 1 a k | . | a k + 1 a k | = | ( x + 2 ) k + 1 2 k + 1 ln ( k + 1 ) ( x + 2 ) k 2 k ln k | Take lim k → ∞ on both sides, lim k → ∞ | a k + 1 a k | = lim k → ∞ | ( x + 2 ) k + 1 2 k + 1 ln ( k + 1 ) ( x + 2 ) k 2 k ln k | = lim k → ∞ | ( x + 2 ) k + 1 2 k + 1 ln ( k + 1 ) ⋅ 2 k ln k ( x + 2 ) k | = | x + 2 | 2 lim k → ∞ ( ln k ln ( k + 1 ) ) = | x + 2 | 2 lim k → ∞ ( 1 k 1 k + 1 ) ( by L'hospitals rule ) Simplify further, lim k → ∞

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ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter 9, Problem 23RE

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The radius of convergence and interval of convergence.

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The radius of convergence and interval of convergence.

Solution Summary: The author explains how to determine the radius of convergence, and the interval, using the Ratio test.

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The radius of convergence and interval of convergence.

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Chapter 9 Solutions