Explanation Given: The series ∑ k = 1 ∞ x 2 k k 2 . Result used: (1) Let the series ∑ a k of non-negative terms and r = lim k → ∞ | a k + 1 a k | (i) If 0 ≤ r < 1 , the series converges. (ii) If r > 1 ( including r = ∞ ) , the series diverges. (iii) If r = 1 , the test is inconclusive”. (2) The p -series ∑ k = 1 ∞ 1 k p converges if p > 1 and diverges if p ≤ 1 . Calculation: The given series ∑ k = 1 ∞ x 2 k k 2 . Here, the k th term is, a k = x 2 k k 2 . Thus, the ( k + 1 ) th term is, a k + 1 = x 2 ( k + 1 ) ( k + 1 ) 2

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 8.5, Problem 86E

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The value of x≥0 for which the series converges.

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Chapter 8.5, Problem 85E

Chapter 8.5, Problem 87E

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Chapter 8 Solutions

Calculus: Early Transcendentals (2nd Edition)

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The value of x ≥ 0 for which the series converges.

Solution Summary: The author explains that the series converges for xin left[0,1right].

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The value of x≥0 for which the series converges.

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