Explanation Given: The series is ∑ k = 1 ∞ 1 k 3 − k + 1 . Result used: (1) “Let ∑ a k and ∑ b k are the series with positive terms and let lim n → ∞ a n b n = L , where L is a finite number. (i) If 0 < L < ∞ , then ∑ a k and ∑ b k either both series converge or both diverge.” (ii) If L = 0 and ∑ b k converges, then ∑ a k converges. (iii) If L = ∞ and ∑ b k diverges, then ∑ a k diverges”. (2) The p -series ∑ k = 1 ∞ 1 k p is convergent if p > 1 . Calculation: Consider k 3 + 1 < k 3 + k for all k ≤ 1 . k 3 − k + 1 < k 3 1 k 3 < 1 k 3 − k + 1 1 k 3 < 1 k 3 − k + 1 ∑ k = 1 ∞ 1 k 3 − k + 1 > ∑ k = 1 ∞ 1 k 3 Consider the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ 1 k 3 , which must be smaller than ∑ k = 1 ∞ a k = ∑ k = 1 ∞ 1 k 3 − k + 1

Calculus, Single Variable: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134766850

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

Publisher: PEARSON

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Chapter 10.5, Problem 47E

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Whether the series ∑k=1∞1k3−k+1 converges or diverges.

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Chapter 10.5, Problem 46E

Chapter 10.5, Problem 48E

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Calculus, Single Variable: Early Transcendentals (3rd Edition)

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Whether the series ∑ k = 1 ∞ 1 k 3 − k + 1 converges or diverges.

Solution Summary: The author analyzes whether the series displaystyle 'underset' is convergent or divergent.

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Whether the series ∑k=1∞1k3−k+1 converges or diverges.

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