Explanation Given: The series is ∑ k = 1 ∞ 1 sin h k . Result used: (1) Let ∑ a k and ∑ b k are the series with positive terms and lim x → ∞ a n b n = L , where L is a finite number. (i) If 0 < L < ∞ , then either both series ∑ a k and ∑ b k converges or both diverges. (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 = 0 ∞ a r k converges if | r | < 1 and diverges if | r | ≤ 1 . Calculation: Consider the series ∑ k = 1 ∞ 1 sin h k . 1 sin h k = 2 e k − e − k 2 e k − e − k < 1 e k Consider the series ∑ k = 1 ∞ b k = ∑ k = 1 ∞ 1 e k , which must be greater than. ∑ k = 1 ∞ a k = ∑ k = 1 ∞ 1 sin h k

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, Problem 40RE

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Whether the series ∑k=1∞1sinhk converges or diverges.

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Chapter 8, Problem 39RE

Chapter 8, Problem 41RE

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Calculus: Early Transcendentals (2nd Edition)

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Whether the series ∑k=1∞1sinhk converges or diverges.

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