Explanation Theorem used: Convergence of the p-series: The p -series ∑ k = 1 ∞ 1 k p converges for p > 1 and diverges for p ≤ 1 . Limit Comparison test: Let ∑ a k and ∑ b k be series with positive terms and let lim k → ∞ a k b k = L . 1. If 0 < L < ∞ ( L is a finite positive number), then ∑ a k and ∑ b k either both converge or both diverge. 2. If L = 0 and ∑ b k converges, then ∑ a k converges. 3. If L = ∞ and ∑ b k diverges, then ∑ a k diverges. Calculation: Simplify the series ∑ k = 2 ∞ k + 1 − k k 2 − k as follows. ∑ k = 2 ∞ k + 1 − k k 2 − k = ∑ k = 2 ∞ ( k + 1 − k k 2 − k ⋅ k + 1 + k k + 1 + k ) = ∑ k = 2 ∞ ( ( k + 1 ) 2 − ( k ) 2 ( k 2 − k ) ( k + 1 + k ) ) [ ∵ ( a + b ) ( a − b ) = a 2 − b 2 ] = ∑ k = 2 ∞ ( k + 1 − k ( k 2 − k ) ( k + 1 + k ) ) = ∑ k = 2 ∞ ( 1 ( k 2 − k ) ( k + 1 + k ) ) Consider the series ∑ k = 2 ∞ b k = ∑ k = 2 ∞ 1 k 3 / 2 . By the p -series, as the value of p = 3 2 > 1 , the series ∑ k = 2 ∞ b k converges

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 10.8, Problem 68E

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Whether the series ∑k=2∞k+1−kk2−k converge or not.

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Chapter 10.8, Problem 67E

Chapter 10.8, Problem 69E

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Chapter 10 Solutions

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

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Whether the series ∑ k = 2 ∞ k + 1 − k k 2 − k converge or not.

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Whether the series ∑k=2∞k+1−kk2−k converge or not.

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Chapter 10 Solutions