Explanation Given: The series is ∑ k = 2 ∞ a k = ∑ k = 2 ∞ 4 k ( ln k ) 2 . Result used: (1) “Suppose f is a continuous, positive, decreasing function, for x ≥ 1 , and let a k = f ( k ) for k = 1 , 2 , 3 , ⋯ , then ∑ k = 1 ∞ a k and ∫ 1 ∞ f ( x ) d x either both converge or both diverge. (2) The function f ( x ) is decreasing function if f ′ ( x ) < 0 . Chain rule: d [ f ( x ) ] n d x = n [ f ( x ) ] n − 1 f ′ ( x ) Definition used: The improper integral ∫ a b f ( x ) d x is convergent if the corresponding limit exists. Calculation: Consider the function from given series 4 k ( ln k ) 2 = 4 ( k − 1 ( ln k ) − 2 ) . The derivative of the function f ( x ) = 4 ( x ( ln x ) − 2 ) is obtained as follows, f ′ ( x ) = 4 [ ( x − 1 ) d d x ( ( ln x ) − 2 ) + ( ln x ) − 2 ⋅ d d x ( x − 1 ) ] = 4 ( 1 x ⋅ ( − 2 ( ln x ) − 3 d d x ( ln x ) ) + 1 ( ln x ) 2 ⋅ ( ( − 1 ) x − 1 − 1 ) ) = 4 ( − 2 x ( ln x ) 3 ⋅ 1 x + 1 ( ln x ) 2 ⋅ ( − x − 2 ) ) = − 4 ( 2 + ln x ) x 2 ( ln x ) 3 Clearly, the function f ′ ( x ) is continuous, positive and decreasing for x ≥ 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 8.4, Problem 57E

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Whether the series convergence or divergence of the series or state that the test does not apply.

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Chapter 8.4, Problem 56E

Chapter 8.4, Problem 58E

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Chapter 8 Solutions

Calculus: Early Transcendentals (2nd Edition)

Ch. 8.1 - Define sequence and give an example.Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Given the series k=1k, evaluate the first four...Ch. 8.1 - The terms of a sequence of partial sums are...Ch. 8.1 - Consider the infinite series k=11k. Evaluate the...Ch. 8.1 - Explicit formulas Write the first four terms of...Ch. 8.1 - Explicit formulas Write the first four terms of...

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Whether the series convergence or divergence of the series or state that the test does not apply.

Solution Summary: The author determines whether the series convergence or divergence or state that the test does not apply. The improper integral is convergent if the corresponding limit exists.

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Whether the series convergence or divergence of the series or state that the test does not apply.

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Chapter 8 Solutions