Explanation Given: The series is ∑ k = 2 ∞ a k = ∑ k = 2 ∞ 1 k ( ln k ) p , where p is a real number. 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 . (3) The series ∑ n = 2 ∞ 1 k ( ln k ) diverges. Chain rule: d [ f ( x ) ] n d x = n [ f ( x ) ] n − 1 f ′ ( x ) . Product rule: d d x [ f ( x ) g ( x ) ] = d f d x g ( x ) + f ( x ) d g d x . Proof: Compute the value of p if the given series converges. For p = 1 , The series ∑ n = 2 ∞ 1 k ( ln k ) p is diverges by the Result (3). For p ≠ 1 , Consider the function from given series f ( x ) = 1 x ( ln x ) p . The derivative of the function is obtained as follows, f ′ ( x ) = d d x ( 1 x ( ln x ) p ) = d d x ( x − 1 ( ln x ) − p ) Apply the product rule stated above, f ′ ( x ) = ( x − 1 ) d d x ( ( ln x ) − p ) + ( ln x ) − p ⋅ d d x ( x − 1 ) = 1 x ⋅ d d x ( ( ln x ) − p ) + 1 ( ln x ) p ⋅ ( ( − 1 ) x − 1 − 1 ) = 1 x ⋅ d d x ( ( ln x ) − p ) + 1 ( ln x ) p ⋅ ( − x − 2 ) = 1 x ⋅ d d x ( ( ln x ) − p ) − x − 2 ( ln x ) p Apply the chain rule stated above, f ′ ( x ) = 1 x [ ( − p ) ( ln x ) − p − 1 ⋅ d d x ( ln x ) ] − 1 x 2 ( ln x ) p = 1 x [ ( − p ) ( ln x ) − ( p + 1 ) ⋅ 1 x ] − 1 x 2 ( ln x ) p = − p x 2 ( ln x ) p + 1 − 1 x 2 ( ln x ) p = − ( p x 2 ( ln x ) p + 1 + ln x x 2 ( ln x ) p ( ln x ) ) Simplify further and obtain the derivative, f ′ ( x ) = − ( p x 2 ( ln x ) p + 1 + ln x x 2 ( ln x ) p + 1 ) = − p + ln x x 2 ( ln x ) p + 1 Since f ′ ( x ) < 0 for x > e − p , the given function is decreasing by Result (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, Problem 60RE

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Chapter 8, Problem 59RE

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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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The series ∑ k = 2 ∞ 1 k ( ln k ) p converges provided p > 1 .

Solution Summary: The author explains how the series displaystyle 'underset' converges, provided p>1, where p is a real number.

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To show: The series ∑k=2∞1k(lnk)p converges provided p>1.

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Chapter 8 Solutions

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The series ∑ k = 2 ∞ 1 k ( ln k ) p converges provided p &gt; 1 .

Solution Summary: The author explains how the series displaystyle 'underset' converges, provided p>1, where p is a real number.

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To show: The series ∑k=2∞1k(lnk)p converges provided p>1.

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Chapter 8 Solutions

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The series ∑ k = 2 ∞ 1 k ( ln k ) p converges provided p > 1 .