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

See similar textbooks

Videos

Question

Chapter 8, Problem 60RE

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8, Problem 59RE

Chapter 8, Problem 61RE

Students have asked these similar questions

Logarithmic p-series a. Show that the improper integral dx (p a positive constant) x (In x)" converges if and only if p > 1. b. What implications does the fact in part (a) have for the con- vergence of the series n (In n)P n= Give reasons for your answer.

Time series

Match each function with its power series representation. Functions: Power Series: Function 1: 2 1 f(x) = 2+ 7x Power series (29): 247 for lel < n=0 Function 2: Power series (13): g(x) = xe. El-1)" (; Power series (83): Function 3: 1 h(x) = r*e=*, E(-1)"; n+1 n=0 Power series (37): Function 4: 1 E(-1)"; p(x) = a² In(1 + x*) for |æ| < 1. 2n +1 Power series (7): 3" Function 5: n! q(x) = Vr tan¬(VT) for |x| < 1.

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...

Additional Math Textbook Solutions

Find more solutions based on key concepts

The equation relating the cost c to the number of yards n for each material.

Glencoe Math Accelerated, Student Edition

The intercepts of the equation 9 x 2 +4y=36 are ______. (pp.18-19)

Precalculus Enhanced with Graphing Utilities (7th Edition)

In Exercises 5–12, find and sketch the domain for each function. 11.

University Calculus: Early Transcendentals (3rd Edition)

In Exercises 9–18, write an iterated integral for over the described region R using (a) vertical cross-section...

Thomas' Calculus: Early Transcendentals (14th Edition)

Precalculus Enhanced with Graphing Utilities

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises ...

University Calculus: Early Transcendentals (4th Edition)

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

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.

Videos

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

Want to see the full answer?

Chapter 8 Solutions

Additional Math Textbook Solutions

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.

Videos

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

Want to see the full answer?

Chapter 8 Solutions

Additional Math Textbook Solutions

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