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For wht values of p does converge

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 4E Chapter 7.7

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 4E

For wht values of p does converge?

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Problem 4EFor what values of p does convergeAnswer;Step-1; Definition ; The partial sum of the series is given by = + ++..............+ . If the sequence of these partial sums {Sn} converges to L, then the sum of the series converges to L. If {Sn} diverges, then the sum of the series diverges. = is convergent . Conversely , a series is divergent if the sequence of partial sums is divergent. If and are convergent series , then + ) and - ) are convergent . If C , then is convergent series.NOTE : The terms grow without bound , so the sequence does not converge. Integral test ; The series can be compared to an integral to establish convergence or divergence. Let f(n) = be a positive monotone decreasing function . If f(x) dx = f(x) dx < , then the series converges. But if the integral diverges, then the series does so as well.Step-2 Now , we have to evaluate for what values of p does dx converges. The P- integrals Consider the function f(x) = (where p> 0) for all xLooking at this function closely we see that f(x) presents an improper behaviour at 0 and only . In order to discuss convergence or divergence of dx . Now , we need to convert this integral as dx = dx + dx …………..(1) We have = And dx = dx For both limits, we need to evaluate the indefinite integral dx. We have two cases: if p=1, then we have = dx = ln(x) If p , then we have = +C , since Step-3 ; In order to decide on convergence or divergence of the above two improper integrals, we need to consider the cases : p < 1, p = 1 and p > 1. If p < 1 , then we have = = dx = = ( ) = , since as cthen And = = dx = = ( ) = , since as cthen = Step-4 If p = 1 , then we have = = dx = = ( ln(1) - ln(c)) = (-ln(c)) , since ln(1) = 0. = + And = = dx = = ( ln(c) - ln(1)) = (ln(c)) , since ln(1) = 0. = + Step-5 If p > 1 , then we have = = dx = = ( ) = , since as cthen= And = = dx = = ( ) = , since as cthen = Step-6; The P- test : regardless of the values of the number p , the improper integral dx is always divergent . Moreover , we have dx is convergent if and only if p < 1. dx is convergent if and only if p > 1.

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Chapter 7.7, Problem 4E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This full solution covers the following key subjects: converge, values, wht. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. The answer to “For wht values of p does converge?” is broken down into a number of easy to follow steps, and 7 words. The full step-by-step solution to problem: 4E from chapter: 7.7 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. Since the solution to 4E from 7.7 chapter was answered, more than 341 students have viewed the full step-by-step answer.

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For wht values of p does converge