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Explain why or why not Determine whether the | Ch 8 - 1RE

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

Solution for problem 1RE Chapter 8

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 1RE

Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. The terms of the sequence {an}increase in magnitude, so the limit of the sequence does not exist.

b. The terms of the series  approach zero, so the series converges.

c. The terms of the sequence of partial sums of the series ∑ak approach 5/2, so the infinite series converges to 5/2.

d. An alternating series that converges absolutely must converge conditionally.

Step-by-Step Solution:
Step 1 of 3

STEP_BY_STEP SOLUTIONStep-1 Definition ; A series is said to be convergent if it approaches some limit .Formally , the infinite series a n is convergent if the sequence of partial sums n=1 n S n a k is convergent . Conversely , a series is divergent if the k=1sequence of partial sums is divergent. If U and V are convergent series , then k k( U +kV k ) and ( U - Vk k ) are convergent . If C = / 0 , then U k and V kboth converge or both diverge . Step-2 a) Now , we have to check the result ‘The terms of the sequence {a } incneases in magnitude , so the limit of the sequence does not exist ‘ is true (or) false. For example ; a = 1 - 1 . n n 1 Clearly as n , then the value of n 0 So , we can say that {a } increases in magnitude , and lim ( 1 - 1 ) n n n = lim (1) - lim 1 n n n = 1 - 0 = 1, since as n , 1 0 . n Hence , the limit of the sequence lim {a } = lim ( 1 - 1) = 1 n n n n Therefore , the given statement is false. Therefore , The terms of the sequence {a } nncreases in magnitude , so the limit of thesequence does not exist is false. Step-3\nb) Now , we have to check the result ‘ The terms of the series (1/ k ) approach zero ,so the series converges ‘ is true (or) false. We , know the theorem If a series converges then the limit of the sequence is zero.In this theorem the sequence approaches to zero is necessary but not sufficient forconvergence of the series. So , the above statement is false .Therefore , The terms of the series (1/ k ) approach zero , so the series converges ‘ isfalse. Step-4c) Now , we have to check ‘ The terms of the sequence of partial sums of the series (a k approach to (5/2) , so the infinite series converges to (5/2) ‘ is true (or) false .This is the definition of convergence , since from step(1). Therefore , the above statement is true. Therefore , The terms of the sequence of partial sums of the series (a ) kapproach to (5/2) , so the infinite series converges to (5/2) is true Step-5d) Now , we have to check ‘ An alternating series that converges absolutely must convergeconditionally’ is true (or) false . From the definition , it is clear that ‘If a series converges absolutely , it does notconverge conditionally. Therefore , the given statement is false .\n Therefore , An alternating series that converges absolutely must convergeconditionally’ is false.

Step 2 of 3

Chapter 8, Problem 1RE is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Explain why or why not Determine whether the | Ch 8 - 1RE