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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

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

a. The terms of the sequence \(\left\{a_{n}\right\}\) increase in magnitude, so the limit of the sequence does not exist.

b. The terms of the series \(\sum 1 / \sqrt{k}\) approach zero, so the series converges.

c. The terms of the sequence of partial sums of the series \(\sum a_{k}\) 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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