The sequence {n/(n + 1)} has a least upper bound of 1 Show that if M is a number less than 1, then the terms of {n/(n + 1)} eventually exceed M. That is, if M < 1 there is an integer N such that n/(n + 1) > M whenever n > N. Since n/(n + 1) < 1 for every n, this proves that 1 is a least upper bound for {n/(n + 1)}.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.1: Infinite Sequences And Summation Notation
Problem 72E
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The sequence {n/(n + 1)} has a least upper bound of 1
Show that if M is a number less than 1, then the terms of
{n/(n + 1)} eventually exceed M. That is, if M < 1 there is
an integer N such that n/(n + 1) > M whenever n > N. Since
n/(n + 1) < 1 for every n, this proves that 1 is a least upper
bound for {n/(n + 1)}.
Transcribed Image Text:The sequence {n/(n + 1)} has a least upper bound of 1 Show that if M is a number less than 1, then the terms of {n/(n + 1)} eventually exceed M. That is, if M < 1 there is an integer N such that n/(n + 1) > M whenever n > N. Since n/(n + 1) < 1 for every n, this proves that 1 is a least upper bound for {n/(n + 1)}.
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