Question
Uniqueness of least upper bounds Show that if $M_{1}$ and $M_{2}$are least upper bounds for the sequence $\left\{a_{n}\right\},$ then $M_{1}=M_{2}$ .That is, a sequence cannot have two different least upper bounds.
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Step 1: Assume that $M_{1}$ and $M_{2}$ are two different least upper bounds for the sequence $\left\{a_{n}\right\}$. Show more…
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Show that if $M_{1}$ and $M_{2}$ are least upper bounds for the sequence $\left\{a_{n}\right\},$ then $M_{1}=M_{2}$ That is, a sequence cannot have two different least upper bounds.
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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 $\{n /(n+1)\}$.
Infinite Sequences And Series
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 1 $\{n /(n+1)\}$ 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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