Answer to Question #103928 in Calculus for BIVEK SAH

Question #103928
Use the order completeness property to show that the set
S={n/(n+1): n∈N}

has a
supremum and infimum.
1
Expert's answer
2020-03-06T11:17:57-0500

1 STEP: Search for the supremum of a given sequence



nn+1=(n+1)1n+1=11n+1\frac{n}{n+1}=\frac{(n+1)-1}{n+1}=1-\frac{1}{n+1}



Conclusion,


Sn=11n+1<1,nNsup(Sn)=1S_n=1-\frac{1}{n+1}<1,\quad\forall n\in\mathbb{N}\\[0.5cm] \boxed{\sup(S_n)=1}



2 STEP: Search for the infimum of a given sequence



1+nn+n,nN11+n1n+nnn+1nn+n=121+n\le n+n,\forall n\in\mathbb{N}\longrightarrow\\[0.5cm] \frac{1}{1+n}\ge\frac{1}{n+n}\longrightarrow\frac{n}{n+1}\ge\frac{n}{n+n}=\frac{1}{2}

Conclusion,


Sn12,nNinf(Sn)=12S_n\ge\frac{1}{2},\quad\forall n\in\mathbb{N}\\[0.5cm] \boxed{\inf(S_n)=\frac{1}{2}}


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