Find $\sup\left\{ \frac{m}{|m| + n }\right\} $ where $m \in \Bbb{Z}$ and $n \in \Bbb{N}$

Question

Im trying to do $\sup A $ and $\inf A$ if $A= \left\{ \frac{m}{|m| + n } : m \in \Bbb{Z} , n \in \Bbb{N}\right\} $ rigoriously.

my try:

Let $f(m,n) = \dfrac{m}{|m|+n} $. Notice that $f(1,1) = \dfrac{1}{2}$ and $f(-1,1) = \dfrac{-1}{2}$

If $m \geq 0$, then $f(m,n) = \dfrac{m}{m+n} < \dfrac{m}{m} = 1$

Next, if $m<0$, then $f(m,n) = \dfrac{m}{n-m} > \dfrac{m}{n} $

Claim: $\sup A = 1 $ and $\inf A = - \infty ?$

Proof: To prove $\sup A = 1$ we show no upper bound $( \dfrac{m}{|m|+n} \leq u )$ is less than $1$

if $1 > u$ then there is some integer $N>1$ such that $N(1-u) > 1 $ or equivalently

$$ 1-u > \dfrac{1}{N} \iff 1 - \dfrac{1}{N} > u \iff \dfrac{N-1}{N} > u \iff \dfrac{ N-1}{N-1 + 1 } > u$$

But with $m=N-1$ and $n=1$ we have found an element of $A$ that is greater than upper bound $u$. Thus $\boxed{ \sup A = 1 }$.

Now, how am I still unsure about the infimum? Any suggestion? Am I on the right track?

Answer 1

Infimum:

Note : The Denominator is $>0.$

Let $m <0$.

Then

$\inf ( \dfrac {m}{|m|+n} )= -\sup (\dfrac {-m}{|m| +n} )$.

Find $\sup ( \dfrac{|m|}{|m|+n}).$

$\dfrac{|m|}{|m|+n} <1$, i.e. $1$ is an upper bound.

Fix $n$:

$\lim_{|m| \rightarrow \infty}\dfrac{|m|}{|m|+n}=1.$

Hence $\sup_{|m|>0} (\dfrac{|m|}{|m|+n})=1.$

Can you finish?

Answer 2

The infimum is not $-\infty$. It is $-1$. Note that $-m \leq |m| \leq |m|+n$. This gives $\frac m {|m|+n|} \geq -1$. So $-1$ is lower bound. For showing that the infimum is excactly $-1$: consider the elements $\frac {-n^{2}} {|-n^{2}|+n}$ and let $n \to \infty$. [ Let $a >-1$. Then verify that $\frac {-n^{2}} {|-n^{2}|+n} <a$ if we choose $n >\frac {-t} {1+t} $].