Finding the $\max$ and $\min$ of $\left\{\frac{m+n}{m+2n}\colon n,m \in \mathbb N^*\right\}$ Let's say I have the set $S = \left\{\dfrac{m+n}{m+2n}\colon n,m \in \mathbb N^*\right\}$.
I can then say that: 
$$\frac{m+n}{m+2n} = \frac{m+2n - n}{m+2n} = 1- \frac{n}{m+2n} \tag 1$$
Is there any way I can simplify any more? I know that I could just say that, to find the max of my set, I have to find the min of $\dfrac{n}{m+2n}$, but I'm not sure as to how I should go about doing this. Should I just say that $\dfrac{n}{m+2n}$ is going to zero as $m$ goes to infinity, and thus the max of $S$ doesn't exist, but it does have a supremum of $0$? Is that right?
Similarly, I need to find the max of $\dfrac{n}{m+2n}$ to find the min of $S$, which I don't know how to do at all. 
Any help is appreciated, thank you!
 A: The set $S$ does not have a maximum, nor a minimum. Below are two different approaches to prove this fact.

Approach 1.
Suppose that $S$ has a maximum $M$. Since, by definition, the maximum belongs to $S$, one would have $M=\dfrac{m+n}{m+2n}$, for some $m,n$ in $\mathbb N^*$ and $\forall s\in S(s\leq M)$. However $\dfrac{(m+1)+n}{(m+1)+2n}$ is larger than $M$ (it's enough to use $(1)$ to see this) and it also belongs to $S$. A contradiction.
Similarly, suppose that $S$ had a minimum, which due to $(1)$ would be $1-\dfrac{n}{1+2n}$, for some $n$ in $\mathbb N^*$. But $1-\dfrac{n+1}{1+2(n+1)}$ is even smaller and still in $S$. A contradiction too.
Therefore $S$ has no maxima, nor minima.
Approach 2.
When the maximum and minimum exist, they are respectively equal to the supremum and infimum.
The supremum of $S$ is $1$ and if $1$ was of the form $1-\dfrac{n}{m+2n}$, for some $m,n$ in $\mathbb N^*$ one would need to have $n=0$, which isn't allowed.
The infimum of $S$ is $\dfrac 12$. But
$$
\begin{align}
\dfrac 1 2 \in S&\implies 
\exists m,n\in \mathbb N^*\left(\dfrac{1}{2}=1-\dfrac{n}{m+2n}\right)\\
&\implies
\exists m,n\in \mathbb N^*\left(\dfrac{1}{2}=\dfrac{n}{m+2n}\right)\\
&\implies \exists m,n\in \mathbb N^*\left(m+2n=2n\right)\\
&\implies \exists m,n\in\mathbb N^*(m=0)_.
\end{align}
$$

Edit:
To prove that $\dfrac 1 2$ is the infimum of $S$, one can use the fact (perhaps definition?) which is probably already known to you:

$$\inf(S)=\dfrac 1 2\iff \mathop{\forall}_{\varepsilon >0}\mathop{\exists}_{s\in S}\left(s<\dfrac12+\varepsilon\right)_.$$

Let $\varepsilon$ be a positive number, let $n$ be any positive natural number larger than $\dfrac{1}{4}\left(\dfrac{1}{\varepsilon}-2\right)$, let $m=1$ and let $s=\dfrac{m+n}{m+2n}$. The goal is to prove that $s<\dfrac12+\varepsilon$.
Note that
$$
\begin{align}
n>\dfrac{1}{4}\left(\dfrac{1}{\varepsilon}-2\right)&\implies \dfrac{1}{\varepsilon}<4n+2\\
&\implies \dfrac{1}{4n+2}<\varepsilon\\
&\implies \dfrac{2(n+1)-(2n+1)}{4n+2}<\varepsilon\\
&\implies \dfrac{n+1}{2n+1}-\dfrac 12<\varepsilon\\
&\implies \underbrace{\dfrac{m+n}{m+2n}}_{=\large s}<\varepsilon+\dfrac 1 2
\end{align}
$$
Due to how $n$ was chosen and modus ponens, it follows that $\frac 1 2$ is the infimum of $S$.
It's natural to ask why did I chose $n$ the way I did. I picked such an $n$ because I followed a similar thought process to that of Rob, but went a little deeper into the limits to find a useful $n$.
A: Since $m,n\in\mathbb{N}^\ast$, we have
$$
\frac{m+n}{m+2n}-1=-\frac{n}{m+2n}\lt0\tag1
$$
and
$$
\frac{m+n}{m+2n}-\frac12=\frac{m}{2m+4n}\gt0\tag2
$$
$(1)$ and $(2)$ show that
$$
\frac12\lt\frac{m+n}{m+2n}\lt1\tag3
$$
Note also that
$$
\begin{align}
\lim_{m\to\infty}\left(\frac{m+n}{m+2n}-1\right)
&=\lim_{m\to\infty}-\frac{n}{m+2n}\\
&=\lim_{m\to\infty}-\frac{\frac nm}{1+2\frac nm}\\
&=-\frac0{1+0}\\[9pt]
&=0\tag4
\end{align}
$$
and
$$
\begin{align}
\lim_{n\to\infty}\left(\frac{m+n}{m+2n}-\frac12\right)
&=\lim_{n\to\infty}\frac{m}{2m+4n}\\
&=\lim_{n\to\infty}\frac{\frac mn}{2\frac mn+4}\\
&=-\frac0{2\cdot0+4}\\[9pt]
&=0\tag5
\end{align}
$$
$(4)$ and $(5)$ show that $\frac{m+n}{m+2n}$ can come as close to $\frac12$ and $1$ as we wish.
This means that $\frac{m+n}{m+2n}$ has no min or max on $\mathbb{N}^\ast$, but it does have an infimum of $\frac12$ and a supremum of $1$ on $\mathbb{N}^\ast$.
