My attempt:
Assume $\exists r \in \mathbb{R}$ s.t $1>r\geq\frac{n-1}{n+1} \forall n \in \mathbb{N}$
We know $1-r>0$, by the archimedean property we have $\exists n\in \mathbb{N}$ s.t.:
$$n(1-r)>r+1$$ $$n-rn>r+1$$ $$n-1>rn+r$$ $$\frac{n-1}{n+1}>r$$
so any number less than $1$ can't be an upper bound, $1$ must be the supremum of the set.
Any help in verifying or improving my proof or suggesting better ways to prove this would be appreciated.
On
We see that if $\frac{n-1}{n+1}>1$ and $n\in\mathbb Z^{+}$, then we get a contradiction: $n<-1.$
Thus we have,
$$0≤\frac{n-1}{n+1}≤1$$
Now, I will show that for any $0<\varepsilon<1$ there exist $n\in\mathbb Z^{+}$, such that
$$\frac{n-1}{n+1}>1-\varepsilon$$
Algebra tells us, we have
$$n>\frac 2\varepsilon-1,\thinspace\text {where}\thinspace \thinspace n\in\mathbb Z^{+}$$
To be more precise, we can take
$$n=\left\lfloor\frac 2\varepsilon\right\rfloor+1$$
Finally we conclude that, if $n=\left\lfloor\frac 2\varepsilon\right\rfloor+1$ then:
$$1-\varepsilon<\frac{n-1}{n+1}≤1,\thinspace \forall \varepsilon \in (0,1)$$
This means,
$$\sup\left\{\frac {n-1}{n+1},\thinspace n\in\mathbb Z^{+}\right\}=1.$$
$a_n:=\dfrac{n-1}{n+1}=1 -\dfrac{2}{n+1}\ge 0$, $n=1,2,..;$
$1$ is an upper bound.
Let $1> r>0$. Need to show that $1-r$ is not an upper bound of $a_n$, i.e. there is a $n \in \mathbb{Z^+}$ s.t.
$1>a_n >1-r$, or $\dfrac{2}{n+1} < r$;
Archimedian property:
There is a $n\in \mathbb{Z^+}$ s.t. $n >(2/r)-1$, and we are done.