Here is the question:
Find the supremum of $A=\left\{\frac{n}{n+1}\right\} n \in \mathbb{N^*}$ without using limits
What I am having trouble with is proving my finding is a supremum.
My attempt:
I think/suspected that the supremum is $1$ so I tried working as follows:
Elements of the sequence go as follows:$1/2,2/3,3/4,4/5$,etc...
Since the denumerator is always greater than the numerator then one can say $1$ is an upper bound, we can show it by putting $\frac{n}{n+1}>1$ which results in $0>1$ so this is impossible
Since we showed that $1$ is an upper bound, the first condition of it being a supremum is satisfied, we still need to show that there is not any other upper bound that is smaller than it, usually how this is done is by reaching a contradiction by using something like the Archimedean property but I am not able to reach that.
Assume that $1$ is not a supremum, this means that $\exists x \in \mathbb{R}$ s.t. $1>x\geq \frac{n}{n+1} \forall n \in \mathbb{N^*}$
so logically, we need to show that if $x<1$ then $\frac{n}{n+1}>x$
some draft thoughts:
to reach $\frac{n}{n+1}>x$, we need $n>x(n+1)$, here is where I get stuck. I tried doing some manipulation to see where I would be able to use the archimedean property and work things out backward but I am not able to. Any help would be appreciated.
Take $r\in(-\infty,1)$; I will prove that $r$ is not an upper bound of$$A=\left\{\frac n{n+1}\,\middle|\,n\in\Bbb N\right\}.$$If $r\leqslant0$, that's clear, since, in fact, the every element is greater than $r$. Otherwise, suppose that $r\in(0,1)$. Since $r<1$ and $r>0$, $\frac{1-r}r>0$. So, by the Archimedean property, there is some $n\in\Bbb N$ such that $\frac1n<\frac{1-r}r$. But\begin{align}\frac1n<\frac{1-r}r&\iff r<n-nr\\&\iff(n+1)r<n\\&\iff r<\frac n{n+1}.\end{align}But $\frac n{n+1}\in A$ and so this proves that $r$ is not an upper bound of $A$. Therefore, $1$ is the least upper bound.