Let be $A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$ prove there exists $\sup A$ and find said supreme.

Question

$A=\left\{\frac{n}{n+1}: n \in \mathbb{N}\right\} \subset \mathbb{R}$

I did: How $1\geq n$ because $n\in\mathbb{N}$

then $n+1\geq n \rightarrow \frac{n+1}{n}\geq1 \rightarrow \frac{n}{n+1}\leq1$

then $\forall n\in\mathbb{N} \frac{n}{n+1}\leq1 $ then $\sup A = 1$ is right?

Answer 1

Yes it is. Let’s prove it. We already know that $1$ is an upper bound of $A$. We need to show it’s the least upper bound. Thus for any $\epsilon > 0$ we need to find an $n$ such that : $\dfrac{n}{n+1} > 1 - \epsilon \implies n > \dfrac{1}{\epsilon} - 1$. Those natural $n$ satisfies this inequality does it. This proves that $1$ is the least upper bound for $A$.

Answer 2

With your justification, you get that sup $A$ exists and is $\leqslant 1$, but you don't get the equality.

To get it, you just need to remark that

$$\frac{n}{n+1} \rightarrow 1 \text{ when } n\rightarrow +\infty $$

And since as you said $\frac{n}{n+1}\leqslant 1$ for all $n$, you have that :

$$\sup A = 1$$

Answer 3

No, it is not correct. First of all, if $n\in\Bbb N$, you have $n\geqslant1$, instead of $1\geqslant n$. And, from the fact that $(\forall n\in\Bbb N):\frac n{n+1}\leqslant1$, you get that $A$ is bounded above and that $\sup A\leqslant1$. But you still have to prove that $\sup A=1$. That's not hard, in fact: if $a<1$, then, by the Archimedian property, there is some $n\in\Bbb N$ such that $n+1>\frac1{1-a}$. But\begin{align}n+1>\frac1{1-a}&\iff 1-a>\frac1{n+1}\\&\iff a<1-\frac1{n+1}=\frac n{n+1}.\end{align}So, $a$ is not an upper bound of $A$ and therefore $1$ is indeed the least upper bound. In other words, $\sup A=1$.