How to use the definition of supremum to prove that $\sup{S} = 2$?

Question

Let $S$ $=$ {$2-\frac{3}{n+1} | n\in \mathbb{N}$}. Use the definition of supremum to prove that $\sup{S} = 2$.

Here's what I have so far. We want to show that $2$ is an upper bound, and it is also the least upper bound.

We know $2$ is an upper bound because, $2 - \frac{3}{n+1} < 2 \text{, for all } n \in \mathbb{N}$

But how do I show it's the least upper bound? Someone tried explaining it to me, and I still don't really understand.

Answer 1

HINT: Use method of contradiction :assume there is some number less than 2 which is supremum and contradict by showing a number in set can always be found greater than this assumed supremum.

Answer 2

Assume that there is a LOWER bound, $2-\epsilon$, $\epsilon>0$. Then for all $n\in\mathbb{N}$

$$2-\frac{3}{n+1}<2-\epsilon$$ or $$\epsilon<\frac{3}{n+1}$$ for ALL $n$. Can you take it from here?

Edit, let me do more algebra for you. This is equivalent to:

$$n<\frac{3}{\epsilon}-1$$

You're assuming this is true for ALL natural numbers. What's the problem?