Proof that Sup$(\{\frac{n}{n+1}| n \in \mathbb N \})=1$ I am practising for an exam and wish to show:

Sup$\Big(\Big\{\frac{n}{n+1}\,\Big|\, n \in \mathbb N \Big\}\Big)=1$

We usually start out with noticing that for all $n \in \mathbb N$: 
$$ \frac{n}{n+1}<1$$ So $1$ is certainly an upper bound for this sequence.  We now need to show it is the least upper bound, also known as the supremum. We can use the the second characterisation of the supremum, we know that for any $\epsilon >0$, we have that 
$$ \frac{n}{n+1} > 1 - \epsilon$$
for some $n$.
I've been trying to find such an $n$, but haven't really seen how one usually does this. You can't just 'rearrange'  the inequality - if you can I do not see how this would help.

Alternative thought: we have a sequence that is bounded (just shown), I can also derive that $a_{n+1}-a_n >0$ so the sequence is increasing, then the limit of this sequence is the supremum - the limit happens to be 1.
 A: $\frac{n}{n+1} > 1 - \epsilon \iff 1- \frac{n}{n+1} < \epsilon \iff \frac{1}{n+1} < \epsilon \iff n > \frac{1}{\epsilon} - 1$
A: Given $\varepsilon > 0$ we can find $n \in \mathbb{N}$ such that $ 0 > 1 - \varepsilon - \varepsilon \cdot n$
and 
$ 0 > 1 - \varepsilon - \varepsilon \cdot n  \Rightarrow $
$ n > n + 1 - \varepsilon - \varepsilon \cdot n  \Rightarrow $
$ n > (1 - \varepsilon) \cdot (n + 1) \Rightarrow$
$ \frac{n}{n+1} > 1 - \varepsilon$
and that proves what was required.
A: Option:
$a_n=\dfrac{n+1-1}{n+1}= 1-\dfrac{1}{n+1}.$
$1$ is an upper bound .
Assume $a =1 - \epsilon (>0)$ is a lower upper bound.
Archimedes principle:
There is a $n_0+1$, $n_0 \in \mathbb{Z^+}$ s.t. 
$n_0 +1 > 1/\epsilon$.
Then $\epsilon >\dfrac{1}{n_0+1}$, and
$a_{n_0} = 1-\dfrac{1}{n_0+1} >1- \epsilon.$
Contradiction.
Hence $1$ is the least upper bound.
A: Another way to do this: Find upper bound for your set and then find subsequence of $$ \frac{n}{n+1} $$ which converges to your upper bound. And it will be your supremum
