Two Proofs on a Simple Set Concerning Suprema and Maxima Suppose we have the set 
$$\Gamma = \{\frac{n}{n+1} : n \in \mathbb{N}\}$$
We wish to determine what is $\sup \Gamma$ and $\max \Gamma$. We can clearly see that $\sup \Gamma = 1$ and that $\max \Gamma$ does not exist, but to prove this, I would like to know which of the following two proofs is better than the other and why, please.
Proof 1
We construct the following string of inequalities
$$0 < \frac{1}{n} \leq 1$$
$$1 < \frac{n+1}{n} \leq 2$$
$$\frac{1}{2} \leq \frac{n}{n+1} <1$$
Therefore, $\sup \Gamma = 1$ and $\max \Gamma$ does not exist.
Proof 2
Since $n \leq n+1$ for all $n \in \mathbb{N}$, we have
$$0 < n < n+1$$
implying
$$0 < \frac{n}{n+1} < 1$$
rendering the same results as the first proof.
I know that this is trivial, but verification and insight on demonstrating these simple results is what I value to learn more about and christen.
 A: If you say $\dfrac n {n+1} <1 $, that doesn't prove that $\sup\Gamma=1$, but only that $1$ is an upper bound of $\Gamma$, so that $\sup\Gamma\le 1$.  The supremum is the smallest upper bound.  Here's one way.  Suppose there is some smaller upper bound than $1$, so that $\sup \Gamma<1$.  Notice that
$$
\frac n {n+1} = 1 - \frac 1 {n+1} <\sup\Gamma,
$$
so
$$
1-\sup\Gamma < \frac 1 {n+1}
$$
and thus
$$
\frac 1 {1-\sup\Gamma} > n+1.
$$
Thus $1/(1-\sup\Gamma)$ is an upper bound of all integers, so there is a least upper bound $\sup\mathbb N$ of all integers.  The number $-1+\sup\mathbb N$, being less that $\sup N$, is not an upper bound of $\mathbb N$. For any number $a$, some integer $N$ must be in $(-1+a,a)$.  So $N+1>a$.  Applied to $a=\sup N$ this says $N+1>\sup\mathbb N$.  But no integer can be greater than an upper bound of all integers.  Hence our assumption that there is a smaller upper bound than $1$ is false.
A: *

*I would not call this trivial.

*I don't like any of those two since there's missing at least one crucial argument (the sequence is strictly increasing). Just listing true statements (here: inequalities) without arguments or connections is not a proof at all to me.

*That said, I like the first one more since the inequality is stronger (and providing a lower bound which on the other hand is not mentioned) and otherwise they're pretty much identical.
A: $f(x)=\frac{x}{x+1}\Rightarrow f'(x)=\frac{1}{(x+1)^2}>0$ so it is increasing function, thus min in natural number happens when $n=1\Rightarrow \frac{n}{n+1}=\frac{1}{1+1}=\frac{1}{2}$ and increasing bounded sequence converges to their sup. Thus $\sup \frac{n}{n+1}=\lim \frac{n}{n+1}=1$
A: Note that the following are equivalent:
$$\begin{align}\frac{n}{n+1}&<\frac{n+1}{n+2}\\n(n+2)&<(n+1)^2\\n^2+2n&<n^2+2n+1\\0&<1\end{align}$$
Hence, the sequence $n\mapsto\frac{n}{n+1}$ is increasing.
As you've shown in both proof attempts, every element of $\Gamma$ is strictly less than $1.$ Hence, if we can show that $\sup\Gamma=1,$ it follows that $\max\Gamma$ does not exist.
Now, take any $\epsilon>0,$ and note that the following are equivalent: $$\begin{align}1-\epsilon&<\frac{n}{n+1}\\1-\frac{n}{n+1}&<\epsilon\\\frac1{n+1}&<\epsilon\end{align}$$ Readily, there is an $n$ that make these equivalent statements true, and so numbers less than $1$ are not upper bounds of $\Gamma.$ Thus, since $1$ is an upper bound of $\Gamma,$ then $\sup\Gamma=1,$ as desired.
