how to prove sup(A) where A={(n+1)/n|n∈N}?

Question

I am looking for proof sup(A) = 2 where A={(n+1)/n|n∈N}

since if n>0. (n+1)/n<2 for all n ∈ natural numbers. it has an upper bound - 2. s.t. a≤2, for all a ∈ A.

poof that it is the least upper bound?

proof though Archimedes Principle leaves with u < 2 - 1/n, am i allowed to state that u is therefore in the set A as stated as (n+1)/n<2?

Answer 1

$$g(x) = \frac{x+1}{x} = 1 + \frac{1}{x}$$

and we know that $f(x) = 1/x$ is decreasing for $x > 0$. Thus $g$ is maximised by choosing the smallest value in your set (as $n \in \mathbb{N} \Rightarrow n > 0$). Thus

$$ \sup(A) = g(\min\{n \in \mathbb{N}\}) = g(1) = 1 + 1/1 = 2.$$

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Note further that since this supremum is obtained, it is equal to $\max(A)$.