I have a question regarding trying to prove that a number is supremum of the set: $ A=\left \{ \frac{4n}{n+1}:n\in \mathbb{N} \right \} $. Clearly 4 is an upper bound to the set. But how can i prove for example that the number $ 10$ is not the supremum of $A $?
My attempt: If $s = sup(A) $ there exist no number $10 - \varepsilon$ where $\varepsilon > 0$ such that $a > 10 - \varepsilon$ where $a \in A$. I am trying to prove this by contradiction where I am assuming that this number exists and arrive at the following:
$\frac{4n}{n+1}>10-\varepsilon\\\\\Leftrightarrow 4n>(10-\varepsilon)(n+1)\\\\\Leftrightarrow -6n+\varepsilon n > 10 - \varepsilon\\\\\Leftrightarrow n(\varepsilon-6)>10-\varepsilon\\\\\text{case:} \ \varepsilon\geq 6:n>\frac{10-\varepsilon}{\varepsilon-6}\\\\\text{case:}\ \varepsilon <6:n<\frac{10-\varepsilon}{\varepsilon-6}$
I dont understand how to interpret this result really and would appreciate any explanation.