Let $(b_n)_{n\geq 1}$ be a sequence of real numbers whose terms are defined as follows :
$b_1 :=x+\frac{22}{13^2} -\frac{2}{13}y$, $b_2 :=x+\frac{22}{14^2} -\frac{2}{14}y$ and so on, where $x \in \mathbb R$ and $ y \in \mathbb N$ and $y \geq 2$ is arbitrarily fixed.
I need to calculate $\text{Sup}_{n \in \mathbb N}(b_n)$.
Claim: $\text{Sup}_{n \in \mathbb N}(b_n) =x$.
Attempt : It's not difficult to guess by inspection that the sequence is strictly increasing. For that purpose it's enough to check that the function $f(m)= \frac{22}{m^2}-\frac{2y}{m}$ is strictly decreasing for $m \geq 13$. For this we need that $2m^2 > 20m +11$ for all $m \geq 13$.
Once we have shown that the sequence is increasing we note that the terms $\frac{22}{m^2}-\frac{2y}{m}$ goes to $0$ as $m$ increases (Here is the place where I need more rigour). Therefore the sequence has supremum $x$
Please correct me if the above argument is not correct. Also it will be valuable to me if someone points out a more rigorous solution.
Any help from anyone is welcome.
Here is a proof that for large enough $m$ , $22/m^2- 2y/m$ is arbitrarily close to $0$ . It suffices to prove this for $1/m$ since if $1/m$ approaches $0$ then $2y/m$ approaches $0$ and $22/m^2$ approaches $0$ . So we have to prove for all $\epsilon>0$ there exists a natural N such that for for all $m>N$ we have $1/m<\epsilon$ (Since $m$ is always postive ) rearranging we can see that picking $N>1/\epsilon$(Which exists thanks to the Archimedean property ) concludes the proof .
Form here it’s easy to conclude that $x$ is the supremum .