Proving that $A=\left\{\frac{n}{2n+1}:n \in \mathbb{N}\right\}$ is bounded by $\frac12$ Let $A=\left\{\dfrac{n}{2n+1}:n \in \mathbb{N}\right\}$. I want to prove that $supA=\dfrac{1}{2}$ so I need to show that $$\forall\epsilon\gt0 \exists a\in A:a\gt\dfrac{1}{2}-\epsilon$$
So suppose by contradiction that  $$\exists \epsilon\gt0 \forall a\in A:a\le\dfrac{1}{2}-\epsilon$$
which essentially means that $$n \le \left( \dfrac{1}{\epsilon}-1\right)\cdot \dfrac{1}{2}$$
and this is not possible as the natural numbers are not bounded from above.
Am I right? This seems to easy - I could have plugged any other number (smaller then $\frac{1}{2}$) and prove the same.

The idea is right, the calculations I'm not that sure as I get $$\exists \epsilon >0\,\,s.t.\,\,\forall n\in\mathbb{N}\,\,,\,\frac{n}{2n+1}\leq\frac{1}{2}{-\epsilon}\Longleftrightarrow \rlap{/}{n}\leq \rlap{/}{n}+\frac{1}{2}-2n\epsilon-\epsilon\Longleftrightarrow$$$$\Longleftrightarrow2n\epsilon\leq \frac{1}{2}-\epsilon\Longleftrightarrow n\leq\frac{1}{4}\left(\frac{1}{\epsilon}-2\right)$$
