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.

First of all, check that for every $n \in \mathbb{N}$
$$
\frac{n}{2n+1} \leq \frac12.
$$
This is trivial. The fact that $1/2=\sup A$ in this case follows from the fact that
$$
\lim_{n \to +\infty} \frac{n}{2n+1} = \frac12.
$$
This indeed implies that $\sup A$ cannot be smaller than $1/2$.
