Let $x\lt \frac12$, and assume that $x$ is an upper bound for $A$. This implies that \[ \forall n\in\N: \quad \frac{n}{2n+1} \le x \] Hence \[ \forall n\in\N: \quad n \le 2nx + x, \] and thus \[ \forall n\in\N: \quad (1-2x)n \le x. \] Since $x\lt \frac12$ we have $1-2x\gt0$ so we may divide both sides of the inequality by $1-2x$ and conclude that \[ \forall n\in\N:\quad n \le \frac{x}{1-2x}. \] This implies that $x/(1-2x)$ is an upper bound for the set of natural numbers, in direct contradiction with the archimedean property.
The contradiction shows that no number less than $\frac 12$ can be an upper bound for $A$.
Hence $\frac{1}{2}$ is the least upper bound for $A$.