 Solution

Write each member as $3-2/(n+1)$. Every member is strictly below $3$, so $3$ is an upper bound. If $u<3$, the Archimedean property supplies $n\in \mathbb{N}$ with $n+1>2/(3-u)$; then $3-2/(n+1)>u$. Thus no smaller number is an upper bound, proving $\sup S=3$. Since $3\notin S$, there is no greatest element. Equivalently, each member is followed by a larger member because $2/(n+2)<2/(n+1)$.