I am looking for the upper bound and the lower bound of the following set: $$ S=\left\{\frac{\lfloor n\sqrt{2}\rfloor}{n}:n\in\mathbb{N}^*\right\} $$ we have $\sqrt{2}-\frac{1}{n}< \frac{\lfloor n\sqrt{2}\rfloor}{n}\le\sqrt{2} $ for all $n\ge 1$. for $\epsilon>0$ there exists $n_0\ge 1$ sach that $\frac {1}{n_0}<\epsilon$ then $-\epsilon <-\frac {1}{n_0}$. Thus $\sqrt{2}- \epsilon < \frac{\lfloor n_0\sqrt{2}\rfloor}{n_0}\le\sqrt{2} $. then we have $\sup (S)=\sqrt{2}$.
But I am unable to find the lower bound. An idea please.
Note that $\frac{\lfloor \sqrt{2} \rfloor}{1} = 1$ and $\frac{\lfloor 2\sqrt{2} \rfloor}{2} = 1$. For $n \geq 3$ we have
$$\frac{\lfloor n\sqrt{2} \rfloor}{n} > \frac{n\sqrt{2}-1}{n} > \sqrt{2} - \frac{1}{3} > 1$$
Thus the lower bound is $1$.