find $\limsup$ and $\liminf$ of $a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor$ $n\in\mathbb{N}\}$

Question

I have to find $\limsup$ and $\liminf$ of $a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor : n \in\mathbb{N}\}$

I suppose that I have to relate to subsequences . First one is $a_{n_k}$ for all n that maintains $\sqrt n\in\mathbb {N}$ , so the limit is $0$.

Second one $a_{n_j}$ includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it.

Answer 1

We have that

$$a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor \ge 0$$

and for $n=m^2 \implies a_n=0$ then $\liminf a_n =0$.

For $\limsup$ we have

$$a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor \le 1$$

and for any $\epsilon>0$ we can find $n=m^2-1$ such that $\sqrt{n} - \lfloor\sqrt{n}\rfloor=1-\epsilon$.