Set of subsequence limits of $\{\sqrt n\}$

Question

Question:
$$a_n= \{\sqrt n\}$$ 1.Find $\liminf_{n\to\infty} a_n$
2.Find $\limsup_{n\to\infty}a_n$
3.Find the Set of subsequence limits of $a_n$
My Answer:

1. used the fact that the sequence is zero at $a_{n^2}$
2. used the fact that the sequence converges to one at $a_{n^2-1}$
3. I'm pretty sure the set is $[0,1]$ and I tried to prove it by finding a subsequence converging to any number between 0 and 1.

Answer 1

Note that $a_n = \sqrt{n} - [\sqrt{n}]$. Morever, for any $n$ we have $[\sqrt{n^2}] = [n] = n$ and $\sqrt{(n+1)^2} = n+1$ meaning that if $\alpha \in (0,2)$ then we have $[\sqrt{n^2 + [\alpha n]}] = n$ (because $n^2 + [\alpha n] \le n^2 + 2n < (n+1)^2$). Taking $m_n = n^2 + [\alpha n]$, then $$ a_{m_n} = \sqrt{n^2 + [\alpha n]} - n = \frac{n^2 + [\alpha n] - n^2}{\sqrt{n^2 + [\alpha n]} + n} = \frac{[\alpha n]}{n}\cdot \frac{1}{\sqrt{1+ \frac{[\alpha n]}{n^2}} + 1} \xrightarrow[n \to \infty]{}\alpha \frac{1}{2} = \frac{\alpha}{2}$$ Since $\alpha \in (0,2)$ was arbitrary, we've found a subsequence converging to arbitrary $\frac{\alpha}{2} \in (0,1)$.

Answer 2

The $n^2-1$ was a good idea. Similarly, you can take $a_{n^2 + \lfloor 2nx\rfloor}$ so that $$ \sqrt{n^2 + \lfloor 2nx\rfloor} = n + \frac{ \lfloor 2nx\rfloor}{\sqrt{n^2 + \lfloor 2nx\rfloor} + n} = n + \frac{ \lfloor 2nx\rfloor}{2 n} + o(1) $$