How to show that $ \langle \sqrt n \rangle $ is not converging

Question

I am trying to show that

$(a_n)_{n=1}^{\infty} = (\langle\sqrt n \rangle)_{n=1}^{\infty} $ is not a converging sequence ($n \in \mathbb N$).

Where $\langle x \rangle = x - [x]$, and [x] is floor(x). ( e.g. [5.5] = 5, $\langle 5.5 \rangle = 5.5 - 5 = 0.5 $).

I know intuitively that if $a_n = \langle \sqrt n \rangle $ then there is a subsequence of $(a_n)$, say $(a_{n_i})$, such that $\frac{1}{2}<\langle a_{n_i} \rangle<1$. However, I don't know how to state a mathematic expression this kind of subsequence.

Thanks!

Answer 1

Moreover this sequence is dense on $[0;1]$.

Indeed let $ 0\leqslant p\leqslant q,\,\, p,q\in\mathbb Z. $ Then $qn\leqslant \sqrt{q^2n^2+2pn}<qn+1.$

Therefore $ \{\sqrt{q^2n^2+2pn}\}=\sqrt{q^2n^2+2pn}-qn=\frac{2pn}{\sqrt{q^2n^2+2pn}+qn}\to \frac pq.$

Answer 2

For each $k\in \Bbb N$ there will be an integer $n$ whose square root is approximately $k+1/2$. As $(k+1/2)^2=k^2+k+1/4$ then $\sqrt{k^2+k}$ will be just under $k+1/2$ and $\sqrt{k^2+k+1}$ will be just over it.