I saw this question somewhere and was having a hard time proving it so I would love to get some help.
Prove using $\varepsilon$ and $N$ that the following sequence is divergent
$$(a_n)=\sqrt n - \lfloor \sqrt n \rfloor$$ $$\lim_{n\to\infty} (a_n) \neq L$$
I have managed to prove it for $ L \neq 0 $ but can't manage to do so for $L=0$.
If $\lim\limits_{n\to\infty} (a_n)=L$, then for every $\varepsilon>0$ there is an $N>0$, such that $\vert a_n-L\vert <\varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $L\neq0$ is impossible. We can see this by choosing $0<\varepsilon<\vert L\vert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=\beta+2+2\sqrt{2\beta},\quad\text{where }\beta=2k^2\text{, for some odd $k>N$.}$$
This way, we have $N<m\in\mathbb N$ and $a_m=a_{\beta} + a_2$.
If $(a_n)\to 0$ then we must also have that $(a_m)\to 0$. But if the subsequence $(a_m)\to 0$ then $(a_ {\beta})\to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.