How can I find an $\varepsilon$ environment to disprove $\lim_{n\to \infty}(\{{\sqrt{n}}\}) = 0$? Prove:
$$
\lim_{n\to \infty}(\{{\sqrt{n}}\}) \neq 0
$$
It's obvious that $\{\sqrt{n}\} = 0$ when $\sqrt{n} \in \mathbb{N}$ and
$1\gt$ $\{\sqrt{n}\}$ $\gt 0$ when $\sqrt{n} \notin \mathbb{N}$.
I have to find an infinite group of numbers that each one of them is greater than $\ 0+\varepsilon$. That's how I define an  $\varepsilon$ environment for $\\0$ and disprove it. How can I find it? thanks.
 A: This is a bit more subtle than it looks at first glance. Here's how I would approach a problem like this: I know that $\{\sqrt{n}\}=0$ when $n=k^2$ and when $n=(k+1)^2$.  Since I want to maximize it, I should probably look at the 'worst-case scenario': $\{\sqrt{n}\}$ is increasing between squares, then drops back to zero only to increase again; so what if I look at $n=(k+1)^2-1$?  This is an infinite subsequence of numbers trending to $\infty$, and you should be able to show a bound away from $0$ for $\{\sqrt{(k+1)^2-1}\}$.  (Note that $\lfloor\sqrt{(k+1)^2-1}\rfloor=k$, so you just have to find a lower bound on the distance $\sqrt{(k+1)^2-1}-k$.)
A: We can try the following: since $\;\{\sqrt n\}=\sqrt n-\left\lfloor\sqrt n\right\rfloor\;$, we can take the subsequence for all the indexes of the form $\;n^2+n+1\;$ , i.e.: $\;3,7,13,\ldots\;$ , and thus we have
$$\{\sqrt{n^2+n+1}\}=\sqrt{n^2+n+1}-\left\lfloor \sqrt{n^2+n+1}\right\rfloor\;$$
But we always have that $\;\left\lfloor\sqrt{n^2+n+1}\right\rfloor=n\;$ (why?) , so 
$$\;\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor=\sqrt{n^2+n+1}-n\;$$
and thus this partial limit is
$$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-n\right)=\lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n+1}+n+1}=$$
$$=\lim_{n\to\infty}\frac{1+\frac1n}{\sqrt{1+\frac1n+\frac1{n^2}}+1+\frac1n}=\frac1{1+1}=\frac12$$
