Proving divergence of the fraction of the square root - without using proof by contradiction

Question

I need to solve the following question:

Prove using $\varepsilon$ and N that the following sequence is divergent: $$a_n = \sqrt{n} - ⌊\sqrt{n}⌋$$ $$\lim_{n\rightarrow \infty} (a_n) \ne L$$

Without using proof by contradiction -

which means I need to show that for any $L\in \mathbb{R}$ exists $\varepsilon>0$ such that for any $N\in \mathbb{N}$ exists n such that $n>N \,$ and $\, |a_n-L| \ge \varepsilon$.

I've tried to split it into $L\le 0$ and $0<L\le 1$ and $L>1$

or $L\le 0 $ and $L>0$ - but nothing helped...

Answer 1

What can you say about $a_{k^2}$ and $a_{k^2 - 1}$ as $k$ ranges through the positive integers? What's the limit of their difference?

Answer 2

You can try finding two subsequences that converge to two different numbers.

Check these limits.

$$\lim_{n\to +\infty} \,(\sqrt {n^2} - n) = 0$$

$$\lim_{n\to +\infty} \big(\,\sqrt {(n^2-1)} - (n-1)\,\big) = 1$$