Prove that a sequence is divergent (By definition - Epsilon-N Way)

Question

First, this is the question:

Prove (using epsilon-N definition) that the sequence $ a_n = \left<\sqrt{n}\right> $ is divergent.

Note: $ \left<x\right> = x- \lfloor x \rfloor$

My question:

I proved it by splitting it into cases: $L=0$ and $ L \neq 0 $

I wonder if there's a simpler and more beautiful proof to this question?

Thanks!

Answer 1

hint

$$(\forall x\in \Bbb R) \; 0\le x-\lfloor x\rfloor <1$$

the possible limit would be in $[0,1]$.

$$\lim_{n\to +\infty}a_{n^2}=0$$

Answer 2

For each $n \ge 1$ you have $$\left( n + 1 - \frac 1n \right)^2 < n^2 + 2n < (n+1)^2$$ so that $$ n + 1 - \frac 1n < \sqrt{n^2 + 2n} < n+1$$ and consequently $$1 - \frac{1}{n} < \langle \sqrt{n^2 + 2n} \rangle < 1$$ for all $n$. Thus $L = \lim a_n$, if it exists, must equal $1$.

Answer 3

$\langle \sqrt {n^2 -1}\rangle > \frac {n-1}{n}\\ \langle \sqrt {n^2}\rangle = 0$

For any $L$, there exist $n>4$ such that $| \sqrt {n} - \lfloor n\rfloor - L| > \frac 14$

Or you could use Cauchy sequences.

A sequence is Cauchy if:

$\forall \epsilon > 0, \exists N>0: n,m>N \implies |a_n - a_m|<\epsilon$

So to show that it is not Cauchy say that there exist $m,n > N$ such that $|a_n-a_m|>\epsilon$