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

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 $1$.
