How to prove that $\lim_{n\to \infty } \left(\sqrt{2} n-\left\lfloor \sqrt{2} n\right\rfloor \right) $ doesn't exist? I've tried to show that every $L\in\Re$ is not a limit. It's pretty easy for $L>1$ and $L<-1$. The main problem is when $-1\leq L\leq 1$ .
It is an homework question, so i should use the basic limit definition ($\epsilon$).
Thanks!
Let $\{x\} = x - \lfloor x\rfloor = x $ (mod 1) be the fractional part of $x$.
Suppose such a limit L exists. Then given any $\epsilon > 0$, there would have to exist an $N$ s.t. for all $n \ge N$, we would have $|\{n\sqrt{2}\} - L| < \epsilon$. Now, as Aaron said, calculate
$|\{(n+1)\sqrt{2}\} - \{n\sqrt{2}\}| = |\sqrt{2} - (\lfloor (n+1)\sqrt{2}\rfloor - \lfloor n\sqrt{2} \rfloor)| > 1/3$
This leads us to consider $\epsilon = 1/6$, say. Supposing our inequality holds for $n = N$, the above shows that even the very next term $\{(N + 1)\sqrt{2}\}$ must land farther away from $L$ than $\epsilon$. So no such $N$, and therefore no such limit, can exist.
Explicitly calculating the first few $\{n\sqrt{2}\}$ leads to the intuition that the sequence "moves around a lot"; this can be made precise and is a special case of the equidistribution theorem, which is itself an instance of the ergodic theorem.