Proving Via delta epsilon $(n\sqrt2- \lfloor n\sqrt2\rfloor )$

Question

$a_n = (n\sqrt2- \lfloor n\sqrt2\rfloor )$

Any hints on how to prove there is no limit?

Answer 1

For $ n\in\mathbb{N},$ it is

$$\frac{x}{10}< \sqrt{2}\cdot 10^n-[\sqrt{2}\cdot 10^n]<\frac{x+1}{10},$$

where $x$ is the digit that is in the position $n+1$ of the decimal expansion of $\sqrt{2}.$ Since $\sqrt{2}$ is irrational we can find in its decimal expansion two digits $x\ne y$ in arbitrary large positions. Assume $x<y.$ We should have

$$\frac{x}{10}< \sqrt{2}\cdot 10^n-[\sqrt{2}\cdot 10^n]<\frac{x+1}{10}\le \frac{y}{10}< \sqrt{2}\cdot 10^m-[\sqrt{2}\cdot 10^m]<\frac{y+1}{10}.$$ Thus, if the limit exists it must be $l=\frac{x+1}{10}=\frac{y}{10}.$ In particular, in the decimal expansion of $\sqrt{2}$ (from some decimal place) there are exactly two different digits: $x$ and $x+1.$ (If there is a different decimal $z\notin\{x,x+1\}$ then

$$\frac{z}{10}< \sqrt{2}\cdot 10^k-[\sqrt{2}\cdot 10^k]<\frac{z+1}{10},$$ for infinitely many $k$'s contradicts $l=\frac{x+1}{10}.$ But, assuming that there are only two different digits (from some decimal position) we get that

$$\frac{x}{10}< \sqrt{2}\cdot 10^n-[\sqrt{2}\cdot 10^n]<\frac{x}{10}+\frac{x+1}{100},$$ for infinitely many $n$'s, in contradiction with the fact that the limit should be $l=\frac{x+1}{10}.$

As @columbus8myhw says in a comment we can replace $\sqrt{2}$ with any irrational number. The proof is exactly the same. As @AndreNicolas says in another comment this closely related to the fact that the fraccional part of $n\alpha$ is dense on $[0,1]$ for any irrational number ($\sqrt{2}$ in this case).