I would like to prove the following fact:
$\displaystyle\lim_{n \to \infty } (nt – \lfloor nt \rfloor) =0 $, where $\lfloor x \rfloor := \max \{m \in \mathbb Z: \ m \le x \} $ denotes the floor function.
My idea to prove this was the following:
first, prove it for $t\in\mathbb Q$ and then use that the rationals are dense in the reals.
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Thus,$(t=\frac {p } {q }\in\mathbb Q)\;\land\;(n=q^j,j\in\mathbb N)\implies nt\in\mathbb Z\implies\lfloor nt \rfloor =nt\iff nt – \lfloor nt \rfloor =0 $.
But what happens with the difference $ nt – \lfloor nt \rfloor$ when $q^{j-1 } < n < q^j $?
Does it decrese as $j \to \infty $?
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Secondly if $\exists\displaystyle\lim_{n\to\infty}(nt-\lfloor nt\rfloor)$ may we argue that if $|x-t|< \epsilon, \ x \in \mathbb R, \ t \in \mathbb Q $ then: $$\lim_{n \to \infty } nt – \lfloor nt \rfloor =0 \implies\lim_{n \to \infty } nx – \lfloor nx \rfloor < \epsilon ?$$
Of course, any other approach is fine if it proves or disproves the limit!
Most grateful for any help provided!
Best Answer
It's not true. If $t=\dfrac12$, then $a_n=nt-\lfloor nt\rfloor$ alternates between $0$ and $\dfrac 12$,
so the limit as $n\to\infty$ does not exist.