Convergence of floor-function

Question

I am given an expression using the Gauss-brackets [] (assuming this is meant as floor-function). Let $0 \leq s < t\leq 1$ and $n\in\mathbb{N}$. I want to know against what the following expression converges: $$[nt]-[ns],~n\rightarrow\infty.$$ I have a hunch that it could be $t-s$ but no idea how to prove that. Additionally, I have a similar expression involving a sequence $\vartheta_n\rightarrow 0$ but $n\vartheta_n^2\rightarrow\infty$, and whish to know about the convergence of $$[ns+\vartheta_n^{-2}L]-[ns],$$ $L>0$ being a constant. I am greatful for any hints on how to solve this.

Answer 1

Assuming $[]$ is the floor function as you mentioned (even though $\lfloor \rfloor$ is the typical notation), $[nt]-[ns]$ diverges as $n\rightarrow\infty$. Simple proof is to consider $s,t$ that are rational numbers between $0$ and $1$, then consider $n\rightarrow\infty$ while being a multiple of the LCM of their denominators.

However, the following converges as $n\rightarrow\infty$. $$\frac{[nt]-[ns]}{n} \tag{1}\label{eq1}$$

To evaluate the above, consider the following breakdown of a general number into it's whole and fractional parts, $$x = [x] + r, \; 0 \leq r < 1$$

With this form, try to obtain inequality relationship between $x-y$ and $[x] - [y]$. Apply these inequalities for Eq. \eqref{eq1}, apply $n\rightarrow\infty$, and you should get the result. You can do a similar analysis to answer the ratio version of your second part, noting that $\frac{\vartheta_n^{-2}}{n}\rightarrow 0$.