# Proving limits involving floor function.

Let $$\theta$$ be real, and $$0<\theta<1$$, define $$g_n = 0$$ if $$\lfloor n\theta\rfloor=\lfloor(n-1)\theta \rfloor$$ and $$g_n=1$$ otherwise. For example, take $$\theta = \frac{3}{4}$$, then $$g_2 =1$$, since $$\left\lfloor \frac{3}{2}\right\rfloor = 1 \ne 0 =\left\lfloor \frac{3}{4} \right\rfloor$$. Prove that $$\lim_{n \to \infty} \dfrac{g_1+g_2+\cdots + g_n}{n} = \theta$$ Note $$\lfloor x \rfloor$$ denotes the greatest integer function/ floor function.

This is what I did: $$g_1$$ is always $$0$$, since $$\lfloor\theta\rfloor=0$$, and $$g_2 = 0$$ if $$\theta \in \left(0,\frac{1}{2}\right)$$. Now consier $$g_4$$, $$\begin{array}{cccc} & \left( 0 ,\frac{1}{4}\right) & \left[ \frac{1}{4} , \frac{1}{3} \right) & \left[ \frac{1}{3} , \frac{1}{2} \right) & \left[ \frac{1}{2} , \frac{2}{3} \right) & \left[ \frac{2}{3} , \frac{3}{4} \right) & \left[ \frac{3}{4} , 1 \right) \\ \hline \lfloor 4\theta \rfloor & 0 & 1 & 1 & 2 & 2 & 3\\ \lfloor 3\theta \rfloor & 0 & 0 & 1 & 1 & 2 & 2\\ \end{array}$$

Hence $$g_4 = 0$$ when $$\theta \in\left( 0 ,\frac{1}{4}\right) \cup \left[ \frac{1}{3} , \frac{1}{2} \right) \cup \left[ \frac{2}{3} , \frac{3}{4} \right)$$ and $$g_4 = 1$$ otherwise. Similarly, $$g_n = 0$$ when $$\theta \in \left( 0 ,\frac{1}{n}\right) \cup \left[ \frac{1}{n-1} , \frac{2}{n} \right) \cup \left[ \frac{2}{n-1} , \frac{3}{n} \right) \cup \cdots \cup \left[\frac{n-2}{n-1}, \frac{n-1}{n} \right)$$ and $$g_n=1$$ otherwise. Consider the case when $$\theta = \frac{p}{q}$$ is a rational number with $$p,q>0 \in \mathbb{Z}$$ and $$p,q$$ are copime. Then we see a periodic behaviour, $$g_1 = g_{mq+1} = 0$$ where $$m \in \mathbb{Z_{\ge 0}}$$ and $$g_i = g_{mq+i}$$ for $$i \in \{ 1,2,\cdots,q-1\}$$. But for irrational $$\theta$$ I don't even know how to start. Any hints on how to further progess in this problem ?

Hint: Show that $$g_n=\lfloor n\theta \rfloor-\lfloor (n-1)\theta \rfloor$$, so that $$\sum_{i=1}^n g_n=\lfloor n\theta\rfloor$$ by a telescoping sum argument. So what you need to do is show that $$\lim_{n\to\infty}\frac{\lfloor n\theta\rfloor}{n}=\theta$$ which shouldn't be too difficult for you since that is a series of rational approximations that you can show that $$|\theta-a_n|<\frac1n$$ for all $$n$$
• Thank you so much. After this, we can write the limit as : $\lim_{n \to \infty} \frac{n\theta - \{ n \theta \} }{n} = \theta - 0$, where $\{ x \}$ represents the fractional part of $x$, and $\{ x \} \in [0,1)$