Evaluate: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n\lfloor kx\rfloor}{n^2}$$
My approach:
I simply use this. $$ \boxed{x-1 < \lfloor x\rfloor \leq x} $$ to apply squeeze theorem and finally come to this conclusion, $$\frac{\left\lfloor x\left(\left(\frac{n(n+1)}2\right)-n\right)\right\rfloor}{n^2}< \frac{\left(\begin{array}{c} \text { Required number } \\ \end{array}\right)}{n^2} \leq \frac{x\left(\frac{n(n+1)}2\right)}{n^2}$$
taking limit($n\to\infty$) to LHS & RHS I am getting $\frac{x}2$
I am looking for any other method better than this.Any suggestion or hint would be greatly appreciated.
$$\lfloor kx\rfloor=kx-\{kx\},$$ where $0\le\{kx\}<1$.
The summation gives
$$\frac{n(n+1)}{2n^2}x-\frac1{n^2}\sum_{k=1}^n\{kx\}=\frac x2+\frac1n\left(\frac12-\overline{\{kx\}}\right)\to\frac x2$$
because $0\le\overline{\{kx\}}<1$.