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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.

Matcha Latte
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    Other than squeeze , i dont see any other method. U cud convert to fraction and apply sandwich, but its the same method – maveric Jul 01 '20 at 14:52
  • yeah! that's why i am looking for. –  Jul 01 '20 at 14:54
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    For greatest integer, you could use $\lceil x \rceil$ (\lceil and \rceil). – K.defaoite Jul 01 '20 at 15:09
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    @K.defaoite: the informal "greatest integer" denotes the "greatest integer not larger than the argument" and is the $\text{floor}$ function. –  Jul 01 '20 at 15:41

1 Answers1

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$$\lfloor kx\rfloor=kx-\{kx\},$$ 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 $0\le\overline{\{kx\}}<1$.