Once floored, always floored

From the definition of the floor function, it follows that $\displaystyle y -1 \leq \lfloor y \rfloor \leq y$ $\forall y$ $\in \mathbb{R}$ . Therefore for each term in $S_n (x)$ we can write, $\displaystyle rx -1 \leq \lfloor rx \rfloor \leq rx$ . Therefore, on summing up all the terms, we get, $\sum_{r=1}^n (rx-1) \leq \sum_{r=1}^n \lfloor rx \rfloor \leq \sum_{r=1}^n rx$ $\Rightarrow \frac{nx(n+1)}{2} - n \leq S_n(x) \leq \frac{nx(n+1)}{2}$

Dividing throughout by $n^2$ and taking the limit we get, $\lim_{n \to \infty} \frac{x}{2}\left( 1 + \frac{1}{n} \right) - \frac{1}{n} \leq \lim_{n \to \infty} \frac{S_n(x)}{n^2} \leq \lim_{n \to \infty} \frac{x}{2}\left( 1 + \frac{1}{n} \right)$ Realizing that the limit of the lower estimate is same as that of the upper one, we can state the required limit is the equal value by invoking the Squeeze Theorem .

$\displaystyle \therefore \lim_{n \to \infty} \frac{S_n(x)}{n^2} = \frac{x}{2}$ .

Thus the required value is $\displaystyle \bigg \lfloor \frac{2015.20162017}{2} \bigg\rfloor = \boxed{1007}$ .

Note:
The actual inequality for the floor function is $\displaystyle y -1 < \lfloor y \rfloor \leq y$ . Using the $\leq$ sign instead of $<$ keeps the essence of inequality intact but makes it easier to relate with the squeeze theorem and hence I have written the inequality with the $\leq$ sign.

Break the gif into two parts. One is real part and other fractional part. Since fractional part will return a value between zero and one, it's contribution to the limit becomes zero. The question is now reduced to calculate a basic limit, which can be calculated easily using limit as a sum technique.

cosnidering the limit $\lim_{n \to \infty}\frac{S_n(x)}{n^2}$ when x is fixed

then by stolz cesaro $\lim_{n \to \infty}\frac{S_n(x)}{n^2}=\lim_{n \to \infty}\frac{S_{n+1}(x)-S_n(x)}{2n+1}=\lim_{n \to \infty}\frac{[(n+1)x]}{2n+1}=\frac{x}{2}$

by squeeze

so $\left[\lim_{n \to \infty}\frac{S_n(x)}{n^2}\right]=\left[\frac{x}{2}\right]=1007$