Limit of expression involving the floor function

Question

How could I solve this series limit:$$ x_n=\frac{[x]+[3^2x]+[5^2x]+\ldots+[(2n-1)^2x]}{n^3}$$ where $[\cdot]$ is the floor function. I need to find the limit of $x_n$ as $n\to\infty$. Tried using $x-1<[x]\leq x$ but it didn't help much.

Answer 1

$$x-1 < \lfloor x \rfloor \le x\\ 3^2x-1 < \lfloor 3^2x \rfloor \le 3^2x\\ 5^2x-1 < \lfloor 5^2x \rfloor \le 5^2x\\\vdots\\ (2n-1)^2x-1 < \lfloor (2n-1)^2x \rfloor \le (2n-1)^2x\\$$sum of them

$$x(1+3^2+5^2+...+2n-1)^2-n < \lfloor x \rfloor +\lfloor 3^2x \rfloor+...+\lfloor (2n-1)^2x \rfloor \le x(1+3^2+5^2+...+2n-1)^2\\$$ now find $$1^2+2^2+3^2+...+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}\\ 1^2+3^2+...+(2n-1)^2+(2^2+4^2+...(2n)^2)=\frac{2n(2n+1)(4n+1)}{6}\\ 1^2+3^2+...+(2n-1)^2+2^2(1^2+2^2+...(n)^2)=\frac{2n(2n+1)(4n+1)}{6}\\$$so $$1^2+3^2+...+(2n-1)^2=\frac{2n(2n+1)(4n+1)}{6}-4(\frac{n(n+1)(2n+1)}{6})\sim \frac{16n^3}{6}-4\frac{2n^3}{6}\sim \frac{8}{6}n^3$$ by using squeeze theory $$\lim_{n\to \infty}\frac{x(1+3^2+5^2+...+2n-1)^2-n}{n^3}<\lim_{n\to \infty}\frac{[x]+[3^2x]+[5^2x]+\ldots+[(2n-1)^2x]}{n^3}<\lim_{n\to \infty}\frac{x(1+3^2+5^2+...+2n-1)^2}{n^3}\\ \lim_{n\to \infty}\frac{x(1+3^2+5^2+...+2n-1)^2-n}{n^3}<\lim_{n\to \infty}\frac{[x]+[3^2x]+[5^2x]+\ldots+[(2n-1)^2x]}{n^3}<\lim_{n\to \infty}\frac{x(1+3^2+5^2+...+2n-1)^2}{n^3}\\$$ $$\lim_{n\to \infty}\frac{\frac{8}{6}n^3-n}{n^3}\leq\lim_{n\to \infty}x_n\leq\lim_{n\to \infty}\frac{\frac{8}{6}n^3}{n^3}\\\lim_{n\to \infty}x_n=\frac{8}{6} $$

Answer 2

The identity function overestimates floor function by $<1$. Since $\sum_{j=1}^n(2j-1)^2\in\frac{4}{3}n^3+o(n^3)$, the $n$ terms in the numerator of $x_n$ have a sum $\in\frac{4}{3}n^3x+o(n^3)-\mathcal{0}(n)$. Dividing out $n^3$, only $\frac{4x}{3}$ survives as $n\to\infty$, regardless of $x$.