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Question Number 205307 by universe last updated on 15/Mar/24

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{\lfloor a\rfloor +\lfloor 2a\rfloor +...+\lfloor na\rfloor }{n^2}\mathrm{where}a\in \mathbb{R}$$$$\mathrm{and}\lfloor x\rfloor \mathrm{is}\mathrm{the}\mathrm{floor}\mathrm{of}\mathrm{x}\in \mathbb{R}$$

Commented by Frix last updated on 15/Mar/24

$$\mathrm{Just}\mathrm{guessing}:$$$$-\mathrm{\infty }\mathrm{for}a<0$$$$\frac{a}{2}\mathrm{for}a⩾0$$

Answered by Mathspace last updated on 15/Mar/24

$$\left[a\right]⩽a<\left[a\right]+1\Rightarrow \left[a\right]⩽aeta-1<\left[a\right]\Rightarrow$$$$a-1<\left[a\right]⩽a$$$$ka-1<\left[ka\right]⩽ka\Rightarrow$$$$\sum _{k=1}^n(ka-1)<\sum _{k=1}^n\left[ka\right]⩽\sum _{k=1}^{n}ka$$$$\Rightarrow a.\frac{n(n+1)}{2}-n<\sum _{k=1}^n\left[ka\right]⩽a\frac{n(n+1)}{2}$$$$\Rightarrow \frac{n(n+1)}{n^2}a-\frac{1}{n}<\frac{\sum _{k=1}^n\left[ka\right]}{n^2}⩽\frac{n(n+1)}{2n^2}a$$$$ona{lim}_{n\to +\mathrm{\infty }}\frac{n(n+1)}{2n^2}a-\frac{1}{n}=\frac{a}{2}$$$${lim}_{n\to +\mathrm{\infty }}\frac{n(n+1)a}{2n^2}=\frac{a}{2}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}\frac{\sum _{k=1}^n\left[ka\right]}{n^2}=\frac{a}{2}$$

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