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Question Number 109462 by 150505R last updated on 23/Aug/20

Answered by mathmax by abdo last updated on 24/Aug/20

$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}[{\mathrm{k}}^2\mathrm{x}+{\mathrm{k}}^2]\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]+\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2$$$$\mathrm{we}\mathrm{have}\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2=\frac{1}{{6\mathrm{n}}^3}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)\to \frac{1}{3}(\mathrm{n}\to +\mathrm{\infty })$$$$\mathrm{we}\mathrm{have}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽{\mathrm{k}}^2\mathrm{x}<\left[{\mathrm{k}}^2\mathrm{x}\right]+1\Rightarrow {\mathrm{k}}^2\mathrm{x}-1<\left[{\mathrm{k}}^2\mathrm{x}\right]⩽{\mathrm{k}}^2\mathrm{x}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{k}}^2\mathrm{x}-1)<\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2\mathrm{x}\Rightarrow$$$$\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\mathrm{x}-\mathrm{n}<\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\mathrm{x}\Rightarrow$$$$\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{{6\mathrm{n}}^3}\mathrm{x}-\frac{1}{{\mathrm{n}}^2}<\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]⩽\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{{6\mathrm{n}}^3}\mathrm{x}$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{1}{{\mathrm{n}}^3}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left[{\mathrm{k}}^2\mathrm{x}\right]=\frac{\mathrm{x}}{3}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=\frac{\mathrm{x}+1}{3}$$

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