Ensino Superior ⇒ Parte inteira

[rcompany]

[tex3]\text{Seja }n\in\mathbb{N^*},\,x\in\mathbb{R}\\
\text{Determinar }\lim_{n\to\infty}\frac{\lfloor x\rfloor+\lfloor 2x\rfloor+...+\lfloor nx\rfloor}{n^2}[/tex3]

Resposta

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[tex3]\frac{x}{2}[/tex3]

[rcompany]

[tex3]\forall k\in\mathbb{N^*},\,x\in\mathbb{R},\,\,kx−1 < \lfloor kx\rfloor \leqslant kx\\
\therefore \sum_{k=1}^n(kx-1)<\sum_{k=1}^n\lfloor kx\rfloor\leqslant \sum_{k=1}^n kx\quad (1)\\
(1)\implies x\cdot \frac{n(n+1)}{2}-n<\sum_{k=1}^n\lfloor kx\rfloor\leqslant x\cdot\frac{n(n+1)}{2}\\\\
\implies x\cdot \frac{n(n+1)}{2n^2}-\frac{1}{n}<\frac{\displaystyle\sum_{k=1}^n\lfloor kx\rfloor}{n^2}\leqslant x\cdot\frac{n(n+1)}{2n^2}\quad (2)\\\\
\left.\begin{array}{rl}
\lim_{n\to\infty}x\cdot \dfrac{n(n+1)}{2n^2}-\dfrac{1}{n}=\dfrac{x}{2}\\
\lim_{n\to\infty}x\cdot\dfrac{n(n+1)}{2n^2}=\dfrac{x}{2}\\
(2)
\end{array}\right\}
\implies \lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^n\lfloor kx\rfloor}{n^2}=\dfrac{x}{2}
[/tex3]