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Is it possible to simlify the follwing expression with a floor function:

$$\lim_{n\to \infty}\frac{1}{n}\left(\left\lfloor \frac{n\tau}{T}\right\rfloor-1\right)$$

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Why the tag "probability-theory"? –  Davide Giraudo Oct 11 '12 at 11:20

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up vote 2 down vote accepted

For a real number $x$, $\lim_{n\to +\infty}\frac{\lfloor nx\rfloor}n=x$. Indeed, $$nx-1\leq\lfloor nx\rfloor\leq nx,$$ then we divide by $n$ and apply squeeze theorem.

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