How can I prove that $$ \lim_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x, $$ whenever $x>1$. Here $\left\lfloor \cdot\right\rfloor$ denotes the floor function, or the integer part function.

The integer part $\lfloor z\rfloor$ of $z$ is the largest integer, which does not exceed $z$.

Thanks for your answer.


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    $\begingroup$ $\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}$ is nearly $\frac{x^{n+1}}{x^n}$. $\endgroup$ – Hurkyl Mar 10 '14 at 22:05
  • $\begingroup$ (-1) "This question does not show any research effort" $\endgroup$ – TMM Mar 17 '14 at 21:23

Since $y-1< \lfloor y\rfloor\le y$, for every $y\in\mathbb R$, then $$ \frac{x^{n+1}-1}{x^{n}}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}< \frac{x^{n+1}}{x^n-1}, $$ and hence $$ x-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}<x+\frac{x}{x^n-1}, $$ or $$ -\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}-x<\frac{x}{x^n-1}. $$ Since both $ -\frac{1}{x^n},\,\frac{x}{x^n-1}\to 0, \quad\text{as}\quad n\to\infty, $ then $\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}\to x$.

  • $\begingroup$ Ok! Thank you Yiorgos! $\endgroup$ – Chiquilin Mar 10 '14 at 22:55

Hint we have $$y-1\le\left\lfloor y \right\rfloor\le y$$ then use the squeeze theorem.

  • $\begingroup$ Doesn't a function need to be continuous to use that? Or is x <= y <= z enough? $\endgroup$ – Alexander Cogneau Mar 11 '14 at 8:45
  • $\begingroup$ This inequality is a propriety of the floor function.@AlexanderCogneau $\endgroup$ – user63181 Mar 11 '14 at 8:48
  • $\begingroup$ Looks like a "nice answer" to me! ;-) $\endgroup$ – Namaste Mar 11 '14 at 12:43

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