1
$\begingroup$

This is a problem which emerged from trying to prove that a sequence of random variables on $[0,1]$ does not converge almost surely to $0$.

In the process of trying to prove the above, I encountered the infinite sequence constructed as follows.

  • $x_1 = 1$
  • $x_2 = 1 - \frac{1}{2} - \lfloor (1 - \frac{1}{2}) \rfloor = \frac{1}{2}$, $\qquad \qquad$ i.e. the fractional part of $(1 - \frac{1}{2})$
  • $x_3 = 1 - \frac{1}{2} - \frac{2}{3} - \lfloor (1 - \frac{1}{2} - \frac{2}{3}) \rfloor = \frac{1}{2} + \frac{1}{3} $, $\qquad$ i.e. the fractional part of $(1 - \frac{1}{2} - \frac{2}{3})$
  • $\vdots$
  • $x_n = 1 - \sum_{i=2}^n \frac{n-1}{n} - \lfloor (1 - \sum_{i=2}^n \frac{n-1}{n}) \rfloor$, $\qquad$ i.e. the fractional part of $(1 - \sum_{i=2}^n \frac{n-1}{n})$
  • $\vdots$

I would like to know whether this sequence is dense in $[0,1]$. I have a strong intuition that it is, but I have no idea how to tackle the proof. Any idea on how I could prove or disprove it?

$\endgroup$
1
  • 1
    $\begingroup$ Fractional, rather than decimal. $\endgroup$ Sep 2 '14 at 22:51
2
$\begingroup$

Hint: Denoting $\mathrm{frac}(x)$ to mean the fractional part of x, stating

$$ x_{i+1} = \mathrm{frac}\left( x_i - \frac{n-1}{n} \right) $$

is equivalent to stating

$$ x_{i+1} = \mathrm{frac}\left( x_i + \frac{1}{n} \right) $$

$\endgroup$
2
  • 1
    $\begingroup$ ... and so you are looking at whether the fractional parts of the Harmonic numbers are dense in $[0,1]$. The Harmonic numbers increase without limit, but with arbitrarily small increments, so ... $\endgroup$
    – Henry
    Sep 3 '14 at 6:54
  • $\begingroup$ Many thanks, very good hint, that should be enough for me to figure out the rest of the argument. $\endgroup$ Sep 3 '14 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.