# Is $\{1,\frac{1}{2}, (\frac{1}{2}+\frac{1}{3}), \frac{1}{12}, (\frac{1}{12} + \frac{1}{5}), \dots\}$ dense in $[0,1]$?

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?

• Fractional, rather than decimal. Sep 2, 2014 at 22:51

Hint: Denoting $\mathrm{frac}(x)$ to mean the fractional part of x, stating
$$x_{i+1} = \mathrm{frac}\left( x_i - \frac{n-1}{n} \right)$$
$$x_{i+1} = \mathrm{frac}\left( x_i + \frac{1}{n} \right)$$
• ... 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 ... Sep 3, 2014 at 6:54