# Decimal representation series

Given the following sequence $d(n) = x \cdot n - \lfloor x \cdot n \rfloor$ with $n \in \mathbb{N}$, $x \in \mathbb{R} \backslash \mathbb{Q}$ .

I want to show that for every $z \in [0,1)$ there is a subsequence of $d(n)$ that converges against $z$.

Notes:

I am having serious problems of finding such a subsequence, first I see what the sequence does. If you take for example the subsequence $a(n) = d(10^n)$ you see that it gives you the decimal representation of $x$ starting with the $n'th$ digit. Example: let $x=\pi$, then we get $a(0)=0.141592..., a(1)=0.41592..., a(2)=0.1592...$ and so on.

Also this is a plot of $d(n)$ for $x=\pi$:

You see that for irrational x the height of the function at integers is not constant, nor periodic.

Also this question I asked several weeks ago seems to be related: Proving that an equation is solvable, Floor function.

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Your definition needs a minus in place of the plus, I believe. –  Matthew Conroy Jan 2 '11 at 21:42
Thanks for the correction, of course you are right. –  Listing Jan 2 '11 at 21:54
Our Analysis Teacher gave us a holiday sheet we could make if we wanted and this has prooved to be the only tough question on it. I investigated a bit further and found this -> at.yorku.ca/cgi-bin/… on a forum about topology. I guess it uses math.stackexchange.com/questions/10280/… to show that there is a sequence that has a limit in every open interval. Then you could choose a interval of length $2^{-n}$ to prove it i think? But I don't fully understand it –  Listing Jan 3 '11 at 10:47