On the density of the sequence $\{n \{n \xi \} \}_n$ I have a question that I can't manage to answer myself. It comes from some work in PDE theory, but it is related to analytic number theory.
Let us say that we have an irrational number $\xi$. The question is basically the following: is it true that the numbers $n \{n \xi\}$, where $\{\cdot\}$ stands for the fractional part, are dense in $\mathbb{R}^+$ as $n \in \mathbb{N}$?
If I make some numerical experiments, the answer seems to be affirmative. I know that $\{n\xi\}$ is uniformly distributed in $[0,1]$ and hence everywhere dense, but I need to know if any positive numer can be approximated by terms of the sequence $n \mapsto n \{n \xi\}$.
Any help or reference is really welcome.
Edit
As I said in the comments, I am interested in the behavior of the sequence
$$
n \mapsto \left| 2\{n\xi\}-1 \right| n. \tag{1}
$$
If I plot
$$
n \mapsto  \left( 2\{n\xi\}-1 \right)n,\tag2
$$
it seems to me that the sequence is uniformly distributed in $(-\infty,+\infty)$. Now (1) is just the absolute value of (2). Can we say anything about the absolute value of a u.d. sequence? I've found nothing in the most popular books.
 A: For a random process that produces a countable sequence of real numbers, there are two possibilities:
1) With probability one, the sequence is dense in $\mathbb R$.
2) There is some fixed interval such that the probability that there are any elements of the sequence in the interval is less than $1$.
This is obvious because it is sufficient to work with intervals with rational endpoints, and then we can apply countable additivity of measure.
So if you want to show the sequence is dense, you just have to rule out case 2. Fix an interval, and consider the random variable that is the number of $n \leq N$ such that the $n$th element of your sequence is in that interval. We want to show that the probaiblity this random variable vanishes goes to $0$ as $N$ goes to $\infty$.
You seem to care about two different sequences, $n\{n \xi\}$ and $2n\{n \xi\}-n$. For either one, the expected value of this variable is proportional to $\log n$ and goes to $\infty$. Just a little bit more information on this variable could be sufficient to upper bound the probability of vanishing. One natural approach is to try to show the variance is $o\left((\log n)^2\right)$.
To do that, we need to upper bound the probability that simultaneously $|\{n \xi\} -a/n | < b/n$ and $|\{m \xi \} - a/m | < b/m$. I'm not yet sure how to do this.
A: Just some loosely connected plots of interest.  First of all $\sum_{n < N} \{ n \xi\}^2 = \frac{1}{12}N + O(1)$ which can be seen from the mean ergodic theorem without the $O(1)$ error term.

If we change one letter and plot $\sum_{n < N} \{ n^2 \xi\}^2 = \frac{1}{12}N + O(?)$ a much noisier plot emerges.

Then I tried to plot your last equation $\sum_{n \leq N} 2n \{ n \xi\} - n $ and a pleasant figure emerges.  Personally I don't understand why those partial sums are bounded (almost everywhere) in $\xi$.

