Let $Q=\mathbb Q \cap(0,1)= \{r_1,r_2,\ldots\}$ be the rational numbers in $(0,1)$ listed out so we can count them. Define $x_n=\frac{1}{n}\sum_{k=1}^nr_n$ to be the average of the first $n$ rational numbers from the list.

**Questions:**

What is required for $x_n$ to converge? Certainly $0< x_n < 1$ for all $n$.

Does $x_n$ converge to a rational or irrational?

How does the behavior of the sequence depend on the choice of list? I.e. what if we rearrange the list $\mathbb Q \cap(0,1)=\{r_{p(1)},r_{p(2)},\ldots\}$ with some one-to-one permutation $p: \mathbb N \to \mathbb N$? How does the behavior of $x_n$ depend on $p$?

**My thoughts:**

Intuitively, I feel that we might be able to choose a $p$ so that $x_n\rightarrow y$ for any $y\in[0,1]$. However, it also makes intuitive sense that, if each rational appears only once in the list, that the limit is required to be $\frac{1}{2}.$ Of course, intuition can be very misleading with infinities!

If we are allowed to repeat rational numbers with arbitrary frequency (but still capturing every rational eventually), then we might be able to choose a listing so that $x_n\rightarrow y$ for any $y\in(0,\infty)$.

This last point might be proved by the fact that every positive real number has a sequence of positive rationals converging to it, and every rational in that list can be expressed as a sum of positive rationals less than one. However, the averaging may complicate that idea, and I'll have to think about it more.

**Example I:**

No repetition: $$Q=\bigcup_{n=1}^\infty \bigcup_{k=1}^n \left\{\frac{k}{n+1}\right\} =\left\{\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\ldots\right\}$$ in which case $x_n\rightarrow\frac{1}{2},$ a very nice and simple example. Even if we keep the non-reduced fractions and allow repetition, i.e. with $Q=\{\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\boxed{\frac{2}{4},}\frac{3}{4},\frac{1}{5},\ldots\},$ then $x_n\rightarrow\frac{1}{2}.$ The latter case is easy to prove since we have the subsequence $x_{n_k}=\frac{1}{2}$ for $n_k=\frac{k(k+1)}{2},$ and the deviations from $1/2$ decrease. The non-repetition case, I haven't proved, but simulated numerically, so there may be an error, but I figure there is an easy calculation to show whether it's correct.

**Example II:**

Consider the list generated from the Stern-Brocot tree: $$Q=\left\{\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{5},\frac{3}{5},\frac{3}{4},\ldots\right\}.$$

I'm sure this list could be studied analytically, but for now, I've just done a numerical simulation. The sequence of averages $x_n$ hits $\frac{1}{2}$ infinitely often, but may be oscillatory and hence not converge. If it converges, it does so much slower than the previous examples. It appears that $x_{2^k-1}=0.5$ for all $k$ and that between those values it comes very close to $0.44,$ e.g. $x_{95743}\approx 0.4399.$ However, my computer code is probably not very efficient, and becomes very slow past this.