TITLE: Find an example of series which converges only absolutely on $\mathbb Q$
QUESTION [8 upvotes]: I am currently working on the completeness of metric spaces, so I studied the following theorem:

If $E$ is a Banach space then any absolutely convergent series is convergent.

Since $\mathbb Q$ is not complete, I wanted to illustrate this theorem with an absolutely convergent series on $\mathbb Q$ which wouldn't converges:
$$\sum_{n=0}^\infty \vert r_n\vert\in\mathbb Q\quad\text{ and }\quad \sum_{n=0}^\infty r_n\in\mathbb R\setminus\mathbb Q \quad (\text{ with $r_n\in \mathbb Q$}).$$
Is was not able to find such series. Is there an example like it (as simple as possible ?).

REPLY [2 votes]: The relationship between convergence and absolute convergence in an ordered field is studied in this note, which has its genesis in a question that was asked on this site (by me; it was answered by my coauthor, N.J. Diepeveen).
The case of subfields of $\mathbb{R}$ makes for a nice undergraduate exercise (as seems to be happening here!): see Theorem 1.  Or let me explain: let $x \in [-1,1] \setminus \mathbb{Q}$.  Then there is a sequence (in fact a unique sequence, since $x \notin \mathbb{Q}$, but this is not needed for what follows) $\epsilon_n \in \{ \pm 1\}$ such that 
$x = \sum_{n=1}^{\infty} \frac{\epsilon_n}{2^n}$.
The way you get $\epsilon_n$ is by a simple inductive process: for $n \geq 0$, 
having chosen $\epsilon_1,\ldots,\epsilon_n$ already, we choose $\epsilon_{n+1}$ to be $1$ if $\sum_{k=1}^n \frac{\epsilon_k}{2^k} < x$ and 
$-1$ if $\sum_{k=1}^n \frac{\epsilon_k}{2^k} > x$.  A little thought shows that this converges to $x$.  Now notice that no matter what the sign sequence $\epsilon_n$ is, the absolute series is 
$\sum_{n=1}^{\infty} \frac{1}{2^n} = 1 \in \mathbb{Q}$.
Note that this argument works equally well with $\mathbb{Q}$ replaced by any proper subfield $F \subsetneq \mathbb{R}$, or in fancier terms, any incomplete Archimedean ordered field.  
Note also that this is essentially the same answer as @Martín-Blas Pérez Pinilla, but I wanted to give my take on it.