TITLE: An absolutely convergent series of rational numbers which does not converge to a rational number
QUESTION [19 upvotes]: A standard theorem concerning series of real numbers states that every absolutely convergent series of real numbers converges. I would like to know a counterexample to this statement when we are dealing only with rational numbers. More precisely, I would like to know an example of a series $\sum_{n=0}^\infty q_n$ of rational numbers such that $\sum_{n=0}^\infty|q_n|$ converges to a rational number and that $\sum_{n=0}^\infty q_n$ converges to an irrational number. Furthermore, I want that the reason why the example works is understandable by someone who is only aware of basic statements concerning series.
If it wasn't for the last requirement, I would know how to do it. One possibility would be to consider the power series$$\sum_{n=0}^\infty\binom{-1/2}nx^n,$$which converges to $1/\sqrt{1+x}$ in $(-1,1)$. In particular,$$\sum_{n=0}^\infty\binom{-1/2}n\left(\frac34\right)^n=\frac2{\sqrt7}\notin\Bbb Q.$$But\begin{align}\sum_{n=0}^\infty\left|\binom{-1/2}n\left(\frac34\right)^n\right|&=\sum_{n=0}^\infty(-1)^n\binom{-1/2}n\left(\frac34\right)^n\\&=\sum_{n=0}^\infty\binom{-1/2}n\left(-\frac34\right)^n\\&=2\in\Bbb Q.\end{align}Another possibility consists in using a counting argument (although this only proves that a counter-example exists, rather than exhibiting one). The numbers of the form$$\sum_{n=0}^\infty\frac{\varepsilon_n}{3^n},$$where $(\varepsilon_n)_{n\in\Bbb Z_+}$ is a sequence which takes only the values $1$ and $-1$, form an uncountable set. So, for some sequences $(\varepsilon_n)_{n\in\Bbb Z_+}$, the sum is irrational. But$$\sum_{n=0}^\infty\left|\frac{\varepsilon_n}{3^n}\right|=\frac32\in\Bbb Q.$$

REPLY [5 votes]: Here is a rather interesting (and rather trivial) observation about which Martin touched upon.
Any number $x\in(0,1)$ admits a unique expression as $x=\sum^\infty_{k=1}2^{-k}\varepsilon_k$ where $\varepsilon_k\in\{0,1\}$, and $\sum_k\varepsilon_k=\infty$. Let $\delta_k:=2\varepsilon_n-1$. Then $\delta_k\in\{-1,1\}$ and
$$x=\sum^\infty_{k=1}\frac{\delta_k+1}{2}2^{-k}=\frac12+\sum^\infty_{k=1}\delta_k2^{-k-1}$$
Then, if $(X_n:n\in\mathbb{N})$ is a sequence of Bernoulli $B(\pm1,1/2)$ random variables, then with probability $1$ is $\sum_n2^{-n}X_n$ irrational, yet $\sum_n|2^{-n}X^n|=1$ is rational.
That is, the series that satisfy the conditions oulined by JC are typical.