Absolute convergence to a rational number Let's recall the not so popular/familiar form of completeness of real numbers:

Theorem: Absolute convergence of a series implies its convergence.

Since $\mathbb{Q} $ is not complete there should exist a series $\sum_{n=1}^{\infty} u_n$ with rational terms such that $\sum_{n=1}^{\infty} |u_n|$ converges to a rational number and $\sum_{n=1}^{\infty}u_n$ converges to an irrational number.

I could not think of an obvious example of such a series. Please provide one such example.

 A: Every irrational number in Balanced ternary has a non-repeating expansion, and viceversa. Therefore, if we take a non-repeating sequence $(e_n)_{n\in\mathbb Z^+}$ with $e_n\in\{-1,1\}$, $$\sum_{k=1}^\infty\frac{e_k}{3^k}$$ will be irrational, while $$\sum_{k=1}^\infty\left|\frac{e_k}{3^k}\right|= \sum_{k=1}^\infty\frac1{3^k}=\frac12$$ will be rational.
A: An abstract example: choose $|u_n|=2^{-n}$ so that $\sum_n|u_n|=1$; now there are $\mathfrak{c}$ choices of signs for the $u_n$ but only $\aleph_0$ rationals, so almost all choices you can make lead to irrational numbers. In particular, $\sum_n(-1)^{sq(n)}2^{-n}$, where $sq(n)$ is $0$ if $n$ is a square and $1$ if it isn't, must be irrational (prove this!) but the sum of absolute values is $1$.
A: In this article the author mentions the following example.
Let $b$ be an irrational number in $(0,1)$ whose decimal representation $$b=0.b_1b_2\dots b_k\dots=\sum_{k=1}^{\infty} \frac{b_k} {10^k}$$ consists of only the digits $0$ and $1$. Then the series $$\sum_{k=1}^{\infty} (-1)^{b_k}\cdot\frac{1}{10^k}$$ has the desired property since $$\sum_{k=1}^{\infty} \left|\frac{(-1)^{b_k}}{10^k}\right|=\sum_{k=1}^{\infty} \frac{1}{10^k}=\frac{1}{9}$$ and $$\sum_{k=1}^{\infty} \frac{(-1)^{b_k}}{10^k}=-\sum_{k=1}^{\infty}\frac{1-(-1)^{b_k}}{10^k}+\sum_{k=1}^{\infty} \frac{1}{10^{k}}=-2b+\frac{1}{9}$$
A: Let $a_n=\frac{1}{n(n+1)}$. Then
$$\sum_{n=1}^\infty a_n=1$$
$$\sum_{n=1}^\infty a_{2n-1}=\log2$$
$$\sum_{n=1}^\infty a_{2n}=1-\log2$$
$$\implies a_1-a_2+a_3-…=2\log2-1$$

A more general conclusion: This is similar to the other two answers. Consider $a_n=2^{-n}$. Choose any irrational number $x$ in $(0,1)$ and consider its binary representation, i.e. find a subsequence $\{a_{n_k}\}$ such that $x=\sum_{k=1}^\infty a_{n_k}$. Now define $b_m=a_m$ if $m$ is one of the $n_k$ and $b_m=-a_m$ if not. You can check that $\sum b_m=2x-1$, where $x$ is the irrational number chosen at first.
A: If the series $a_1 + a_2 + a_3 + ...$ sums to an irrational number and all terms are positive and rational, consider $$\frac{a_1}{2} - \frac{a_1}{2} + \frac{a_2}{2} - \frac{a_2}{2} + ...$$
The should converge absolutely to the same irrational number and converge normally to zero (rational). This is the opposite of what you want, so you can add e.g. one plus the negative of the fractional part of the absolute sum to the first term to switch it.
A: Let $a_n^+$ be $\dfrac{a_n+|a_n|}{2}$ and $a_n^-$ be $\dfrac{a_n-|a_n|}{2}$. It is obvious that $a_n^++a_n^-=a_n$ and $a_n^+-a_n^-=|a_n|$. Since both $\sum_{n=1}^{\infty} a_n^+$ and $\sum_{n=1}^{\infty} a_n^+$ are converged. It is easy to construct a sequence, whatever the given rational number and the irrational number are. 
