Absolute convergence to a rational number

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $1$.

Solution 4:

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 $$\sum_{k=1}^{\infty} \frac{(-1)^{b_k}}{10^k}=-\sum_{k=1}^{\infty}\frac{1-(-1)^{b_k}}{10^k}+\sum_{k=1}^{\infty} \frac{1}{10^{k}}=-2b+\frac{1}{9}$$