An absolutely convergent series of rational numbers which does not converge to a rational number

Question

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$$\sum_{n=0}^\infty\left|\frac{\varepsilon_n}{3^n}\right|=\frac32\in\Bbb Q.$$

Answer 1

Here’s one that’s perhaps more in the spirit of direct demonstration that you were looking for:

$$ \sum_{k=1}^\infty\frac1{k(k+1)}=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)=1\;, $$

$$ \sum_{k=1}^\infty\frac{(-1)^k}{k(k+1)}=\sum_{k=1}^\infty(-1)^k\left(\frac1k-\frac1{k+1}\right)=1+2\sum_{k=1}^\infty\frac{(-1)^k}k=1-2\log2\;. $$

Answer 2

Inspired by Gareth Ma's answer:

Let $x \in (0, 1)$ be any irrational number and $$ x = \sum_{n=1}^\infty \frac{a_n}{2^n} \,, \quad a_n \in \{ 0, 1 \} $$ its base-2 representation Then $\epsilon_n = 2a_n-1 \in \{ -1, +1 \}$, $$ \sum_{n=1}^\infty \frac{\epsilon_n}{2^{n+1}} = x - \frac 12 $$ is irrational, whereas $$ \sum_{n=1}^\infty \frac{1}{2^{n+1}} = \frac 12 $$ is rational.

Answer 3

I believe your example with $\epsilon_n$ is fine. In particular, we can use the sequence $(\epsilon_n) = (\color{red}{1}, -1, \color{red}{1}, -1, -1, \color{red}{1}, -1, -1, -1, \color{red}{1}, -1, -1, -1, -1, \color{red}{1}, \ldots)$, which is irrational by looking at the base-$3$ expansion of the number, as it is aperiodic.

Answer 4

I suspect this might draw the same comment from you as the other two existing answers, but like Martin R’s answer it allows you to construct an example for any irrational number $x\in(0,1)$: Choose $q_n=\epsilon_n2^{-n}$ with $\epsilon_n\in\{-1,1\}$, so $\sum_n|q_n|=1\in\mathbb Q$, and choose $\epsilon_n=\operatorname{sgn}\left(x-\sum_{k=1}^{n-1}q_k\right)$.

Since $\sum_{k=n+1}^\infty2^{-k}=2^{-n}$, there is always enough sum left to correct for the overshoot, much like in the proof of the Riemann rearrangement theorem.

Answer 5

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.