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.

• @trisct I know what my space is and is not. It is exactly what I am saying. That should is not justified by the two properties mentioned. Dec 25, 2019 at 2:43
• The author of the linked question provided one in his article.
– user239203
Dec 25, 2019 at 2:54
• @MoonLightSyzygy: Ok consider this: as proved in the linked answer "(A) if every absolutely convergent series in $\mathbb {Q}$ converges in $\mathbb {Q}$ then (B) every Cauchy sequence in $\mathbb {Q}$ converges in $\mathbb {Q}$". Since (B) is false it follows that (A) is also false and hence there must be some absolutely convergent series in $\mathbb {Q}$ which does not converge in $\mathbb{Q}$. Dec 25, 2019 at 6:32
• @MoonLightSyzygy: the set in your example is not an ordered field while $\mathbb {Q}$ is one. Dec 25, 2019 at 6:34
• @Gae.S.: I am adding that example as an answer here for completeness. Dec 25, 2019 at 6:43

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.

• Yours is the simplest example so far. +1 Dec 25, 2019 at 2:27
• @ParamanandSingh It was at first a lucky guess. But on second thought, perhaps it's just like Steven Stadniciki' answer, that there are uncountably many choices of signs and only countably many rationals. So taking any positive series convergent to a rational and switching the signs has a large probability of obtaining an irrational sum. I think to prove his argument one has to prove that there are still uncountably many results modulo the relation that different choices lead to the same sum. In view of that, a concrete example might be more suitable. Dec 25, 2019 at 2:37
• @ParamanandSingh Made a litle improvement. You can now find such a series that converges to any irrational number you want. Dec 25, 2019 at 4:35
• How can you prove the values of your $\log2$ sequences? Dec 25, 2019 at 8:27
• Dec 25, 2019 at 9:02

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.

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 $$1$$.

• Since $(-1)^{n^2}=(-1)^n$ your second series is also a geometric progression with sum $(-1/2)/(1+(1/2))=-1/3$. Dec 25, 2019 at 2:13
• Don't you need to prove first that different choices of signs lead to a different sum? Dec 25, 2019 at 2:16
• @ParamanandSingh I am sure the intention was to choose only the $n^2$-th term to be negative. Or any other non-periodic sparse choice. Dec 25, 2019 at 2:42
• Or maybe a weaker condition like: choices modulo the relation of giving the same sum still give uncountably many results. Dec 25, 2019 at 2:43
• @MoonLightSyzygy: I think you are right. Dec 25, 2019 at 6:22

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

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.

• I suppose this doesn't quite work because the first term is no longer rational. Oh well, still an interesting idea maybe. Dec 25, 2019 at 2:29

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.