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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.

Paramanand Singh
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  • It is ok if the theorem motivated asking your question, but the 'there should' is not justified by $\mathbb{Q}$ not being complete. Potentially all series that converge absolutely to rationals also converge to rationals without the absolute values, while only some series that didn't converge absolutely converge to irrationals both with the absolute values and without. For example, ${x\in\mathbb{R}:\ x\geq 0}\setminus{1}$ is not complete and every series that converges absolutely also converges and both to the same element in the set. – MoonLightSyzygy Dec 25 '19 at 02:35
  • Only series that don't converge absolutely, like $\sum_{n=1}^{\infty}2^{-n}$, have their sum of absolute values and their sum without them in the complement. – MoonLightSyzygy Dec 25 '19 at 02:36
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    @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. – MoonLightSyzygy Dec 25 '19 at 02:43
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    The author of the linked question provided one in his article. –  Dec 25 '19 at 02:54
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    @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} $. – Paramanand Singh Dec 25 '19 at 06:32
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    @MoonLightSyzygy: the set in your example is not an ordered field while $\mathbb {Q} $ is one. – Paramanand Singh Dec 25 '19 at 06:34
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    @Gae.S.: I am adding that example as an answer here for completeness. – Paramanand Singh Dec 25 '19 at 06:43

6 Answers6

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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.

trisct
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    Yours is the simplest example so far. +1 – Paramanand Singh Dec 25 '19 at 02:27
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    @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. – trisct Dec 25 '19 at 02:37
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    @ParamanandSingh Made a litle improvement. You can now find such a series that converges to any irrational number you want. – trisct Dec 25 '19 at 04:35
  • How can you prove the values of your $\log2$ sequences? – ViHdzP Dec 25 '19 at 08:27
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    @URL https://math.stackexchange.com/q/948542/669152 and https://math.stackexchange.com/q/716/669152 – trisct Dec 25 '19 at 09:02
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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.

ViHdzP
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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 $1$.

Steven Stadnicki
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    Since $(-1)^{n^2}=(-1)^n$ your second series is also a geometric progression with sum $(-1/2)/(1+(1/2))=-1/3$. – Paramanand Singh Dec 25 '19 at 02:13
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    Don't you need to prove first that different choices of signs lead to a different sum? – trisct Dec 25 '19 at 02:16
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    @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. – MoonLightSyzygy Dec 25 '19 at 02:42
  • Or maybe a weaker condition like: choices modulo the relation of giving the same sum still give uncountably many results. – trisct Dec 25 '19 at 02:43
  • @MoonLightSyzygy: I think you are right. – Paramanand Singh Dec 25 '19 at 06:22
  • Eep — yes, that was what I meant; mea culpa! (And @triact, the fact that different choices of signs lead to a different sum modulo eventual-constantness of the sequence is straightforward to prove; it's just a slightly scaled version of the binary expansion...) – Steven Stadnicki Dec 25 '19 at 14:47
  • @StevenStadnicki I do believe it is true. But it does seem obvious. Is it possible for you to provide a reference? – trisct Dec 25 '19 at 16:25
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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 $$\sum_{k=1}^{\infty} \frac{(-1)^{b_k}}{10^k}=-\sum_{k=1}^{\infty}\frac{1-(-1)^{b_k}}{10^k}+\sum_{k=1}^{\infty} \frac{1}{10^{k}}=-2b+\frac{1}{9}$$

Community
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Paramanand Singh
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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.

tehtmi
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  • I suppose this doesn't quite work because the first term is no longer rational. Oh well, still an interesting idea maybe. – tehtmi Dec 25 '19 at 02:29
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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.

Drinzjeng Triang
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