Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational. Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?

By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.

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    Does the series $2=\mathrm{e}^{\ln(2)}=\sum_{k=0}^{+\infty}\frac{\ln(2)^n}{n!}$ count? – gniourf_gniourf Feb 09 '16 at 11:58
  • Oh sorry, we may take $S \neq e^{\ln x}$ for the sake of convenience. – User1 Feb 09 '16 at 12:01
  • @Andreas, sorry i can't quite get you..? – User1 Feb 09 '16 at 12:09
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    Beware that assuming $S\neq e^{\ln x}$ for some rational $x$ is the same as assuming $S\neq x$ for some rational $x$. – Nate River Feb 09 '16 at 12:11
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    Would $e^{ln(2)}+1$ be acceptable? Note: excluding this whole category of examples seems a little odd. – lulu Feb 09 '16 at 12:18
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    Take the sequence of inverse square roots of primes. For any positive target value `S` (rational or not), you can find an infinite subsequence whose sum is `S`. – Mark Dickinson Feb 09 '16 at 12:31
  • @exclude "looking trivial": Then you might want to exclude that any term is polynomial in the previous terms/addens. To make it even more impossible to find constructive examples, exclude that any term is algebraic over the previous terms. (all with rational coefficients) – Lutz Lehmann Feb 09 '16 at 12:33
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    If we assume that $S$ is rational, and that $\forall x. S \neq e^{\ln x}$, then Obama is a lizard. Or if that edit meant $\exists x. S \neq e^{\ln x}$ then that seems like a remarkably pointless thing to assume. – user253751 Feb 10 '16 at 02:46
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    An infinite sum of rational numbers can be irrational ! – Saikat Feb 10 '16 at 13:36
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    This is a very bad, unclear question, that somehow generated interesting answers, IMO. – Lynn Feb 10 '16 at 22:50
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    @Lynn, that's your own opinion, which seems not to be shared by 4500 people who have viewed it before you ! – User1 Feb 10 '16 at 22:55
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    No, you *really really* need to clear up the thing where you write $S$ both is and isn't rational, and explain why $e^{\ln 2}$ was unacceptable. You seem to be just swatting away an answer that is *too easy* — asking us to play a game without telling us the rules... – Lynn Feb 10 '16 at 23:02
  • I can't seem to see where i ever wrote that. Can you state it ? – User1 Feb 10 '16 at 23:05
  • @user26857, what is ''abstract'' can be subject to debate, but irrationality theorems are inseparable with abstract algebra. A good example is Lioville's theorem. You may also check Apery's proof for the irrationality of $\zeta(3)$. – User1 Feb 12 '16 at 20:44
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    @Lynn I agree the answers are super cool, but shouldn't we close this as 'unclear what you are asking'? I can't b.c. rep, but it seems like the right thing to do... – Please stop being evil Feb 14 '16 at 07:28
  • It can. Actually, it always is! –  Feb 14 '16 at 11:23
  • Telescope: Take any series of numbers were (a + b) + (c + d) + ... is an infinite series summing to a rational number. Then, by having chosen numbers to telescope, answer where (a) + (b + c) + (d + e) + ... is an infinite series for which every number is irrational. But it has been constructed in a way that can only give a rational sum. – Christos Hayward Feb 14 '16 at 22:27
  • How about this: The sum of the two irrational numbers $\pi$ and $10-\pi$ is rational. To that add $\pi/2$ and then $(10-\pi)/2.$ Then to that add $\pi/4$ and $(10-\pi)/4,$ and then $\pi/8$ and $(10-\pi)/8,$ and so on. The sum is $20.$ – Michael Hardy Nov 21 '17 at 19:13
  • The answer is trivial: take any infinite sum of irrational numbers. If it converges to a rational, you are done. Otherwise, modify a single term by adding the difference to a rational. –  Feb 13 '20 at 07:46

24 Answers24


EDIT: Pardon me, but it has been shown in the comments by robjohn and Michael that these are not linearly independent. Indeed:$$91a_1-10a_2=10$$     — Akiva Weinberger

Think of a series of real numbers with decimal expansions like


That is, a given digit is only 1 in one the numbers in the series, and 0 everywhere else, and distributed like the above.

All those numbers are irrational because their decimal expansion never repeats, they are linearly independent, and their sum is 1/9 = 0.111111...

EDIT: Ángel Valencia proposes the following, unfortunately also without proof. It seems likely to work to me, but I (RemcoGerlich) am working on my own fix with proof.



Akiva Weinberger
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    This is extraordinarily clever; that is, finding a logical inverse of the non-algebraic sums that yield transcendental numbers. – Joshua Feb 09 '16 at 16:16
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    This sequence does not satisfy the rational independence: $10a_1-a_2=1$ – robjohn Feb 09 '16 at 18:14
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    @robjohn: crap, you're right of course. Does my new pattern fix it, you think? – RemcoGerlich Feb 09 '16 at 18:53
  • I don't see a problem right off. However, I cannot look more until later. – robjohn Feb 09 '16 at 19:51
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    @Joshua: thank you! I'm just a programmer / CS guy and strings of 0s and 1s seem natural to me :-) – RemcoGerlich Feb 09 '16 at 21:11
  • Strings of 0s and 1s seems like a very promising way to approach a problem like this. – David K Feb 09 '16 at 21:30
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    @RemcoGerlich: although I don't see anything obvious, it would really be nice to see a proof that the terms in the sequence are rationally independent. – robjohn Feb 10 '16 at 08:56
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    I wonder why this answer got downvoted (without the downvoter leaving a comment). – gniourf_gniourf Feb 10 '16 at 10:28
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    @gniourf_gniourf If I had to guess, it would be because of that lack of proof about the linear independence. – Akiva Weinberger Feb 10 '16 at 12:03
  • @AkivaWeinberger: quite, I wish I could think of a way to prove it, but I don't believe I can. I am amazed this is getting so many upvotes. – RemcoGerlich Feb 10 '16 at 12:33
  • Well, it's very elegant. – Akiva Weinberger Feb 10 '16 at 12:48
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    @gniourf_gniourf I downvote it because of the missing proof about the rationnal independance. There are very nice valid proofs bellow without such a number of upvotes, so I admit I use it as a corrective bias as well (although it doesn't matter much now). – matovitch Feb 10 '16 at 13:10
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    @RemcoGerlich I think you should ask this as a separate question on this site. "Are the following real numbers linearly independent? etc." – Akiva Weinberger Feb 10 '16 at 13:56
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    $3a_1-2a_2=2$, I think, as the other 1's cancel. (if that is binary, otherwise $91a_1-10a_2=10$ – Empy2 Feb 10 '16 at 14:59
  • @matovitch I fully understand your point! thanks for the comment. – gniourf_gniourf Feb 10 '16 at 16:06
  • @Michael, I think it is not exactly it but it works $14a_1-4a_2=9$ in binary. – matovitch Feb 10 '16 at 16:10
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    @RemcoGerlich: again, these are not rationally independent. $$a_1 =11\sum_{n=0}^\infty 10^{-n^2-3n-2}$$ and $$a_2 =-1+1001\sum_{n=0}^\infty 10^{-n^2-3n-3}$$ This means that $$91a_1-10a_2=10$$ – robjohn Feb 10 '16 at 16:29
  • @Michael: I see that you have already pointed this out. – robjohn Feb 10 '16 at 16:43
  • @RemcoGerlich: if you are going to change this to something even more complicated, it would be extremely nice if you would provide a proof of the rational independence, since the first two attempts did not work. – robjohn Feb 10 '16 at 16:45
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    We can (nonconstructively!) fix rational independence while maintaining the spirit of this example. For $a_1$, pick any irrational number that has infinitely many 0s and infinitely many 1s in its decimal expansion. Now, we will successively let $a_k$ have these properties: all digits 0 or 1; all digits where an earlier $a$ was 1 are 0; first location that's zero in all earlier $a$ is 1; infinitely many of both 0,1. There are obviously uncountably many such $a_k$. Excluding rational $a_k$ and ones satisfying a rational linear relation with earlier ones leaves uncountably many, hence >0. – Gareth McCaughan Feb 10 '16 at 16:49
  • @robjohn I have taken the liberty of writing a small notice at the top. RemcoGerlich, I hope you're OK with that. – Akiva Weinberger Feb 10 '16 at 17:35
  • Sorry, I commit some mistakes in my example, but I hope you would see the pattern. I will work in a proof of it. – Ángel Valencia Feb 10 '16 at 20:01
  • @ÁngelValencia I think with your proposition you have $11011000000a_1-a_2 = 1102201100$. – matovitch Feb 10 '16 at 21:06
  • @ÁngelValencia Nope, in fact it is wrong...sorry. :/ – matovitch Feb 10 '16 at 21:29
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    @ÁngelValencia $110110a_1 - 1000a_2=11011$ (this times it is checked...sorry for flooding) – matovitch Feb 10 '16 at 21:35
  • Errata: Where it says "Well, my last try, ..." it must say "Well, it is my last try, ...". – Ángel Valencia Feb 11 '16 at 01:23
  • @matovitch Don't worry, I edit the answer with a better "example". I think it is a real example, but I can't prove it. – Ángel Valencia Feb 11 '16 at 01:25
  • @matovitch downvoting because you think an answer is lacking is okay. Downvoting something you wouldn't ordinarily, to balance out "too many upvotes" is pretty well established as not-okay. Please don't, or at least don't suggest that's what you're doing (votes are private after all) – hobbs Feb 12 '16 at 18:42
  • You need no specific series. Just show that there can exist some rationally independent irrational numbers with the desired properties and leave finding such a series as an exercise to the reader. – Please stop being evil Feb 14 '16 at 07:26
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    Just to support matovitch's comment, $$ a_1=\sum_{k=0}^\infty 10^{-\left.\left(2n^3+9n^2+7n+6\right) \middle/6\right.} $$ and $$ a_2=-11.011 +11011\sum_{k=0}^\infty 10^{-\left.\left(2n^3+9n^2+7n+18\right)\middle/6\right.} $$ So we get $$110110a_1 -1000a_2 =11011$$ – robjohn Feb 14 '16 at 08:58
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    @hobbs We all know the vote game is biased, the highest voted questions getting more views and more votes. At the time I downvoted this one, it had about the same number of upvotes as the answer bellow. In my mind, it was making an unfair game fairer. It's totally fine IMHO for people to vote for reasons I don't agree with and it's better if they are transparent about it so one can confront our views. – matovitch Feb 15 '16 at 19:13
  • A variation of this idea. Take $a_n$ to be the number whose $i$th digit is $1$ if $i$ is a power of the $n$th prime and otherwise $0$. And $a_0$ has the digits which are missing. The numbers $a_n$ are irrational and seem to be linearly independent to me. But I haven't wrote it down. – user60589 Feb 16 '16 at 16:29

If $e^{\ln x}$ is not allowed, we can use another function's Maclaurin series. For example $$\tan \frac{\pi}4=\sum_{n=0}^\infty \frac{(-1)^n 2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}\left(\frac{\pi}4\right)^{2n+1}=1.$$ Note that $(-1)^nB_{2n+2}$ is positive for all $n\in \mathbb{N}$. That guarantees that all terms in the series are positive.

Rasmus Erlemann
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  • Note that $B_{2n+2}<0$ for $n\in \{2,4,6,\ldots \}$. – Rasmus Erlemann Feb 09 '16 at 12:29
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    I don't recognize those coefficients B2n+2. Can you provide a link? – Pieter Geerkens Feb 10 '16 at 08:51
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    @PieterGeerkens [Bernoulli numbers](https://en.m.wikipedia.org/wiki/Bernoulli_number) – Akiva Weinberger Feb 10 '16 at 12:04
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    And the linear independence of the terms follows from the transcendence of $\pi$, which is the cleverest part of the answer. In fact any power series with rational coefficients, evaluated at a transcendental value but having a rational sum, would have done. It makes me wonder if there are "super-transcendental" numbers, which are not roots of any power series with rational coefficients. – Jack M Feb 10 '16 at 15:37
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    @JackM See [here](http://mathoverflow.net/q/42449/20948) or [here](http://math.stackexchange.com/a/896902) – Yai0Phah Feb 11 '16 at 19:22

$$ \begin{align} 1 &=\log(e)\\ &=-\log\left(1-\left(1-\frac1e\right)\right)\\ &=\sum_{k=1}^\infty\frac1k\left(1-\frac1e\right)^k \end{align} $$ Since $e$ is transcendental, no finite rational combinations of the terms can be $0$.

Suppose that some finite rational combination of the terms were $0$, then for some $\{a_k\}\subset\mathbb{Q}$ $$ \begin{align} 0 &=\sum_{k=1}^n\frac{a_k}k\left(1-\frac1e\right)^k\\ &=\sum_{k=1}^n\sum_{j=0}^k\frac{a_k}k\binom{k}{j}(-1)^je^{-j}\\ &=\sum_{k=1}^n\left[\frac{a_k}k+\sum_{j=1}^k\frac{a_k}k\binom{k}{j}(-1)^je^{-j}\right]\\ &=\sum_{k=1}^n\frac{a_k}k+\sum_{j=1}^n\left[\sum_{k=j}^n(-1)^j\frac{a_k}k\binom{k}{j}\right]e^{-j} \end{align} $$ Therefore, $$ 0=\left[\sum_{k=1}^n\frac{a_k}k\right]e^n+\sum_{j=1}^n\left[\sum_{k=j}^n(-1)^j\frac{a_k}k\binom{k}{j}\right]e^{n-j} $$ which is impossible since $e$ is transcendental.

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    How does that follow directly from $e$ being transcendental? – user253751 Feb 11 '16 at 09:55
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    If a finite rational combination of the terms was equal to 0 then we would have a rational polynomial to which 1/e was a soloution. That would mean that 1/e is algebraic and hence that e is algebraic. Since e is transcendental this would be a contrauction. (someone please check this reasoning is correct an if it is feel free to add it to the answer) – Peter Green Feb 11 '16 at 10:20
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    @PeterGreen: other than multiplying by a power of $e$ to get a non-trivial element of $\mathbb{Q}[e]$ equal to $0$, that is the argument I was considering. If you think it needs to be added to the answer, I will do so. – robjohn Feb 11 '16 at 12:44

\begin{align} 1&=\frac{\sqrt2}2+\frac{\sqrt3}6+\frac{\sqrt5}{531}+\frac{\sqrt7}{376169}+\dotsb\\ &=\sum_{p\text{ prime}}\frac{\sqrt p}{a_p} \end{align} where $a_p$ is a certain sequence; it's not hard to show that there exists a sequence $a_p$ that satisfies the above. (In fact, there are infinitely many that work.)

Each of those terms are linearly independent.

Akiva Weinberger
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    What is this (type of) series called? Links would make your answer better. – Foo Bar Feb 09 '16 at 14:49
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    Does it need a name? It's just an infinite sum. I didn't use any links. – Akiva Weinberger Feb 09 '16 at 14:50
  • Out of curiosity, how would you prove your claim that there exist infinitely many sequences $a_p$? – user3002473 Feb 09 '16 at 17:33
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    @user3002473 For one thing, we can set $a_2$ to be any integer greater than or equal to $2$, and change the rest of the sequence accordingly. (I think there's actually _uncountably_ many such sequences, but that would require a more delicate argument.) – Akiva Weinberger Feb 09 '16 at 17:57
  • @user3002473 The sequence that I used was defined inductively; given $a_{p_i}$ for all primes $p_i – Akiva Weinberger Feb 09 '16 at 17:59
  • @AkivaWeinberger so partial sums are bounded by 1 and monotonically increasing but how will you show the limit is actually 1? – Paladin Feb 09 '16 at 18:37
  • @Debanil I haven't filled in the details, but I'm pretty sure it's doable. I think that the if sum (call it $S$) is less than $1$, you can find a prime $p$ such that $S-\frac{\sqrt p}{a_p}+\frac{\sqrt p}{a_p-1}<1$, implying that $a_p$ isn't actually the smallest number satisfying the condition in my last comment. EDIT: Yeah, I'll finish the proof in a little bit, but I can't talk now. – Akiva Weinberger Feb 09 '16 at 18:43
  • @Debanil The square root ensures that the summands are irrational. Picking distict primes ensures that there is no rational relation among the summnds. – Hagen von Eitzen Feb 09 '16 at 18:52
  • oh sorry totally lost myself in that example! :P – Paladin Feb 09 '16 at 18:53
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    We might make the summands "even more" independent: $\sqrt{p_n}$ could be replaced by any $\alpha_n$ provided $\alpha_n>0$ and $\sum \alpha_n=\infty$. We might for instance pick $\alpha_n\in[1,2]$ such that it is transcendental over $\Bbb Q(\alpha_1,\ldots,\alpha_{n-1})$. – Hagen von Eitzen Feb 09 '16 at 19:02
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    @Debanil The inductive definition I wrote above only referred to the specific sequence I used in the original post ($(2,6,531,\dots)$). To get another sequence that has the series sum to $1$, we can let $a_2$ be any integer great than $2$, and only use the inductive definition to define $a_{p_i}$ for $p_i>2$. – Akiva Weinberger Feb 09 '16 at 19:21
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    @Debanil To show that it tends to $1$: Note that $\frac{\sqrt p}{a_p}<1$ for all $p$. Also, note that clearly the sequence $a_p$ is unbounded (or else $S$ would diverge). Now, assume $S\ne1$, and let $p$ be such that $a_p>1+\frac1{1-S}$. We have:\begin{align}S-\frac{\sqrt p}{a_p}+\frac{\sqrt p}{a_p-1}&=S+\frac{\sqrt p}{a_p(a_p-1)}\\& – Akiva Weinberger Feb 09 '16 at 20:16
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    @AkivaWeinberger your example is really stunning...... how did you get the intuition? I mean how were you so sure about the result before proving it? Was it just the optimality condition and the divergence of the prime numers? Thanks! – Paladin Feb 10 '16 at 04:42
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    @Debanil Well, after computing the first four terms, and seeing that $\frac{\sqrt2}2+\frac{\sqrt3}6+\frac{\sqrt5}{531}+\frac{\sqrt7}{376169}={}$$ 0.99999 99999977866\dots$, it seemed _very_ unlikely that it wouldn't converge to $1$. – Akiva Weinberger Feb 10 '16 at 11:47
  • As @HagenvonEitzen noticed, this example can become very general; we can impose conditions much stronger than "linear independence" if we want. I think that's basically why I thought of this; I was trying to get at an idea that's as general as possible. – Akiva Weinberger Feb 10 '16 at 11:52

Select some rational number $S$ and any sequence of linearly independent irrational numbers $x_k$. $x_k=\sqrt{p_k}$ with primes $p_k$ is one example. Then start with $S_0=0$.

The iteration assumption is $S_n=\sum_{k=1}^na_k<S$. Since $\Bbb Q·x_{n+1}$ is dense in $\Bbb R$ and disjoint from $\Bbb Q+\Bbb Q·x_1+…+\Bbb Q·x_n$ by linear independence, there is a rational number $r_{n+1}$ so that $r_{n+1}·x_{n+1}$ is between $(S-S_n)/2$ and $S-S_n$. Set $a_{n+1}=r_{n+1}·x_{n+1}$, then by linear independence $S_{n+1}<S$. This gives one example of the requested sequence of positive irrational numbers whose series converges to the rational number $S$.

Lutz Lehmann
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    This may fail the "linear combination" condition, although that condition needs explication. – Gerry Myerson Feb 09 '16 at 12:14
  • By ''linear combination'', i mean, say the numbers $\sqrt 2$ and $1$: $ \sqrt 2 + (1-\sqrt 2)=1$, which is rational. Not yet sure if Lutz's answer satisfies this. – User1 Feb 09 '16 at 12:20
  • Changed the construction to universal non-constructive "construction". If any such sequence of independent irrationals exists, then it can be made into a series whose terms satisfy the independence conditions. – Lutz Lehmann Feb 09 '16 at 12:39
  • Here you also need to prove that $S_n \ne S$ for every $n$. Whilst you may be able to do so, it is easier to require: $$(S-S_n)*1/3 < x_{n+1} < (S-S_n)*2/3$$ So, to the original poster, yes for every positive rational. – Max Murphy Feb 12 '16 at 22:12
  • No, not really. Since $S$ is rational and every partial sum irrational. – Lutz Lehmann Feb 12 '16 at 22:15
  • Using the $1/3$, $2/3$ rule this part of the proof is immune to a failure in your independence. I think that `any sequence of linearly independent irrational numbers` doesn't cut it. Ah, I see @User1 has mentioned that. Are you going to update your answer to reflect that? – Max Murphy Feb 12 '16 at 22:24
  • Not if you use the usual definition of linear independence of a set, that no finite linear combination results in zero. -- One can make that even more strict by demanding that $x_{k+1}$ is transcendental over $\Bbb Q[x_1,…,x_k]$. – Lutz Lehmann Feb 13 '16 at 08:22

Take the Taylor development of $$\sin\left(\frac\pi6\right),$$ where the terms are taken in pairs (to avoid negatives)

All terms are irrational by the transcendence of $\pi$.


Taking $a_k = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}}$

we have $\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}) +\dots = 1$

As per Mario Carneiro's suggestion in the comments, let us instead take

$a_k = \frac{1}{\sqrt{p_{k-1}}} - \frac{1}{\sqrt{p_{k}}}$,

where $p_0 = 1$ and $p_k$ for $k > 0$ is the k-th prime number ($p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$ etc.), so

$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}) +(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}) +\dots = 1$

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    This fails the rational independence test: $a_1+a_2+a_3=1-\frac1{\sqrt4}=1/2$. Try using only $\sqrt p$ for prime $p$ instead. – Mario Carneiro Feb 10 '16 at 22:15
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    @MarioCarneiro I had actually considered something like that, but notice that the OP's condition (as stated in the mathematical expression) applies only to any two terms. I would agree that restating the condition more strongly as: "any finite set of terms must be rationally independent" would probably be better. That said, I'll look into your suggestion later. – Aky Feb 11 '16 at 02:58

$$ 1=\sin\Big(\frac{\pi}{2}\Big)=\sum_{n=1}^\infty (-1)^{n-1}\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!} $$ Note that as $\pi$ is transcedental, the powers $1,\pi,\pi^2,\ldots,$ are linearly independent over $\mathbb Q$.

Unfortunately, some of the terms are negative. We then replace $-\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!}$ by $-\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!}+\frac{1}{2^n}$.

Yiorgos S. Smyrlis
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Set $q_1 = 1$ and pick a sequence of points $q_2, q_3, \dots$ from the sequence of sets $R_n$ where $R_n = \left( \frac{1}{n}, \frac{1}{\sqrt{n}} \right) \setminus \mathrm{span}_\Bbb{Q}\{q_1, \dots, q_{n-1}\}$ for $n>1$. The set $R_n$ is always nonempty because the interval contains uncountably many points and the span contains only countably many. The $q_i$ are all $\Bbb{Q}$-independent and positive. The sequence produces a sum $\sum_{n=1}^\infty q_n$ that diverges (by comparison with $1/n$). The terms decrease to zero. Therefore, there is a subsequence converging to any positive rational number (greater than one) you care to pick.

The choice of $q_1$ is not essential. If you wish to find a sum to any rational number less than $1$, pick any $q_1$ less than your rational number.

The same argument goes through replacing the rationals, $\Bbb{Q}$, with the algebraics, $\Bbb{A}$.

I also observe that while others manage to find one such sequence, we have found rather many more (about $2^\mathfrak{c}$, I think) by this argument. :-)

Eric Towers
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  • I know one should not use comment to just thank someone. But I was going to write something similar but without the elegant subsequence trick to generalise. Finding an example is so hard in comparison. Nice ! ;) – matovitch Feb 09 '16 at 18:54

Take a transcendental $\alpha\in\left(0,\,1\right)$. The geometric series with $k$th term $\alpha^{k-1}\left(1-\alpha\right)$ is independent in the linear-over-rationals sense demanded. But the sum is $1$, which is rational.

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This is really very simple. I think none of the current answers is quite equivalent to mine (or I wouldn't have posted it), even though some come close.

First fix the positive rational sum $S$ you want to get. Now inductively choose terms $a_k$ as follows. Let $S_k=\sum_{1\leq i<k}a_i$ be the sum of the previously chosen terms (so $S_1=0$ initially), for which we shall ensure that $S_k<S$. The previous choices exclude countably many values, so the interval $(\frac {S-S_k}2,S-S_k)$ contains some (uncountably many) values that are not excluded. Choose any such value as $a_k$ (invoking the axiom of dependent choice, if you need to be specific about this). With $a_k<S-S_k$ it is ensured that $S_{k+1}<S$, and since $S-S_{k+1}<\frac12(S-S_k)$ it is ensured that $\lim_{k\to\infty}S_k=S$, in other words $\sum_{k=1}^\infty a_k=S$.

Marc van Leeuwen
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  • This is just a slight simplification to my answer, using a larger dense set in $((S-S_k)/2, S-S_k)$. Restrict to the transcendentals over $\Bbb Q[a_1,…,a_k])$ to cover all bases. – Lutz Lehmann Feb 11 '16 at 11:03
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    @LutzL: Well it is pretty close, as I remarked. For me the main difference is that rather than trying to force independence from the start by some clever set-up, just avoid dependence while making choices (which are happening anyway). This construction would work for any notion of "rational independence" that can be ensured by successive choices, with only countably many values excluded by some initial portion (I don't have anything beyond algebraic independence over $\Bbb Q$ in mind, but it might exist). – Marc van Leeuwen Feb 11 '16 at 13:40


Just begin with the sequence $$a_n = \frac{\sqrt{2}}{n}$$ or any sequence with irrational elements and the properties $$ \lim_{n \to \infty} a_n = 0 \\ \sum_{n=1}^{\infty}{a_n} = \infty $$

and for whatever positive number $y$ you want, let $$\begin{align} r_0 &= y \\ i_0 &= \min\{k \in \mathbb{N} \mid a_k < r_0\}\\ b_n &= a_{i_n} \\ i_{n+1} &= \min\{k \in \mathbb{N} \mid k > i_n \land a_k < r_n\} \\ r_{n+1} &= r_n-b_n \\ \end{align} $$ or more compact $$ b_n = \max \left\{ a_k : a_k < b_{n-1} \land \sum_{i=1}^{n-1}{b_i}+a_k < y \right\} $$

As $\sum a_n$ tends to infinity, you can always get enough terms for $\sum b_n$ to reach $y$, and as $a_n$ tends to zero, you can always get arbitrarily close.

So by this definition, $$\sum_{n=0}^{\infty}{b_n} = y$$ for any positive choice of $y$, rational or irrational, and $b_n$ has only irrational elements.

Edit: As a bonus, my choice for $a_n$ also gives you that every partial sum is irrational and can easily be rewritten to work for transcendent numbers.


First, any countable set has a dense complement (in $\Bbb R$). So any rational linear combination of countable many reals has dense complement.

Let $S_n = \sum_{i=0}^{n} a_i$ be an strictly increasing sequence of real numbers with rational limit. Then you will find for every $n$ an irrational number $S'_n$ with $S_{n-1} \le S_n' \le S_n $ and since we choose $S'_n$ out of the uncountable interval $(S_{n-1} ,\le S_n)$ and only exclude countably many choices, namely rationals and rational linear combinations of the choices before we also can choose the $S'_n$ linearly independent.

Then set $a'_n=S_{n+1}-S_{n}$ which is irrational since the $S'_n$ were linearly independent and by construction we have $\sum_{i=0}^{\infty} a'_i = \sum_{i=0}^{\infty} a_i $.

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How about \begin{align} a_1&=42-\frac\pi{4-\pi}\\\\ a_n&={\Big(\frac\pi4\Big)}^{n-1} &\forall n\ge 2 \end{align}

Clearly $\sum_{n=2}^\infty a_n$ is a power series, converging to  $\frac{\pi/4}{1-\frac{\pi}4}=\frac{\pi/4}{(4-\pi)/4}=\frac\pi{4-\pi}$,  so $\sum_{n=1}^\infty a_n$ converges to 42.  The terms are linearly independent because they are constructed from different powers of $\pi$.

We could replace

  • $\pi$ with any (positive) transcendental number,
  • 4 with any number $>\pi$ (that is not a rational multiple of $\pi$), and
  • 42 with any rational number that is $>\sum_{n=2}^\infty a_n$.
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Take any series that leads to a rational value.

$$\sum\limits_{k=0}^{\infty} a_{k} \in \mathbb{Q}, a_{k}\neq 0$$

Let us create a rational field extension over some transcendental number $z$, $\mathbb{Q}[z]=\{p+zq: p,q \in \mathbb{Q} \}$. We know that we can create an infinite number of different extensions of this type. For example, we can take $z_{k}=\pi^{\frac{1}{k}}, k \in \mathbb{N}, k > 0$.

Every extensions $\mathbb{Q}[z_{m}]$ contains values that are as close as we want to any $a_{n}$. We can associate one of $\mathbb{Q}[z_{m}]$ to each $a_{n}$. With that we have got a ground for possible substitution.

We take one $$b_{k} \in \mathbb{Q}[z_{k}]$$ and replace $a_{k}$ with that value.

With that we have replaced our series with

$$\sum\limits_{k=0}^{\infty} b_{k}$$

We need to prove that we can make $$\sum\limits_{k=0}^{\infty} b_{k} = \sum\limits_{k=0}^{\infty} a_{k}$$

However, $b_{k}$ can be as close as we want to $a_{k}$. With that we can make $\sum\limits_{k=0}^{\infty} a_{k}-b_{k}$ as small as we want, which means that for every extension $\mathbb{Q}[z_{k}]$ we can find $p_{k}$ and $q_{k} \neq 0$ so that $p_{k}+q_{k}z_{k}$ is a sufficiently good replacement for $a_{k}$ which will keep $\sum\limits_{k=0}^{\infty} b_{k}=\sum\limits_{k=0}^{\infty} a_{k}$

$p_{k}+q_{k}z_{k}$ from different extensions, where $q_{k} \neq 0$, are independent, because all values that are dependent with $q \neq 0$ are contained within each $\mathbb{Q}[z_{k}]$.

This means that $\sum\limits_{k=0}^{\infty} b_{k} \in \mathbb{Q}$ each $b_{k}$ is irrational (specifically transcendental) and any $b_{k}$ is independent as required.

(There would be no difference to make the extension over irrational numbers, transcendentals are making it all more obvious.)


Call $p_k$ the $k$-th prime number. Then $\{\sqrt{p_k}\}_{k\in \mathbb N}$ is a set of independent irrational numbers.

Let's define positive rational coefficients $q_1$, $q_2$, ... such that $\sum_{k=1}^\infty q_k\sqrt{p_k}=1$.


  • $q_1>0$ rational such that $0<1-q_1 {\sqrt{2}}<\frac 1 2$
  • $q_2>0$ rational such that $0<1-(q_1 {\sqrt{2}}+q_2 {\sqrt{3}})<\frac 1 4$


  • $q_k$ rational such that $0<1-\sum_{i=1}^k q_k \sqrt{p_k}<\frac 1 {2^{k+1}}$


Then we have $\sum_{i=1}^k q_i \sqrt{p_i} \to 1$.

Marco Disce
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This would also work:

$$\sum_{k=0}^{\infty} \frac6{\pi^2k^2}+\frac{90}{\pi^4k^4} = 2$$

Each term is irrational because of the transcedence of $\pi$.

Suppose that two terms are a linear combination of each other. $$a\left(\frac6{\pi^2k^2}+\frac{90}{\pi^4k^4}\right)+b=\left(\frac6{\pi^2l^2}+\frac{90}{\pi^4l^4}\right)$$ $$a\left(\frac6{k^2}+\frac{90}{\pi^2k^4}\right)+b=\left(\frac6{l^2}+\frac{90}{\pi^2l^4}\right)$$


$$\pi^2 = \frac{\frac{90a}{k^4}+\frac{90}{l^4}}{\frac{6a}{k^2}-\frac6{l^2}+b}$$

However, the left hand side is irrational while the right hand side is rational.

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Consider that $$\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \dots \text{ (for } -1 < x < 1).$$

Then since $\frac{\pi}{5}$ is in $(-1,1)$, we have:

$$\underbrace{\frac{1}{1 - \frac{\pi}{5}}}_{\dfrac{5}{5-\pi}} = 1 + \underbrace{\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots}_{\text{infinite sum of irrational numbers}}$$

Subtracting $1$ from both sides of the above equation gives:

$$\underbrace{\dfrac{5}{5-\pi} - 1}_{\dfrac{\pi}{5-\pi}} = \frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots$$

Now the right hand side is an infinite sum of terms with each term an irrational number. Unfortunately, the left hand side is also irrational. Let's multiply both sides of the above equation by the reciprocal of the left hand side to induce rationality of that side, i.e., multiply both sides by $\frac{5-\pi}{\pi}$. This gives:

$$\frac{5-\pi}{\pi}\cdot \dfrac{\pi}{5-\pi}= \frac{5-\pi}{\pi} \cdot \left [\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots \right ]$$

Simplifying gives:

$$\underbrace{1}_{\text{rational number}} = \underbrace{\frac{5-\pi}{5} + \frac{(5-\pi)\pi}{5^{2}} + \frac{(5-\pi)\pi^{2}}{5^{3}} + \dots}_{\text{infinite sum of irrational terms}} $$

EDIT: Further justification/verification that the sum in the right hand side of the equation above equals $1$: Recognize it as a geometric series! The factor we multiply at each step is $r:=\frac{\pi}{5}$. Then we know the series converges to $\frac{\text{first term}}{1 - r} = \dfrac{\frac{5-\pi}{5}}{1 - \frac{\pi}{5}} = \dfrac{\frac{5-\pi}{5}}{\frac{5-\pi}{5}} = 1$.

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  • I believe the terms in my final sum are also "linearly independent" in the way you've described, though I didn't read that condition when I answered this. – layman Nov 21 '16 at 00:58
  • This is a valid answer (I guess), but it is kind of a loop hole since I would classify it as a [telescopic series](https://en.wikipedia.org/wiki/Telescoping_series). I can think of far easier (i.e. more obvious) examples of this kind: $$\sum_{n=1}^{\infty}\left[\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right]\qquad\text{or}\qquad \sum_{n=1}^\infty\left[\frac1{1+\ln(n)}-\frac1{1+\ln(n+1)}\right].$$ – M. Winter Jan 25 '18 at 10:18

For all rational $r$ there is a (many indeed) sequence of irrational $\{\theta_n\}$ converging to $r$.

Take $w_n = \theta_n+ \sum_0^n a_k$ where the $a_k$ are rational and the series converging to $0$; clearly $w_n$ is irrational and you have successive sums.

Furthermore $w_n\to r$.

In other words, for all rational it is verified the question.

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  • Yeah but how can you be sure your terms satisfy the "linearly independent" condition in OP's post? – layman Nov 21 '16 at 03:36

We will show that for any positive rational number (or real number) we can find such a series, which has the extra property that the terms are algebraically independent.

The following lemma is trivial to prove.

Lemma Let $\{x_n\}$ be any sequence of positive numbers and $x \in (0, \infty)$. Then, there exists some rational numbers $a_n >0 $ such that

$$x-\frac{1}{n}< \sum_{k=1}^n a_kx_k \leq x$$

Proof Induction. We need $$x-\frac{1}{n}-\sum_{k=1}^{n-1} a_kx_k < a_nx_n \leq x- \sum_{k=1}^{n-1} a_kx_k $$ which follows from the density of the rationals.

The exercise Now pick $\{ x_n \}$ to be any sequence of positive algebraically independent, transcendent numbers and $x$ to be any rational. Use the above Lemma. Then $a_kx_k$ are irrational, algebraically independent and their sum is the desired natural.

N. S.
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To construct a positive series $\sum_{i=1}^\infty u_i=1$ where the $u_i$ are linearly independent start with $u_1=\frac{e}{10}$ and define $u_{n+1}$ recursively by setting $u_{n+1}=\frac{e^n}{10^n}\left\lfloor \frac{10^n}{e^n}\left(1-\sum_{i=1}^n u_i\right)\right\rfloor$. Since $e$ is transcendental the terms are linearly independent over the rationals.

Mikhail Katz
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Yes, for example $\tan(\frac{\pi}{4})=1$ Write $\tan$ as an infinite Taylor expansion in powers of $\pi$. Other trigonometric functions of irrational arguments can also have rational outputs, eg. $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

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Arif Burhan
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    Both of your examples have already been suggested in other answers: [this](http://math.stackexchange.com/a/1647422/147873) and [this](http://math.stackexchange.com/a/1647632/147873) – Winther Feb 12 '16 at 21:08

YES. An infinite sum of irrational numbers can be rational. PROOF: Let the set A be all the positive irrational numbers and the set B be the negative irrational numbers. Take each positive irrational number and add it to the matching negative irrational number to get 0. The sum of all these 0 numbers is 0 which is a rational number.

  • This doesn't satisfy the rational independence condition. – User1 Feb 13 '16 at 09:22
  • You can *almost* do this, though. You just need to break up the sets so that the numbers are all rationally independent. Can you show that there exists some way to divide up the set of all irrational numbers so that no chunk is rationally dependent with any other chunk? – Please stop being evil Feb 14 '16 at 07:33

There is a theorem (by Riemann, I think) which states that:
"A convergent series which is not absolutely convergent
can be rearranged to sum to any number we choose"

So all we need to do to show that what the OP asks is possible, is to take
an arbitrary convergent series which is not absolutely convergent,
and which consists of irrational numbers only

Then by applying the theorem, it means we can rearrange its elements
and get any rational (and also any irrational) number we want.

An example of such a series is, let's say:

$\sum_\limits{n=1}^\infty \frac{(-1)^n}{n\sqrt5}$

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  • The terms in the series you give as an example are “not linearly independent” (as defined by OP). – gniourf_gniourf Feb 10 '16 at 16:11
  • Oh, OK... But that requirement was maybe added later, wasn't it? Anyway... I guess my answer is still valid if looking at the question title only. And ... also I guess we can pick another series of irrationals. The idea of using the theorem is what I wanted to share here. This gives us a non-constructive proof. – peter.petrov Feb 10 '16 at 17:53
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    -1 The terms in the series are all positive, so Riemann's theorem does not apply. – Mario Carneiro Feb 10 '16 at 22:19
  • The theorem is named (Levy-)Steinitz and originates with Grassmann. There was some ugliness with Grassmanns manuscripts sent to Cauchy,… – Lutz Lehmann Feb 11 '16 at 11:06