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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    $\begingroup$ Does the series $2=\mathrm{e}^{\ln(2)}=\sum_{k=0}^{+\infty}\frac{\ln(2)^n}{n!}$ count? $\endgroup$ – gniourf_gniourf Feb 9 '16 at 11:58
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    $\begingroup$ Beware that assuming $S\neq e^{\ln x}$ for some rational $x$ is the same as assuming $S\neq x$ for some rational $x$. $\endgroup$ – Nate River Feb 9 '16 at 12:11
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    $\begingroup$ Would $e^{ln(2)}+1$ be acceptable? Note: excluding this whole category of examples seems a little odd. $\endgroup$ – lulu Feb 9 '16 at 12:18
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    $\begingroup$ 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. $\endgroup$ – Mark Dickinson Feb 9 '16 at 12:31
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    $\begingroup$ 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... $\endgroup$ – Lynn Feb 10 '16 at 23:02

24 Answers 24


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.

  • $\begingroup$ Note that $B_{2n+2}<0$ for $n\in \{2,4,6,\ldots \}$. $\endgroup$ – Scippy Feb 9 '16 at 12:29
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    $\begingroup$ I don't recognize those coefficients B<sup>2n+2</sup>. Can you provide a link? $\endgroup$ – Pieter Geerkens Feb 10 '16 at 8:51
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    $\begingroup$ @PieterGeerkens Bernoulli numbers $\endgroup$ – Akiva Weinberger Feb 10 '16 at 12:04
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    $\begingroup$ 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. $\endgroup$ – Jack M Feb 10 '16 at 15:37
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    $\begingroup$ @JackM See here or here $\endgroup$ – Yai0Phah Feb 11 '16 at 19:22

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.



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    $\begingroup$ This is extraordinarily clever; that is, finding a logical inverse of the non-algebraic sums that yield transcendental numbers. $\endgroup$ – Joshua Feb 9 '16 at 16:16
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    $\begingroup$ This sequence does not satisfy the rational independence: $10a_1-a_2=1$ $\endgroup$ – robjohn Feb 9 '16 at 18:14
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    $\begingroup$ @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. $\endgroup$ – robjohn Feb 10 '16 at 8:56
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    $\begingroup$ @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). $\endgroup$ – matovitch Feb 10 '16 at 13:10
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    $\begingroup$ $3a_1-2a_2=2$, I think, as the other 1's cancel. (if that is binary, otherwise $91a_1-10a_2=10$ $\endgroup$ – Empy2 Feb 10 '16 at 14:59

$$ \begin{align} 1 &=\log(e)\\ &=-\log\left(1-\left(1-\frac1e\right)\right)\\ &=\sum_{k=1}^\infty\frac1k\left(1-\frac1e\right)^k \end{align} $$ 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.

  • $\begingroup$ How does that follow directly from $e$ being transcendental? $\endgroup$ – user253751 Feb 11 '16 at 9:55
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    $\begingroup$ 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) $\endgroup$ – Peter Green Feb 11 '16 at 10:20
  • $\begingroup$ @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. $\endgroup$ – 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.

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    $\begingroup$ What is this (type of) series called? Links would make your answer better. $\endgroup$ – Foo Bar Feb 9 '16 at 14:49
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    $\begingroup$ Does it need a name? It's just an infinite sum. I didn't use any links. $\endgroup$ – Akiva Weinberger Feb 9 '16 at 14:50
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    $\begingroup$ 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 $\Bbb Q(\alpha_1,\ldots,\alpha_{n-1})$. $\endgroup$ – Hagen von Eitzen Feb 9 '16 at 19:02
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    $\begingroup$ @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! $\endgroup$ – Neel Feb 10 '16 at 4:42
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    $\begingroup$ @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 $1$. $\endgroup$ – Akiva Weinberger Feb 10 '16 at 11:47

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 $S$.

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    $\begingroup$ This may fail the "linear combination" condition, although that condition needs explication. $\endgroup$ – Gerry Myerson Feb 9 '16 at 12:14
  • $\begingroup$ 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. $\endgroup$ – User1 Feb 9 '16 at 12:20
  • $\begingroup$ 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. $\endgroup$ – Dr. Lutz Lehmann Feb 9 '16 at 12:39
  • $\begingroup$ 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. $\endgroup$ – Max Murphy Feb 12 '16 at 22:12
  • $\begingroup$ No, not really. Since $S$ is rational and every partial sum irrational. $\endgroup$ – Dr. Lutz Lehmann Feb 12 '16 at 22:15

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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    $\begingroup$ 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. $\endgroup$ – Mario Carneiro Feb 10 '16 at 22:15
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    $\begingroup$ @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. $\endgroup$ – Aky Feb 11 '16 at 2:58

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. :-)

  • $\begingroup$ 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 ! ;) $\endgroup$ – matovitch Feb 9 '16 at 18:54

$$ 1=\sin\Big(\frac{\pi}{2}\Big)=\sum_{n=1}^\infty (-1)^{n-1}\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!} $$ 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 $-\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!}+\frac{1}{2^n}$.


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.


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 $\sum_{k=1}^\infty a_k=S$.

  • $\begingroup$ 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. $\endgroup$ – Dr. Lutz Lehmann Feb 11 '16 at 11:03
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    $\begingroup$ @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). $\endgroup$ – 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 $.


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 $>\sum_{n=2}^\infty a_n$.

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


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.


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.


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 $\frac{\text{first term}}{1 - r} = \dfrac{\frac{5-\pi}{5}}{1 - \frac{\pi}{5}} = \dfrac{\frac{5-\pi}{5}}{\frac{5-\pi}{5}} = 1$.

  • $\begingroup$ 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. $\endgroup$ – layman Nov 21 '16 at 0:58
  • $\begingroup$ This is a valid answer (I guess), but it is kind of a loop hole since I would classify it as a telescopic series. I can think of far easier (i.e. more obvious) examples of this kind: $$\sum_{n=1}^{\infty}\left[\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right]\qquad\text{or}\qquad \sum_{n=1}^\infty\left[\frac1{1+\ln(n)}-\frac1{1+\ln(n+1)}\right].$$ $\endgroup$ – 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.

  • $\begingroup$ Yeah but how can you be sure your terms satisfy the "linearly independent" condition in OP's post? $\endgroup$ – layman Nov 21 '16 at 3:36

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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    $\begingroup$ Both of your examples have already been suggested in other answers: this and this $\endgroup$ – 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.

  • $\begingroup$ This doesn't satisfy the rational independence condition. $\endgroup$ – User1 Feb 13 '16 at 9:22
  • $\begingroup$ 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? $\endgroup$ – Please stop being evil Feb 14 '16 at 7:33

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.

  • $\begingroup$ this looks like they are linearly dependent $\endgroup$ – supinf Nov 6 '17 at 12:15
  • $\begingroup$ OK, I fixed it. @supinf $\endgroup$ – Mikhail Katz Nov 7 '17 at 10:04

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:

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

  • $\begingroup$ The terms in the series you give as an example are “not linearly independent” (as defined by OP). $\endgroup$ – gniourf_gniourf Feb 10 '16 at 16:11
  • $\begingroup$ 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. $\endgroup$ – peter.petrov Feb 10 '16 at 17:53
  • 1
    $\begingroup$ -1 The terms in the series are all positive, so Riemann's theorem does not apply. $\endgroup$ – Mario Carneiro Feb 10 '16 at 22:19
  • $\begingroup$ The theorem is named (Levy-)Steinitz and originates with Grassmann. There was some ugliness with Grassmanns manuscripts sent to Cauchy,… $\endgroup$ – Dr. Lutz Lehmann Feb 11 '16 at 11:06

protected by Asaf Karagila Feb 13 '16 at 8:31

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