Can an infinite sum of irrational numbers be rational?

Question

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 $a_i = ua_j + v$.

Answer 1

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

0.1100110000110000001100000000110000000000110000000...
0.0011001001001000010010000001001000000001001000000...
0.0000000110000100100001000010000100000010000100000...
0.0000000000000011000000100100000010000100000010000...
0.0000000000000000000000011000000001001000000001000...
0.0000000000000000000000000000000000110000000000100...
0.0000000000000000000000000000000000000000000000011...
...

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.

0.10010000000100000000000000100000000000000000000001000000...
0.01101100011011000000000011011000000000000000000110110000...
0.00000011100000111000011100001110000000000000111000001110...
0.00000000000000000111100000000001111000001111000000000000...
0.00000000000000000000000000000000000111110000000000000000...
...

     —

Answer 2

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.

Answer 3

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

Answer 4

\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.

Answer 5

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 $S$.

Answer 6

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 $\pi$.

Answer 7

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

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

Answer 8

$$ 1=\sin\Big(\frac{\pi}{2}\Big)=\sum_{n=1}^\infty (-1)^{n-1}\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!} $$ Note that as $\pi$ is transcendental, 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 $-\frac{\pi^{2n-1}}{2^{2n-1}(2n-1)!}+\frac{1}{2^n}$.

Answer 9

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. :-)

Answer 10

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.

Answer 11

Yes.

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.

Answer 12

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

Answer 13

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 $.

Answer 14

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.)

Answer 15

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

Answer 16

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 $\sum_{k=1}^\infty q_k\sqrt{p_k}=1$.

Define

• $q_1>0$ 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 $\sum_{i=1}^k q_i \sqrt{p_i} \to 1$.

Answer 17

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. $$a\left(\frac6{\pi^2k^2}+\frac{90}{\pi^4k^4}\right)+b=\left(\frac6{\pi^2l^2}+\frac{90}{\pi^4l^4}\right)$$ $$a\left(\frac6{k^2}+\frac{90}{\pi^2k^4}\right)+b=\left(\frac6{l^2}+\frac{90}{\pi^2l^4}\right)$$

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

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

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

Answer 18

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

Answer 19

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.

Answer 20

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.

Answer 21

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.

Answer 22

From $\sin x=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}x^{2n+1}$ and that $\sin\pi=0$, we conclude $$\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}\pi^{2n+1}=0 $$

Answer 23

Below, I will provide three explicit infinite sums with irrational summands that converge to rational numbers. They are examples of rational zeta series, and they all satisfy the additional requirements as stated by the OP.

Let $\zeta(\cdot)$ be the Riemann zeta function. Then the three series are:

$$ \sum_{n=1}^{\infty} \left(\zeta(2n)-1 \right) = \frac{3}{4} \tag{1}\label{1} $$ and

$$ \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n-1}} = 1 \tag{2}\label{2}$$

and finally

$$ \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{2^{2n}} = \frac{1}{6} \ .\tag{3}\label{3} $$

The first identity can be found in equation (1.7) on p. 2 of this article (PDF) by Adamchik and Srivastava. One can find the second identity as equation (52) on p. 274 of this book by Srivastava. The third identity is described as equation (43) on p. 16 of the following article (PDF) by Borwein, Bradley and Crandall.

All three series solely consist of positive terms, because $\zeta(n) > 1$ for all $n > 1$.

Moreover, the series all involve irrational summands, as $$\zeta(2n) = \frac{|B_{2n}| (2 \pi )^{2n}}{2 (2n)!} \tag{4}\label{4} .$$ Here, $B_{2n}$ is the $2n$-th Bernoulli number (see this wiki paragraph for more information). The irrationality of these coefficients follows from the transcendentality of $\pi$.

Finally, all respective terms in equations \eqref{1} -- \eqref{3} are linearly independent over $\mathbb{Q}$, because they involve summands with even-valued powers of $\pi$, as described in equation \eqref{4}. Thus, the three series fullfil the requirements described in the question.

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Added

I've found a fourth (rational zeta) series that satisfies all criteria. It is the following:

$$ \sum_{n=1}^{\infty} \left( \zeta(2n)-1 \right) \left( \frac{3}{2} \right)^{2n} = \frac{23}{10} \tag{5}\label{5} $$

It can be found in equation (219) on p. 285 of Srivastava's book (mentioned above). The procedure of checking it meets all requirements is similar to the one for the other three series.

Answer 24

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.

Answer 25

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. $\sin(\frac{\pi}{6}) = \frac{1}{2}$.