Can an infinite sum of irrational numbers be rational?

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 $a_i = ua_j + v$.

Solution 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...
...

     —

Solution 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.

Solution 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.