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

rationalnumbers u,v such that a_i = ua_j + v.

**Answer**

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 fixwithproof.

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

—

**Attribution***Source : Link , Question Author : User1 , Answer Author : Akiva Weinberger*