Does every sequence of rationals, whose sum is irrational, have a subsequence whose sum is rational

Question

Assume we have a sequence of rational numbers $a=(a_n)$. Assume we have a summation function $S: \mathscr {L}^1 \mapsto \mathbb R, \ \ S(a)=\sum a_n$ ($\mathscr {L}^1$ is the sequence space whose sums of absolute values converges). Assume also that $S(a) \in \mathbb R \setminus \mathbb Q$.

I would like to know if every such sequence $a$ has a subsequence $b$ (infinitely long) such that $S(b) \in \mathbb Q$.

Take as an example $a_n = 1/n^2$. Then $S(a)=\pi^2/6$. But $a$ has a subsequence $b=(b_n)=(1/(2^n)^2)$ (ie. all squares of powers of $2$). Then $S(b)=4/3$. Is this case with every such sequence?

Answer 1

No. For example, take the sequence $a_n=2^{-2^n}$, $n=1$, $2$, $\dots$. An infinite subsequence $(a_{n_k})$ of $(a_n)$ will have sum $$S:=\sum_k a_{n_k}=\sum_k 2^{-2^{n_k}}.$$ So, the binary expansion of $S$ will have $1$s in positions $2^{n_1}$, $2^{n_2}$, $2^{n_3}$, $\dots$, and $0$s everywhere else. This is not a periodic sequence, so $S$ must be irrational.

Answer 2

No; for example, if $(n_i)$ is a strictly increasing sequence of positive integers, then we can imitate the proof of the irrationality of $e$ to see that

$$\sum_{i=1}^\infty \frac{1}{n_1 \dots n_i} \notin \mathbf Q.$$

But every sub-series of this series has the same property (it just amounts to grouping some of the $n_i$ together).

Answer 3

What if you took

$\ \ a_1=.1$

$\ \ a_2=.0101$

$\ \ a_3=.00001001$

$\ \ a_4=.0000000010001$

$\ \ \ \ \ \ \ \ \vdots$

?