As explained in this post, we could build a series such that every subseries has an irrational sum.
Conversely, if we have a series such that all but finitely many terms are zero and the rest are rational, then the sum of every subseries will be rational.
However, if we have a convergent series $\sum a_n$ such that infinitely many $a_n$ are non-zero, we may always select a subseries that converges to an irrational constant. To prove this, we begin by constructing a subsequence:
- Take $a_{n_1}$ to be non-zero.
- For each $k\geq 1$, take $a_{n_{k+1}}$ to be non-zero, satisfying
$$
|a_{n_{k+1}}| < 3^{-k} \cdot |a_{n_k}|
$$
Now, for any sequence $(\xi_k) \in \{0,1\}^{\Bbb N}$, we note that the map
$$
\Phi:(\xi_k) \mapsto \sum_{k=1}^\infty \xi_k \, a_{n_k}
$$
is injective (one-to-one), and that each $\sum_{k=1}^\infty \xi_k \, a_{n_k}$ is the sum of some (absolutely convergent) subseries.
Since the set $\{0,1\}^{\Bbb N}$ is uncountable and $\Phi$ is injective, we may conclude that there are at least uncountably many values that the sum of subseries may attain. However, there are only countably many rationals.
So, there must exist a subseries whose sum is irrational.
