Riemann rearrangement theorem with restricted choices

Question

Note: For convenience, all sequences will be indexed by the positive integers $\mathbb Z_+$.

Definitions and some motivation:

The Riemann rearrangement theorem says that if we have a sequence that is conditionally but not absolutely convergent, we can rearrange it to converge to any desired value. Looking at the proof a little, we can get the following statement:

Let $a_n$ be a sequence of real numbers with $|a_n| \to 0$. If $\sum a_n$ diverges, then there exists a $\{-1, 1\}$-valued sequence $\varepsilon_n$ such that $\sum \varepsilon_n a_n$ converges. What if we only allow a "small" set of sequences $\varepsilon_n$? Is it still possible to get the sum to converge after multiplication by $\varepsilon_n$?

Question set up:

Let $E = \{-1, 1\}$. Define the map $H: (0, 1) \to E^{\mathbb Z_+}$ as follows - expand $x$ in binary form, where we always take the expansion that ends in an infinite number of $1$'s whenever a choice is to be made. Then the $i$'th coordinate of $H(x)$ is $1$ if the $i$'th binary decimal digit of $x$ is $1$, and $-1$ otherwise.

Question:

For every subset $S$ of $(0, 1)$ with Lebesgue measure $0$, does there always exist a sequence $a_n$ of real numbers with $|a_n| \to 0$, and $\sum a_n$ divergent such that for any sequence $\varepsilon_n$ in $H(S)$, $\sum \varepsilon_n a_n$ fails to converge?

Answer 1

No. For example, we can define $S$ by insisting that $\epsilon_{2n+1}=\epsilon_{2n}$. If we have no restrictions, then a good strategy to make $\sum \epsilon_n a_n$ convergent would be to choose the $\epsilon_n$ recursively in such a way that the partial sums stay as close to $0$ as possible. A slightly modified version of this strategy still works with our restriction; we now just let $b_n=a_{2n}+a_{2n+1}$ take the role of the $a_n$'s.