Suppose $\sum_{n\ge 1} |a_n| = A<\infty.$ Under what conditions is $\sum_{n\ge 1} \epsilon_n a_n = [-A,A]$, for $\epsilon_n \in \{-1,1\}$? Consider the space of sequences:
$$
\mathcal{E} = \{\{\epsilon_n\}_{n= 1}^{\infty}: \epsilon_n = \pm 1\}
$$
This can be considered a "random choice of sign" in the probabilistic context, for example. My question: if $\{a_n\}_{n=1}^{\infty}$ is an absolutely summable sequence with $\sum_{n\ge 1} |a_n|=A$, under what conditions on $\{a_n\}$ is the following map a surjection?
$$
f: \mathcal{E} \to \left[-A,A\right],\,
\{\epsilon_n\}_{n=1}^\infty \mapsto \sum_{n\ge 1}{\epsilon_n} a_n
$$
Note: I'm asking this question as a follow-up to a special case where $a_n=n^{-2}$ and have reused some of the language for continuity. In that question, the answer was no because $\pi^2/6 \approx 1.645,$ so one could never 'get back' to zero.
Cases where the question is affirmative include $a_n=0$ and $a_n=2^{-n}$, but I don't think other geometric series work. A necessary condition is $|a_1|\le \sum_{n\ge 2} |a_n|$, and in fact I think its generalization is sufficient: if for all $m\in\mathbb{N}$
$$
|a_m|\le \sum_{n>m}|a_n|,
$$ then $f$ is a surjection. Heuristically, this is because you can 'double back' as much as you'd like, allowing you to reach every number in $[-A,A]$. But maybe a weaker condition suffices, or perhaps even an explicit description of admissible $\{a_n\}$?
 A: To simplify notation, we can of course assume $a_n \geqslant 0$ for all $n$.
The condition you (correctly) think is sufficient becomes the necessary and sufficient condition if we additionally assume that the sequence $(a_n)$ is monotonic. Without the monotonicity assumption, the necessary condition becomes more awkward to state, but nothing essential changes.
First let's show sufficiency, using only the assumption
$$a_m \leqslant \sum_{n > m} a_n$$
for all $m$. Choose any target value $L \in [0,A)$ (for negative targets, just flip all signs $\epsilon_n$, and the targets $\pm A$ are trivial to achieve). Let $n_1$ be the smallest positive integer such that
$$s_{n_1} := \sum_{n = 1}^{n_1} a_n > L\,.$$
Then $L < s_{n_1} \leqslant L + a_{n_1}$. Take $\epsilon_n = 1$ for $n \leqslant n_1$. If
$$\sum_{n > n_1} a_n = s_{n-1} - L\,,$$
then take $\epsilon_n = -1$ for all $n > n_1$, otherwise let $n_2$ be the smallest integer $> n_1$ such that
$$\sum_{n = n_1 + 1}^{n_2} a_n > s_{n_1} - L$$
and put $\epsilon_n = -1$ for $n_1 < n \leqslant n_2$. Then
$$L - a_{n_2} \leqslant s_{n_2} = \sum_{n = 1}^{n_2} \epsilon_n a_n < L\,.$$
Rinse and repeat. We construct a [possibly finite] sequence $0 = n_0 < n_1 < n_2 < n_3 \ldots$, setting $\epsilon_n = (-1)^k$ for $n_k < n \leqslant n_{k+1}$, such that
$$0 < (-1)^{k}\biggl( L - \sum_{n = 1}^{n_k} \epsilon_n a_n\biggr) \leqslant a_{n_k} \tag{$\ast$}$$
holds for all $k$. The sequence is finite if and only if we have
$$(-1)^{k}\biggl( L - \sum_{n = 1}^{n_k} \epsilon_n a_n\biggr) = \sum_{n = n_k+1}^{\infty} a_n$$
at some point $k$, then we put $\epsilon_n = (-1)^k$ for all $n > n_k$, and it is clear that this leads to
$$L = \sum_{n = 1}^{\infty} \epsilon_n a_n\,.$$
Otherwise, we have infinitely many sign changes, but since $a_{n_k} \to 0$, the inequality $(\ast)$ ensures that a subsequence of the partial sums of
$$\sum_{n = 1}^{\infty} \epsilon_n a_n$$
converges to $L$, and since the series converges absolutely, it follows that the whole series converges to $L$.
For the necessity, assume that $(a_n)$ is monotonic, and there is an $m$ such that
$$a_m > \sum_{n = m+1}^{\infty} a_n\,.$$
Then no target strictly between
$$\sum_{n = 1}^{m-1} a_n \qquad\text{and}\qquad \sum_{n = 1}^{m} a_n - \sum_{n = m+1}^{\infty} a_n$$
is reachable. If we have $\epsilon_r = -1$ for some $r \leqslant m$, then
$$\sum_{n = 1}^{\infty} \epsilon_n a_n \leqslant \sum_{n = 1}^{m} a_n - 2a_r + \sum_{n = m+1}^{\infty} a_n < \sum_{n = 1}^{m} a_n - 2a_r + a_m \leqslant \sum_{n = 1}^{m-1} a_n - a_m < \sum_{n = 1}^{m-1} a_n$$
since
$$\sum_{n = m+1}^{\infty} a_n < a_m \leqslant a_r\,.$$
And if $\epsilon_n = 1$ for all $n \leqslant m$, then clearly
$$\sum_{n = 1}^{\infty} \epsilon_n a_n \geqslant \sum_{n = 1}^{m} a_n - \sum_{n = m+1}^{\infty} a_n\,.$$
