Let {a$_n$} be a weakly decreasing sequence of positive reals. $\,$ Assume that the sum $\,$ s = $\sum$ a$_n$ converges. $\,$ Next introduce arbitrary signs $\,$ $\epsilon$$_n$ = +$1$ $\,$ or $\,$ $\epsilon$$_n$ = -$1$ $\,$ and consider the set $\qquad$ $\qquad$ $\quad$ L = { $\sum$ $\epsilon$$_n$a$_n$} $\,$ of all possible signed sums. $\,$ All of these are absolutely convergent and therefore converge themselves. Clearly, L is contained in the interval [-s,s] and is symmetric about the origin.
For example, if a$_n$ = $\frac{1}{n^2}$ (n $\geq$ 1), then $\,$ s = $\zeta$(2) = 1.644... $\,$ and the resulting set L has the following properties: $\,$ (i) $\,$ L is uncountable. This is not immediately obvious because even though the number of sign combinations is uncountable, this is partly balanced out by a considerable amount of collapsing. This comes about from finite coincidences such as $\,$ 1/20$^2$ + 1/15$^2$ = 1/12$^2$ $\,$ as well as infinite ones like $\,$ 1/2$^2$ = 1/3$^2$ + 1/4$^2$ + 1/5$^2$ + 1/6$^2$ +1/11$^2$ + $\cdots$ $\,$. $\,$ However, if you select a thin enough subsequence and change just those signs, then the resulting sums are all distinct $\,$(no collapsing). $\;$ (ii) $\,$ $0$ $\notin$ L : $\,$ The numerical value $\zeta$(2) = 1 + .644 ... $\,$ shows that the first term of the series always dominates the sum and can't be cancelled by any combination of the remaining terms. $\,$ (Of course, L also omits an interval around zero). $\,$ (iii) With more work it can be shown that L is closed (i.e. contains all its limit points).
So far then, we know only that L is a proper uncountable closed subset of [-$\zeta$(2),$\zeta$(2)] , but not what it actually is.
Questions: (1) $\,$ For $\,$ a$_n$ = $\frac{1}{n^2}$ $\,$ as above, $\,$ is it possible to describe the set L explicitly?
(2) $\,$ Let b$_n$ = $\frac{1}{n!}$ $\,$ for n $\geq$ 0 with sum $\sum$ $\frac{1}{n!}$ = e $\,$ and corresponding set $\,$ L $\subset$ [-e,e] . This L has the same properties as before but is probably sparser due to the rapid decrease of the b$_n$'s . $\,$ Is there a precise description of the set L in this case?
(3) $\,$ Can the series be recovered from the sums? $\;$ (a) Are there any other sequences {c$_n$} satisfying $\;$ c$_n$ $\geq$ c$_{n+1}$ > 0 with L({c$_n$}) = L({$\frac{1}{n^2}$}) $\,$? $\;$ (b) $\,$ Is there another sequence $\,$ {d$_n$} $\,$ such that $\;$ L({d$_n$}) = $\;$ L({$\frac{1}{n!}$}) $\,$ ?
Thanks
Partial answer: apologies for my earlier answer which had an error. Instead of looking at $\sum \epsilon _n a_n$ with $\epsilon_n's \in \{-1,1\}$ we can take $\epsilon_n's \in \{0,1\}$ and ask when we get the entire interval $[0,L]$ where $L=\sum a_n$. These two formulations are equivalent because $c=\pm1$ if and only if $(c+1)/2 =0$ or $1$. Now suppose $a_n >0,\sum a_n <\infty$ and $a_n$ decreases to 0 so fast that $\sum_{N+1}^\infty a_n <a_N$. [This is the case when $a_n=1/(n!)$]. Then the entire interval $(\sum_{N+1}^\infty a_n ,a_N)$ can contain no sum of the type $\sum \epsilon _n a_n$ with $\epsilon_n's \in \{0,1\}$ This fact if obvious since we have to add at least one $a_j$ with $j \leq N$ to increase the left end point of the interval.The series $\sum 1/n^{2}$ does not satisfy the condition $\sum_{N+1}^\infty a_n <a_N$ and I don't know the answer for this case.