This was asked earlier at MSE.

Let $A = \{a_0, a_1, a_2, \dotsc\}$ denote a weakly decreasing sequence of positive terms whose sum converges. Next introduce plus minus signs in every possible way, to form the collection $L(A) = \{\sum \limits_{j=0}^\infty \epsilon_j a_j : \epsilon_j = \pm1\}$ of values of the signed sums. All of these series converge since they converge absolutely.

For example, let $a_j = 1/j!$, giving the usual series for $e = 2.718\ldots$. Then one can check that $L(A)$ has the following general properties: it is a proper closed uncountable subset of the interval $[-e,+e]$.

Questions: (1) Is it possible to be more specific and to say exactly what the set $L(A)$ actually is (apart from just recapitulating the definition)?

(2) Can one recover the sequence from the values of its signed sums? Or does there exist some $B = \{b_0, b_1, b_2, \dotsc\}$ satisfying (as above) $b_j \geq b_{j+1} > 0$ and such that $L(B) = L(A)$?


  • 2
    $\begingroup$ I think that it is generally regarded as bad practice to switch out of math mode for things like a$_j$ (as opposed to $a_j$). I have edited accordingly. If you really want upright letters, then I believe that the accepted convention is to use \mathrm, like $\mathrm a_j$ ($\mathrm a_j$). $\endgroup$
    – LSpice
    Jun 28, 2018 at 0:25
  • $\begingroup$ Another (equivalent) way to do this: the set of all sums $\sum_{j\in A} a_j$, where $A$ ranges over all subsets of $\mathbb N$. This one is an affine image of yours and vice versa. $\endgroup$ Jun 28, 2018 at 0:44

3 Answers 3


If $a_{n+1} \ge a_n/2$ for all $n$, then $L(A) = [-s,s]$ where $s = \sum_n a_n$.

On the other hand, if $a_{n+1} < a_n/2$ for all $n$, then $L(A)$ is a nowhere dense Cantor-type set, and you can recover $A$ from $L(A)$.

EDIT: I don't have references, but these are not hard to prove. As Gerald Edgar remarks, it's equivalent to consider sums $S(A) = \sum_{j \in A} a_j$ for $A \subseteq \mathbb N$.

In the case $a_{n+1} \ge a_n/2$, so $\sum_{j > n} a_j \ge a_n$, you can use a greedy approach: to make a sum $s$, where $0 \le S(\mathbb N)$, include $n$ in $A$ iff $S(A \cap [0\ldots n-1]) + a_n \le s$, and note that by induction $$S(A \cap [0\ldots n]) + S([n+1 \ldots \infty)) \ge s \ge S(A \cap [0\ldots n])$$

In the case $a_{n+1} < a_n/2$, consider the intervals $J_n(A) = [S(A), S(A \cup [n+1\ldots \infty)]$ for $A \subseteq [1,\ldots, n]$, so that $S(B) \in J_n(A)$ if $A \subseteq B \subseteq A \cup [n+1\ldots\infty)$, and note that $J_{n+1}(A)$ and $J_{n+1}(A \cup \{n+1\})$ are disjoint closed subintervals of $J_n(A)$.

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    $\begingroup$ A reference here would be great. $\endgroup$
    – Winther
    Jun 28, 2018 at 10:39

Let $A$ be the sequence $1/2,1/4,1/8,1/16,1/32\dots$, and let $B$ be the sequence $1/4,1/4,1/4,1/8,1/16,1/32,\dots$. Then $L(B)=L(A)$.


A fun remark :

If you consider the natural measure $\mu$ on $L(A)$ defined as follow : $L_N(A)=\{\sum_{j=0}^{N-1}\epsilon_j a_j : \epsilon_j=\pm 1 \}$ and $$\mu_N=\frac{1}{2^N}\sum_{\epsilon\in \{\pm 1 \}^N} \delta_{\sum_{j=0}^{N-1}\epsilon_j a_j} = \frac{1}{2^N}\sum_{x\in L_N(A)}\delta_x \text{ (if all the sums are disjoint)}$$ and $\mu=\lim_{N\rightarrow} \mu_N$. Then we have $$\int e^{i\lambda x}d\mu(x) = \prod\cos(\lambda a_j)$$ and the sequence $a_i$ is related to the zeros $\lambda_{i,k}$ of the Fourier transform of $\mu$: $$\lambda_{j,k} = \frac{1}{a_j}(\frac{\pi}{2}+k\pi) $$


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