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Let $a_n$ be a sequence of positive real numbers with $\sum a_n < \infty$. What are the necessary and sufficient conditions for the following to hold?

For any $S \in \mathbb R$ with $-\sum a_n \leq S \leq \sum a_n$, there exists some choice of signs $\epsilon_n \in\{-1, 1\}$ such that $\sum \epsilon_n a_n =S$.

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    $\begingroup$ A sufficient condition is that each $a_i \leq \sum_{j > i} a_j$. Then we can do the usual thing: positive signs until the partial sum is greater than $S$, then negative signs until it's less than $S$, then back to positive, etc. Since the series is absolutely convergent this is guaranteed to converge; since the partial sums go back and forth across $S$, it converges to $S$. I doubt if this condition is necessary, though. $\endgroup$
    – Zach Teitler
    Commented Jan 16 at 0:53
  • $\begingroup$ Btw I happened to ask and answer recently an equivalent question [here][1] (check the statement in italic), with $\epsilon_n\in\{0,1\}$ instead of $\{-1,1\}$ [1]:mathoverflow.net/questions/461646/… $\endgroup$
    – Pietro Majer
    Commented Jan 16 at 1:18
  • $\begingroup$ There is also a somewhat related question on MSE $\endgroup$
    – polfosol
    Commented Jan 17 at 10:01

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Let's rearrange so that $a_1\ge a_2\ge a_3\ge \ldots$. Then this works if and only if $a_n\le \sum_{k>n}a_k$ for all $n\ge 1$.

To see this, let me also rephrase the problem as: Can we obtain all $0\le S\le \sum_{n\ge 1} a_n$ as sums $\sum\delta_n a_n$, with $\delta_n=0,1$?

Clearly the condition is necessary, because if $a_N>\sum_{k>N} a_k$, then we won't be able to reach $S\in (\sum_{k>N}a_k,a_N)$.

On the other hand, with the condition assumed, we reach any $S$ by the obvious procedure: Starting with $n=1$ and increasing $n$, make $\delta_n=1$ unless that makes the partial sum $\sum_{k=1}^n \delta_ka_k$ too large; in that case, set $\delta_n=0$.

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    $\begingroup$ I was in fact aware of the answer to the reformulated question, what I missed was that you could reformulate it that way! Nice. $\endgroup$
    – Nate River
    Commented Jan 16 at 1:03
  • $\begingroup$ As a minor detail, it might be worth pointing out that the rearrangement in the first paragraph is indeed (AFAICT) always possible, but only because the sequence elements are specified to be positive and to have a finite sum. $\endgroup$
    – Ilmari Karonen
    Commented Jan 17 at 13:41
  • $\begingroup$ @IlmariKaronen: Yes, I do use here that since $a_n\ge 0$, all series are invariant under rearrangement. $\endgroup$
    – Christian Remling
    Commented Jan 17 at 19:13
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    $\begingroup$ …and since $\sum a_n < \infty$, for each $a_n > 0$ there can only be a finite number of elements $a_k$ such that $a_k ≥ a_n$. (If a finite sum wasn't required, $a_n = 1-1/n$ would be an example of a non-negative sequence that cannot be rearranged into descending order.) $\endgroup$
    – Ilmari Karonen
    Commented Jan 17 at 22:21

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