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Let $a_0>a_1>\cdots>0$ have the property that, for each positive $a<\sum_{n\in\Bbb N}a_n$ (admitting $\infty$ for the sum), there is $A\subset\Bbb N$ such that $a=\sum_{n\in A}a_n$ . Are there known necessary and sufficient conditions on the $a_n$ (not involving arbitrary partial sums) for this property? To illustrate, $a_n=1/(n+1)$ and $a_n=1/2^n$ possess the required property, but $a_n=1/(2+\varepsilon)^n$ does not for any $\varepsilon>0$.

(This question is adapted from one asked on Mathematics Stack Exchange two months ago which received no answer.)

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    $\begingroup$ Even deleting a single term (other than the first term) from the sequence $a_n = 1/2^n$ makes it no longer have your property. $\endgroup$ – Will Brian May 25 at 18:32
  • $\begingroup$ I think it's necessary and sufficient to have $a_n \leq \sum_{m > n}a_m$ for all $n$. Is this the kind of condition you have in mind? I'm not sure what you mean by "not involving arbitrary partial sums". $\endgroup$ – Will Brian May 25 at 18:39
  • $\begingroup$ @WillBrian : Yes, your sum condition doesn't involve indices comprising arbitrary subsets of $\Bbb N$, only terminal segments of $\Bbb N$. So it's not just a rewrite of the originally stated property. $\endgroup$ – John Bentin May 25 at 18:54
  • $\begingroup$ OK, thanks -- I'll see if my idea works, and write an answer if it does :) $\endgroup$ – Will Brian May 25 at 18:54
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    $\begingroup$ I wrote an answer a while ago on Math Stack Exchange that answered exactly this: here. I think it's entirely in line with Will Brian's answer. $\endgroup$ – Milo Brandt May 26 at 1:56
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It is necessary and sufficient that

$\lim_{n \rightarrow \infty}a_n = 0$, and

$a_n \leq \sum_{m > n}a_m$ for all $n$.

In other words: the terms go to zero, and no term is bigger than the sum of all the following terms.

Necessity: First, it is necessary that $\lim_{n \rightarrow \infty}a_n = 0$ because you cannot form any sum smaller than $\lim_{n \rightarrow \infty}a_n$.

Suppose $a_n > \sum_{m > n}a_m$ for some $n$. Let $\varepsilon$ be some number with $0 < \varepsilon < a_n - \sum_{m > n}a_m$. Then I claim there is no $A$ such that $\sum_{m \in A}a_m = a_1+a_2+\dots+a_{n-1}+a_n - \varepsilon$. To see this, just consider two cases: (1) if $\{a_1,\dots,a_n\} \subseteq A$, then $\sum_{m \in A}a_m$ is too big, because $\varepsilon > 0$, and (2) if $a_i \notin A$ for some $i \leq n$, then $\sum_{m \in A}a_m$ is too small, because $\sum_{m \in A}a_m \,\leq\, (a_1+a_2+\dots+a_n) - a_i +\sum_{m > n}a_m \,\leq\, (a_1+a_2+\dots+a_n) -a_n + \sum_{m > n}a_m < (a_1+a_2+\dots+a_n) - \varepsilon.$

Sufficiency: Suppose $a_n \leq \sum_{m > n}a_m$ for all $n$, and let $c$ be any number with $0 \leq c \leq \sum_{m \in \mathbb N}a_m$. Then we can construct the desired $A \subseteq \mathbb N$ recursively, as follows. If it has already been decided for all $m < n$ whether $m \in A$ or not, then put $n \in A$ if and only if $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq c$. (In other words, put $n \in A$ if and only if putting $n \in A$ does not make the sum too big.) Once we have built $A$ according to this rule, it is clear that $\sum_{m \in A}a_m \leq c$, because none of the finite partial sums exceeds $c$.

Now suppose, aiming for a contradiction, that $\sum_{m \in A}a_m < c$, and let $\varepsilon = c - \sum_{m \in A}a_m$. There is some $N$ such that $a_n < \varepsilon$ for all $n \geq N$. For each such $n$, we have $\sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \leq \sum_{m \in A} a_m = c - \varepsilon$, and hence $a_n + \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m < c$. By our rule for constructing $A$, this means $n \in A$ for all $n \geq N$. In other words, $A$ is a co-finite subset of $\mathbb N$.

Let $n$ denote the largest member of $\mathbb N \setminus A$. (Note that $\mathbb N \setminus A \neq \emptyset$, because $\sum_{m \in A}a_m < c \leq \sum_{m \in \mathbb N}a_m$.) Then $\left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+a_n \leq \left( \sum_{m \in A \cap \{1,2,\dots,n-1\}} a_m \right)+ \sum_{m > n}a_m = \sum_{m \in A}a_m < c$. This is the contradiction we were after, because this tells us that we should have had $n \in A$, although $n$ was supposed to be the largest number not in $A$.

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    $\begingroup$ Your condition holds when e.g. $a_n=1+1/(n+1)$, whereas then $a=1$ is not a partial sum. $\endgroup$ – Iosif Pinelis May 25 at 19:37
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    $\begingroup$ @IosifPinelis: I was assuming that $\lim_{n \rightarrow \infty} a_n = 0$. I just failed to notice that this wasn't part of the question. I'll edit to clarity. As for the rest of it, it looks like maybe we had pretty much the same idea at the same time! $\endgroup$ – Will Brian May 25 at 19:40
  • $\begingroup$ $\lim_{n\to\infty}a_n=0$ is entailed by the original condition, because $a$ can be arbitrarily close to zero. $\endgroup$ – John Bentin May 25 at 19:45
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    $\begingroup$ @JohnBentin: Yes -- this is the "necessity" argument for that half of my condition. But it should still be included in the answer because, as Iosif Pinelis points out in his comment, the other half of my condition is not enough by itself. It's the conjunction of the two statements that is both necessary and sufficient. $\endgroup$ – Will Brian May 25 at 20:05
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    $\begingroup$ I will add a link to my on answer on Mathematics which contains some references: Surjective Function from a Cantor Set. $\endgroup$ – Martin Sleziak May 26 at 19:14
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In order for all $a\in(0,a_0+a_1+\cdots)$ to be representable as partial sums of of the $a_i$'s, it is necessary that \begin{equation*} a_\infty:=\lim_n a_n=0; \tag{1} \end{equation*} otherwise, no $a\in(0,a_\infty)$ is a partial sum of the $a_i$'s. So, assume (1).

A sufficient condition is that for all natural $n$ \begin{equation*} a_{n-1}\le a_n+a_{n+1}+\cdots \tag{2} \end{equation*} for all $n$. Indeed, assume (2) holds. Take any $a\in(0,a_0+a_1+\cdots)$. Successively define \begin{equation*} S_0:=\{k\ge0\colon a_k\le a\},\quad k_1:=\min S_0 \end{equation*} and, for $j\ge2$, \begin{equation*} S_{j-1}:=\{k>k_{j-1}\colon a_k\le a-s_{j-1}\},\quad k_j:=\min S_{j-1}, \end{equation*} where \begin{equation*} s_j:=a_{k_1}+\cdots+a_{k_j}. \end{equation*} If $S_{j-1}=\emptyset$ for some $j=1,2,\dots$, then, by (1), ($j\ge2$ and) $s_{j-1}=a$, so that we are done.

It remains to consider the case when $S_{j-1}\ne\emptyset$ for all $j=1,2,\dots$, so that we have $0\le k_1<k_2<\cdots$. Without loss of generality, \begin{equation*} s_j\le a-h \end{equation*} for some real $h>0$ and all $j$. By the construction, for each $j$ either

(i) $k_j=k_{j-1}+1$ or

(ii) $a_{k_j-1}>a-s_{j-1}$.

In case (ii), $a_{k_j-1}>a-s_{j-1}=a-s_j+a_{k_j}\ge h+a_{k_j}$. So, if case (ii) holds for infinitely may $j$'s, then, letting $j\to\infty$ and recalling (1), we get $0\ge h+0$, a contradiction.

So, case (i) holds eventually, for all large enough $j$. Then for some natural $m$ and $n$ we have $s_{m-1}+a_n+a_{n+1}+\cdots<a<s_{m-1}+a_{n-1}$, which contradicts (2). $\Box$

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  • $\begingroup$ Why the downvote? Is there anything wrong with this answer? $\endgroup$ – Iosif Pinelis May 26 at 12:06
  • $\begingroup$ (I did not downvote.) Is your first condition flipped? It currently says that it's necessary for the limit to be positive. $\endgroup$ – aorq May 26 at 14:26
  • $\begingroup$ @aorq : Oops! Of course, I meant that $a_\infty$ must be $=0$, not $>0$. Thank you for the comment. $\endgroup$ – Iosif Pinelis May 26 at 18:28

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