Is there a known condition for partial sums of a decreasing positive sequence to take all values up to the total sum?

Question

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.)

Answer 1

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 $A$.

Answer 2

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). $\Box$