Sum-free sequence but multiset

Question

The question is:

Show that if $S$ is a set of natural numbers such that no number in S can be expressed as a sum of other (not necessarily distinct) numbers in S, then $\sum_{ s \in S} \frac{1}{s} \leq 1$

So, $S \neq \{1,2\}$ since $2 = 1 + 1$; the same element can be used multiple times in the sum.

I looked online and found a Wikipedia page about sum-free sequences, but it seems those only deal with distinct subset sums of $S$. The bound for those is at best $2.85\dotsc$, so I was wondering if this more restrictive version of the problem was in the literature.

Answer 1

Claim: Given such a set $S$ that satisfies the conditions, if $a$ is the smallest term, then we have the stronger statement that

$$\sum \frac{1}{s} \leq \frac{1}{a} + \frac{1}{a+1} + \ldots + \frac{1}{2a-1} .$$

Equality occurs iff $ S = \{ a, a+1, a+2, \ldots 2a-1 \}$, which clearly satisfies the conditions.
Furthermore, equality holds in OP's case iff $ S = \{1\}$.

Proof of claim: I encourage you to do this yourself.

Hints:

1. Show that $|S| \leq a $.

Working modulo $a$, each residue appears at most once, otherwise the larger term can be written as the sum of $a$ multiple times and the smaller term.

2. How can we bound the sum?

The term with residue $i$ is at least $ a+ i$, since $a$ is the smallest term. Hence, the claim follows.