Is the consecutive sum set large in general?

Question

$\DeclareMathOperator\CSS{CSS}$It is well known that for a set $A$ of integers, if $\gcd(A) = d$, then the set of (integer) linear combinations of $A$ is $d\mathbb{Z}$. I'm looking for a probability generalization of this, namely the following.

Let $\varepsilon>0$, a finite set $A$ of positive integers with $\gcd(A) = d$. Let $N$ be large (depending on $A$, $\varepsilon$) and $\alpha\in A^N$ such that the density of every $a\in A$ in $\alpha$ satisfies $\lvert\alpha^{-1}(a)\rvert/N\geq \varepsilon$. Let $\CSS(\alpha)$ (consecutive sum set) denote the set of $b\in\mathbb{N}$ such that for some $n$, $m$, $b = \alpha(n)+\alpha(n+1)+\dotsb+\alpha(n+m-1)$.

Question: Do we have $\lvert\CSS(\alpha) \rvert/\sum_n\alpha(n)\geq (1-\varepsilon)/d$?

Answer 1

No. Consider for instance $A = \{3,5\}$ (so $d=1$) and take $\alpha = (3,5,3,5,3,5,\dots)$. The partial sums $\alpha(1)+\dots+\alpha(n)$ are always equal to $3$ or $0$ mod $4$, so the partial sums $\alpha(n)+\dots+\alpha(n+m-1)$ always avoid $2$ mod $4$. Hence $|CSS(\alpha)|/\sum_n \alpha_n$ cannot exceed $3/4-o(1)$ as $N \to \infty$, which will be less than $(1-\varepsilon)/d$ when $\varepsilon$ is small and $N$ is large.