The Banzhaf form of Penrose's limit theorem

Let $W\subseteq\mathbb{R}_{\ge 0}$ be a set of weights, let $N^{(0)}\subsetneq N^{(1)}\subsetneq\dots$ be an infinite increasing chain of finite non-empty sets, and put $N=\bigcup_{n=0}^{\infty}N^{(n)}$. Assign each voter $i\in N$ a weight $w_i\in W$. For a fixed quota $q\in(0,1)$, let $\mathcal{W}^{(n)}$ be the weighted majority game with voter set $N^{(n)}$, weights $w_i$ and quota $q\cdot\sum_{i\in N^{(n)}}w_i$. A voter $i$ is regular if there are $n_0\in\mathbb{N}$ and $\varepsilon>0$ such that

$$\left|\left\{h\in N^{(n)}\mid w_h=w_i\right\}\right|\cdot\frac{w_i}{\sum_{h\in N^{(n)}}w_h}\ge\varepsilon$$

for all $n\ge n_0$, and the chain is non-atomic if $\lim_{n\to\infty}w_i/\sum_{h\in N^{(n)}}w_h=0$ for every $i\in N$. Banzhaf–Penrose limit conjecture. Penrose's limit theorem holds for regular players in non-atomic chains for the Banzhaf index with every relative quota $q\in(0,1)$; that is, for every pair $i,j\in N$ with nonzero weights,

$$\lim_{n\to\infty}\frac{\operatorname{Bz}_i(\mathcal{W}^{(n)})}{\operatorname{Bz}_j(\mathcal{W}^{(n)})}=\frac{w_i}{w_j}.$$

The corresponding result is known for the Banzhaf index at relative quota $q=\frac12$, while the arbitrary-quota case remains open.