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Kaneko, Suzuki and Tachiya, arXiv:2601.20743v1, Theorem 2, printed/PDF p. 3.

Use $q,d,\mathrm h,\mathcal N_c,S_c$ and condition (G) from Theorem 1. Let $a,b$ be sequences of algebraic integers of $\mathbb Q(q)$, with $a(n)\ge0$ for every $n\ge1$ and $\mathcal N_a$ infinite.

Theorem. Assume

$$\rho=\limsup_{n\to\infty} \max\{\mathrm h(a(n)),\mathrm h(b(n))\}^{1/n}<q.$$

Suppose real sequences $x_j,y_j,z_j\ge1$ satisfy

$$x_j\to\infty,\qquad S_a(x_j),S_b(x_j)=O(y_j),\qquad \#\mathcal N_a(x_j),\#\mathcal N_b(x_j)=o(x_j/z_j),$$ $$\limsup_{j\to\infty}y_j^{(d-1)/x_j}<q/\rho,$$

and

$$\sum_{1\le m<x_j}a(m),\quad \sum_{1\le m<x_j}|b(m)| =o(q^{z_j}x_j/y_j^{d-1}).$$

The last two sums use the distinguished real embedding, whereas $S_c$ uses the maximum over conjugates. If $\mathcal N_b$ is infinite, also assume (G). Then

$$\sum_{n\ge1}\frac{a(n)+b(n)}{q^n}\notin\mathbb Q(q).$$

The root bound gives convergence. In this setting $\rho\ge1$ because a nonzero algebraic integer has conjugate size at least one and $\mathcal N_a$ is infinite, so $q/\rho$ is well defined.

Proof pointer and standing

The proof is on pp. 12–13, using Lemma 4 on pp. 11–12. That lemma separates coefficients below the cutoff from the entire remainder beyond it. The root and auxiliary growth bounds control the latter; the mass estimate controls the former. Together they supply Theorem 1's averaged complete-tail condition for a suitable fixed $\eta$.

This is a checked author statement extraction with a read proof route, not a complete reconstructed proof or an independent review. For integer bases, $d=1$ removes the auxiliary $y_j$ factors and yields Theorem 3.

Bears on. Problem 249 through possible sparse transformations. No admissible transformation of its full series has been constructed here.