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Let $d$ be an integer greater than $1$ and let $f(z)=\sum_{n\ge0}z^{d^n}$ be the Fredholm series. It is well-known that the Roth-Ridout theorem implies that $f(r)$ is a transcendental number for all non-zero rational $r$. By Mahler’s method, it is even true for any nonzero algebraic $r$. Can one extend the arguments of the proof based on the Roth-Ridout theorem to recover Mahler's result?

Thank you in advance.

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    $\begingroup$ My (limited) understanding is that Fredholm never considered such series. Should they be named after Mahler? $\endgroup$
    – Sam Nead
    Commented May 31, 2021 at 7:41

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It seems that the answer is no. From this paper, the Roth-Ridout theorem is not sufficient to establish transcendence for algebraic $r$. Indeed, it's not enough to completely cover the rational case. To obtain the same result as the Mahler's method, one should invoke the more general ($p$-adic) Subspace Theorem, as shown in the Corvaja-Zannier paper.

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