Ordinary generating functions with finitely many singularities at algebraic numbers are rational I have a proof of the following fact related to ordinary generating functions, and I was curious if it was known, as it seems plausible it is classically known:
"Let $\lambda_1,\ldots, \lambda_k$ be algebraic numbers. Let $f(z)= \sum^\infty_{n=0} c_nz^n$ when each $c_n\in \mathbb{Z}.$ Suppose $f$ analytically continues to $\mathbb{C}\setminus \{\lambda_1,\ldots,\lambda_k\}.$ Then, $f$ is a rational function."
 A: It is not necessary to assume that $\lambda_i$ are algebraic. This is a special case of the result in G. Pólya, Mathematische Annalen (1928), Volume: 99, page 687-706, page 704 in particular:

Der Spezialfall, in dem $\mathfrak A$ abzählbar, also $\tau=0$ ist, liefert folgende Aussage: Wenn die Koerffizieten $a_0, a_1, a_2, \dots$ der Potenzreihe
$$a_0+a_1z+a_2z^2+\dots=F(z)$$
ganze Zahlen sind, so muß für die Funktion $F(z)$ einer der folgenden drei Fälle  zutreffen: Entweder ist $F(z)$ eine rationale Funktion, oder ist $F(z)$ eine mehrdeutige Funktion, oder besitzt $F(z)$ unabzählbar viele singuläre Punkte.

The paper actually works with the function $f(z)=F(1/z).$ This is defined near $\infty,$ and it's either rational, doesn't have a unique analytical continuation, or its analytic continuation is singular on an uncountable set $\mathfrak A.$ The quantity $\tau$ is the capacity of $\mathfrak A$ (Chebyshev constant, transfinite diameter), which is zero for compact countable sets.
A: $\lambda_i$ is a pole of finite degree $n_i$. Therefore $f\prod (z-\lambda_i)^n_i$ is a polynomial.
