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The function $$ f : z \in \mathbb{C} \longmapsto \sum_{i} \frac{a_i}{1-a_iz} $$ is meromorphic on $\mathbb{C}$ and has integral Taylor coefficients. It follows from a theorem of Borel that such a function must be in $\mathbb{Q}(z)$; see for example Richard Stanley's answer herehere. In particular $f$ has only finitely many poles ; this implies that only finitely many of the $a_i$'s are nonzero.

The function $$ f : z \in \mathbb{C} \longmapsto \sum_{i} \frac{a_i}{1-a_iz} $$ is meromorphic on $\mathbb{C}$ and has integral Taylor coefficients. It follows from a theorem of Borel that such a function must be in $\mathbb{Q}(z)$; see for example Richard Stanley's answer here. In particular $f$ has only finitely many poles ; this implies that only finitely many of the $a_i$'s are nonzero.

The function $$ f : z \in \mathbb{C} \longmapsto \sum_{i} \frac{a_i}{1-a_iz} $$ is meromorphic on $\mathbb{C}$ and has integral Taylor coefficients. It follows from a theorem of Borel that such a function must be in $\mathbb{Q}(z)$; see for example Richard Stanley's answer here. In particular $f$ has only finitely many poles ; this implies that only finitely many of the $a_i$'s are nonzero.

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The function $$ f : z \in \mathbb{C} \longmapsto \sum_{i} \frac{a_i}{1-a_iz} $$ is meromorphic on $\mathbb{C}$ and has integral Taylor coefficients. It follows from a theorem of Borel that such a function must be in $\mathbb{Q}(z)$; see for example Richard Stanley's answer here. In particular $f$ has only finitely many poles ; this implies that only finitely many of the $a_i$'s are nonzero.