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Let $\sum_{n\ge0}a_nz^n$ be a power series with coefficients in $\mathbb Q_p$ ($p$ a power of prime). Assume that the series converges for all $z\in\{x\in\mathbb Q_p\mid v(x)>0\}$ and there exists a $z_0\in\{x\in\mathbb Q_p\mid v(x)=0\}$ such that $\sum_{n\ge0}a_n z_0^n$ converges. My question: does the series converge for every $z\in\{x\in\mathbb Q_p\mid v(x)=0\}$ ? Thnaks in advance for any hints or answers.

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1 Answer 1

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Your question is this: given a series $f=\sum_na_nz^n$ that converges for all $z\in\Bbb Q_p$ with $v(z)>0$, and given $z_0\in\Bbb Q_p$ with $v(z_0)=0$ such that $\sum_na_nz_0^n$ converges, does the series $f$ converge for every $\zeta\in\Bbb Q_p$ with $v(\zeta)=0$?

The condition for convergence of $\sum_na_nz_0^n$ is that $\lim_{n\to\infty}\bigl(v(a_nz_0^n)\bigr)=\infty$; but since $v(a_nz_0^n)=v(a_n)$, your hypothesis is that $v(a_n)\to\infty$. This forces $v(a_n\zeta^n)\to\infty$, so the answer to your question is Yes.

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