# Analytic continuation of a $p$-adic function

Let $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ be sequences of $$\mathbb Q_p$$ such that the function $$f:z\in\mathbb Q_p\to\sum_{n\ge0}a_nz^n+b_nz^{n+1}$$converges in $$\{|z|_p<1\}$$. Assume that the series $$\sum_{n\ge0}a_n+b_n$$ converges in $$\mathbb Q_p$$. Can the function $$f$$ be continued in a larger disk in an analytic function?

• You are not assuming that $\sum a_nz^n$ and/or $\sum b_nz^n$ converge, right?
• $\sum_{n\ge}a_nz^n$ and $\sum_{n\ge}b_nz^n$ converge for $|z|_p<1$ but not $\sum_{n\ge}a_n$ and $\sum_{n\ge0}b_n$. Mar 9, 2019 at 4:33
I think the answer is "not necessarily", by the following (counter)example. First, let $$b_n=-a_n$$ for $$n\ge0$$. Then $$\sum(a_n+b_n)=0$$ and $$f(z)=\sum_{n\ge0} a_n(1-z)z^n=\sum_{n\ge0} (a_n-a_{n-1})z^n,$$ where we set $$a_{-1}=0$$. Now, define $$a_n:=\frac{-1}{n+1}.$$ Then, $$\sum a_nz^n$$ converges if and only if $$|z|_p<1$$. Also, $$f(z)=\sum_{n\ge0}\frac{1}{n(n+1)}z^n,$$ which converges if and only if $$|z|_p<1$$.