# $p$-adic sine series

This is a quick question about the domain of convergence of $p$-adic sine series. We define the $p$-adic sine function by the following power series

$\sin_p(X) = \sum\limits_{n=0}^\infty (-1)^n\frac{X^{2n+1}}{(2n+1)!}$

The general term is $a_n = (-1)^n/(2n+1)!$ and the region of convergence for the series is $\frac{1}{\limsup\limits_{n \to \infty} |a_n|_p^{1/n}}$. Now I've computed the radius of convergence to be $p^{-2(p-1)}$, which follows from the fact that for any natural number $n$, $v_p(n!) = \frac{n-S_n}{p-1}$ where $v_p$ is the $p$-adic valuation of $n$ and $S_n$ is the sum of the digits in the base $p$ expansion of $n$.

I have two questions about this. First, one of the resources I have says the radius should be $p^{-1/(p-1)}$, and I don't see why this is true.

Secondly, in $\mathbb{Q}_p$, the region of convergence for this series is

$D = \{x \in \mathbb{Q}_p : |x|_p \leq p^{-2/(p-1)}\} = \{x \in \mathbb{Q}_p : |x|_p < p^{-1/(p-1)}.$

Also, all the resource I can find define $\sin_p$ on this disk. My question is if we pass to say $\mathbb{C}_p$, would we not have to make the distinction because there could be elements $x \in \mathbb{C}_p$, that satisfy

$p^{-2/(p-1)} < |x|_p < p^{-1/(p-1)}$

so that we have to be a bit careful in how we define the sine series initially (if this is an obvious question forgive me, I've never had to deal with extensions of $p$-adic fields and am just delving into this material for the first time).

Your error seems to have been in writing $a_n=(-1)^n/(2n+1)!$ Here, $a_n=0$ if $n$ is even and $a_n=(-1)^{(n-1)/2}/n!$ if $n$ is odd.