Consider the p-adic absolute value $ |.|_p$ defined by $|x|_p=|p^{n} \frac{a}{b}|_p=p^{-n}$, where $p$ does not divide $a,b$ and let
$\mathbb{Z}$=set of integers,
$\mathbb{Q}$=set of rational number,
$\mathbb{Z}_p$=completion of $\mathbb{Z}$ under $|.|_p$,
$\mathbb{Q}_p$=completion of $\mathbb{Q}$ under $|.|_p$,
$\bar{\mathbb{Q}_p}$=algebraic closure of $\mathbb{Q}_p$ which is not complete with respect to $|.|_p$,
$\Omega$=completion of $\bar{\mathbb{Q}_p}$ with respect to $|.|_p$.
Define $D(r^{-})=\{x \in \Omega: |x|_p<r \}$. Then we have Lemma which is stated as below:
$Lemma \ 1:$
Every $ f(x) \in\mathbb{Z}_p[[x]]$ converges in $D(1^{-})$.
The proof of this lemma is given in the book of N. Koblitz, p-adic Numbers.
Now consider the binomial function $ B_{a,p}(x)=(1+x)^a=\sum_{n=0}^{\infty} \frac{a(a-1)(a-2) \cdots (a-n+1)}{n!}x^n$.
If $ a \in \mathbb{Z}_p$, then $B_{a,p}(x) \in \mathbb{Z}_p[[x]]$.
My question:-
By the above Lemma, we conclude that
$B_{a,p}(x)$ converges in $D(1^{-})$.
$ \text{Does it mean that the radius of convergence is equal to}$ $ \ 1 \ $?
If not then which condition imply that the radius of convergence of the series represented by $B_{a,p}(x)$ is equal to $ \ 1 \ $ using ratio test or Cauchy-Hadamard method?
I need any condition on the constant $a$ such that the series has radius of convergence equal to $1$?
Help me