The following question was asked in a talk of P-adic numbers and I was not able to complete the proof of this result.
Question : Show that $log_p(x) =\sum_{n=1}^{\infty} (-1)^{n+1} \frac{ (x-1)^n } {n} $ for $x\in 1+ p\mathbb{Z}_p <=> |x-1|_p <1$.
Attempt: I have proved that $log_p(a) $ converges for $|x-1|_p<1$ but I am not able to show how does $x\in 1+p\mathbb{Z}_p <=> |x-1|_p< 1$.
Edit 1 : I am not able to show how does the condition $|x-1|_p <1$ is equivalent to $x\in 1+ p\mathbb{Z}_p$.
Please help me with this!
Using the definition of $|y|_p$ for $y \in \mathbf Q_p$, we have (i) $y \in \mathbf Z_p \Longleftrightarrow |y|_p \leq 1$ and (ii) $y \in p\mathbf Z_p \Longleftrightarrow |y|_p \leq 1/p$.
Since no $p$-adic absolute value is strictly between $1/p$ and $1$, $|y|_p<1 \Longleftrightarrow |y|_p \leq 1/p$. Thus $y \in p\mathbf Z_p \Longleftrightarrow |y|_p < 1$. Using $x-1$ for $y$, $|x-1|_p < 1 \Longleftrightarrow x-1 \in p\mathbf Z_p$, which is equivalent to $x \in 1 + p\mathbf Z_p$.