I've been trying to solve the following problem:
Show that the series $\log(1 + x) = \sum_{n = 1}^{\infty} (-1)^{n+1} x^n/n$ converges on $p\mathbb{Z}_p$ with respect to $|.|_p$ (p-adic absolute value). Show that for any positive integer $n$, $v_p(n!) < n/(p - 1)$.
I tried to prove that the sequence of residues, i.e., $\{\sum_{n = m}^{\infty} (-1)^{n+1} x^n/n\}_{m \to\infty}$ converges to $0$ but I couldn't do it. I believe I am missing something obvious here. Thanks!
On
Given a P-adic field $k$, with $p=P^e$, write $v$ for $v_p$ such that $v_P$ = $e\cdot v_p$. Then:
The series $\log(1+x)$ converges for $v(x) >0$ : If $p^r \le n <p^{r+1}, v(x^n/n)=nv(x )- v(n) \ge n v(x)-r$, which implies elementarily that $v(x^n/n) \to \infty$ as $n\to \infty$ when $v(x)>0$. This shows the ultrametric convergence.
$v(n!)<n/(p-1)$ : Write $n$ in basis $p$, i.e. $n=a_0+a_1p+...+a_rp^r$, where $a_i\in \mathbf Z$ and $0\le a_i \le p-1$. Then \begin{align} &[n/p]=a_1+a_2p+...+a_rp^{r-1} \\ &[n/p^2]=a_2+...+a_rp^{r-2} \\ &~~~~~~~~~~~~\vdots \\& [n/p^r]=a_r \end{align}But clearly $v(n!)=[n/p]+[n/p^2]+...+[n/p^r]=a_1+(1+p)a_2+...+(1+p+...+p^{r-1})a_r$, so that $(p-1)v(n!)=0+(p-1)a_1+(p^2-1)a_2+...+(p^r-1)a_r=n-(a_0+a_1+...+a_r)$. This shows the desired inequality.
Over the $p$-adics, because of the ultrametric property, to show that a series $\sum_n a_n$ converges all that's required is to prove that $a_n\to0$. It is I hope clear that $v_p(n)\le\log_pn$ (this is the usual real logarithm to base $p$ not a $p$-adic logarithm). Thus $$v_p(x^n/n)\ge n-\log_pn\to\infty$$ for $x\in p\Bbb Z_p$. That is, $x^n/n\to0$ $p$-adically. This then implies that the series for the $p$-adic logarithm converges.