# convergence of the p-adic log in $pZ_p$

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!

• Not a dupe, but related discussion here. See David Loeffler's answer in particular. Well, Lord Shark gives a nice argument for the basic result. – Jyrki Lahtonen Nov 22 '19 at 4:23

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.
Given a P-adic field $$k$$, with $$p=P^e$$, write $$v$$ for $$v_p$$ such that $$v_P$$ = $$e$$ $$v_p$$. Then:
1) The series $$log(1+x)$$ converges for $$v(x) >0$$ : If $$p^r \le n , which implies elementarily that $$v(x^n/n) \to \infty$$ as $$n\to \infty$$ when $$v(x)>0$$. This shows the ultrametric convergence.
2) $$v(n!) : 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 $$[n/p]=a_1+a_2p+...+a_rp^{r-1}, [n/p^2]=a_2+...+a_rp^{r-2}, ..., [n/p^r]=a_r$$. 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.