$\log (1+x)$ when $x$ is $p$-adic It's written when $x$ is $p$-adic integer then $\log (1+x) = \sum (-1)^{n-1}\frac{x^n}{n}$ converges, I don't understand what this statement mean. Can one please explain me ?
 A: In the standard setting we are most used to we say that a sequence of real numbers $\{x_n\}_{n=1}^\infty$ converges to a real number $x$ if 
$$\lim_{n\to\infty}|x_n - x| = 0$$
where $|\cdot|$ is the absolute value norm. Convergence of a sequence of $p$-adic numbers is defined the same way, but with $|\cdot|$ being the $p$-adic norm (and $x_n,x$ being $p$-adic numbers). 
In practice the Cauchy-criterion for convergence is perhaps more useful as a working definition (and again with $|\cdot|$ being the $p$-adic norm).

Before you try to solve this problem I would strongly suggest you make yourself familiar with the $p$-adic numbers by solving some simpler problem. A quick google search gives many introductions, for example this.  Anyway, here are some hints:
One very useful fact to solve the problem at hand is this: a series $\sum_{n=1}^\infty a_n$ of $p$-adic numbers converges iff $\lim_{n\to\infty} |a_n|_p = 0$. This result (try to prove it) is a consequence of the non-achimedian property $|x+y|_p \leq \text{max}(|x|_p,|y|_p)$ of the $p$-adic norm.
Thus you need to show that
$$\lim_{n\to\infty} \left|\frac{(-1)^nx^n}{n}\right|_p = \lim_{n\to\infty}\left(|x|_p \cdot \left|\frac{1}{n}\right|_p^{1/n}\right)^n = 0$$
if $x$ is a $p$-adic integer. If you now can prove the following two things


*

*If $x$ is a $p$-adic integer with $p|x$ then $|x|_p< 1$

*$\limsup_{n\to\infty} |1/n|_p^{1/n} = 1$.


then the desired result follows. Some hints: for 1. try to use the non-archimedian property mentioned above togeather with the definition of a $p$-adic integer and norm.
