# Power series in p-adic integers

How can we show that for $x \in \mathbb{Z}_p$, $\log_p(1+x)$ converges in $\mathbb{Z}_p$ when $|x|_p < 1$?

To clarify, $\log_p(1+x)$ is the power series: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$$

How can we show that $\exp_p(x)$ converges for: $|x|_p < p^\frac{-1}{p-1}$? How can we approximate the largest power of $p$ dividing $n!$?

$$\exp_p(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$

I've shown that if $\lim_{n \rightarrow \infty} |a_n|_p = 0$, then $\sum_{n=1}^\infty a_i$ converges. How do I show the following: $$\lim_{n\rightarrow\infty} \left|\frac{(-1)^{n+1}x^n}{n}\right|_p = 0$$ $$\lim_{n \rightarrow\infty} \left|\frac{x^n}{n!}\right|_p = 0$$ I know that: $$\left|\frac{1}{n!}\right|_p = p^\frac{n-S_n}{p-1}$$

• – Jyrki Lahtonen Apr 15 '15 at 12:42
• What do you mean a function converges? Perhaps you meant some power series? – Timbuc Apr 15 '15 at 12:44
• Yes, $\log_p(1+x)$ in its power series representation, as now described above – James Johnson Apr 15 '15 at 13:59
• This is discussed in nearly every book on $p$-adic analysis. Have you read about convergence of power series in any book on $p$-adics (Gouvea, Koblitz, Robert, ...)? – KCd Apr 15 '15 at 14:25
• No, I have not. – James Johnson Apr 15 '15 at 15:25