# Convergence of infinite product in $\mathbb{C}_p$

In classical analysis over $$\mathbb{C}$$ an infinite product: $$\prod_{n \geq 0}a_n$$ converges iff the series $$\sum_{n \geq 0}\log(a_n)$$converges. In the $$p$$-adic setup unfortunately the radius of convergence of the logarithm is small so I think that this approach can not be used. Ther are so far criterium of convergence or absolute convergence for this type of products in $$\mathbb{C}_p$$ ? For example is it enough that : $$\lim_{n \to \infty}\prod_{i=0}^n |a_i|_p < \infty$$ to prove (absolute ?) convergence? Thanks for advices!

A sequence $$\{s_n\}$$ converges in a complete space if and only if it is a Cauchy sequence. With respect to an ultrametric norm (as in the $$p$$-adic case) a sequence is Cauchy if and only if $$|s_{n} - s_{n-1}| \rightarrow 0$$, because the strong triangle inequality gives

$$|s_n - s_m| \le \max\{|s_{n-i} - s_{n-i-1}|\}, \quad i = 0,\ldots,n+1-m.$$

For infinite products, one often says (by convention) that products tending to zero do not converge. With that convention, and assuming the $$a_n$$ are never $$-1$$, we have

$$\prod (1 + a_i) \ \text{converges if and only if |a_i| \rightarrow 0 if and only if \sum a_i converges.}$$

If $$s_n$$ denotes the partial products, then $$s_n - s_{n-1} = s_n a_n$$ so

$$|s_n - s_{n-1}| = |s_n||a_n|.$$

Assume the product converges. Then $$|s_n|$$ is eventually constant and non-zero, so the fact that $$|s_n - s_{n-1}|$$ tends to zero implies that $$|a_n|$$ tends to zero. Conversely, if $$|a_n|$$ tends to zero, then $$|1+a_n| = 1$$ for large enough $$n$$, and so $$|s_n|$$ is constant (and non-zero since $$a_n$$ is never $$-1$$) for large enough $$n$$, and then $$|s_n -s_{n-1}|$$ also tends to zero.