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

Question

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!

Answer 1

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.