complete ultrametric field

A field $K$ equipped with a non-archimedean valuation $|\cdot|$ is called a non-archimedean field or also an ultrametric field, since the valuation induces the ultrametric $d(x,\,y) = |x\!-\!y|$ of $K$ .

Theorem 1   Let $(K,\,d)$ be a complete ultrametric field. A necessary and sufficient condition for the convergence of the series

(1)

in $K$ is that

(2)

Proof. Let $\varepsilon$ be any positive number. When (1) converges, it satisfies the Cauchy condition and therefore exists a number $m_\varepsilon$ such that surely $$|a_{m+1}| = |\sum_{j=1}^{m+1}a_j-\sum_{j=1}^{m}a_j| < \varepsilon$$ for all $m \geqq m_\varepsilon$ ; thus (2) is necessary. On the contrary, suppose the validity of (2). Now one may determine such a great number $n_\varepsilon$ that $$|a_m| < \varepsilon \quad \forall m \geqq n_\varepsilon.$$ No matter how great is the natural number $n$ , the ultrametric then guarantees the inequality $$|a_m\!+\!a_{m+1}\!+\ldots+\!a_{m+n}| \leqq \max\{|a_m|,\,|a_{m+1}|,\,\ldots,\,|a_{m+n}|\} < \varepsilon$$ always when $m \geqq n_\varepsilon$ . Thus the partial sums of (1) form a Cauchy sequence, which converges in the complete field. Hence the series (1) converges, and (2) is sufficient.