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# 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.

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

$\displaystyle a_{1}\!+\!a_{2}\!+\!a_{3}\!+\ldots$ | (1) |

in $K$ is that

$\displaystyle\lim_{{n\to\infty}}a_{n}\;=\;0.$ | (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

$|a_{{m+1}}|\;=\;\left|\sum_{{j=1}}^{{m+1}}a_{j}-\sum_{{j=1}}^{{m}}a_{j}\right|% \;<\;\varepsilon$ |

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

$|a_{m}|\;<\;\varepsilon\qquad\forall m\,\geqq\,n_{\varepsilon}.$ |

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.

## Mathematics Subject Classification

12J10*no label found*54E35

*no label found*

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