[proofplan] We prove all three parts in sequence. For (1), we telescope the difference $x_m - x_n$ into consecutive differences $x_{k+1} - x_k$ and apply the ultrametric inequality to bound $|x_m - x_n|$ by $\max_{n \leq k < m} |x_{k+1} - x_k|$, which tends to zero. Part (2) is an immediate consequence of (1) and the definition of completeness. Part (3) follows from (2) by identifying the series with a telescoping sequence of partial sums. [/proofplan] [step:Prove (1): consecutive decay implies Cauchy in a non-archimedean field] Let $\varepsilon > 0$. Since $|x_n - x_{n+1}| \to 0$, there exists $N \in \mathbb{N}$ such that $|x_n - x_{n+1}| < \varepsilon$ for all $n \geq N$. Let $m > n \geq N$. Telescope the difference: \begin{align*} x_m - x_n = \sum_{k=n}^{m-1} (x_{k+1} - x_k). \end{align*} Applying the ultrametric inequality (the strong triangle inequality) iteratively to this finite sum: \begin{align*} |x_m - x_n| = \left|\sum_{k=n}^{m-1} (x_{k+1} - x_k)\right| \leq \max_{n \leq k \leq m-1} |x_{k+1} - x_k| < \varepsilon, \end{align*} since each term satisfies $|x_{k+1} - x_k| < \varepsilon$ for $k \geq N \leq n$. Therefore $(x_n)$ is Cauchy. [guided] In an archimedean field like $\mathbb{R}$, the condition $|x_n - x_{n+1}| \to 0$ does not imply that $(x_n)$ is Cauchy. The standard counterexample is the harmonic partial sums $x_n = \sum_{k=1}^{n} 1/k$: consecutive differences $1/n \to 0$, but the sequence diverges. The reason is that the ordinary triangle inequality only gives $|x_m - x_n| \leq \sum_{k=n}^{m-1} |x_{k+1} - x_k|$, and this sum of small terms can be large. In a non-archimedean field, the ultrametric inequality replaces the sum with a maximum: \begin{align*} |x_m - x_n| = \left|\sum_{k=n}^{m-1} (x_{k+1} - x_k)\right| \leq \max_{n \leq k \leq m-1} |x_{k+1} - x_k|. \end{align*} This is a much stronger bound — the maximum of finitely many terms that are all less than $\varepsilon$ is itself less than $\varepsilon$, regardless of how many terms there are. This is why the non-archimedean property makes the Cauchy criterion dramatically easier to verify. To justify the iterated ultrametric inequality: for two terms, $|a + b| \leq \max(|a|, |b|)$. For three terms, $|a + b + c| \leq \max(|a + b|, |c|) \leq \max(\max(|a|, |b|), |c|) = \max(|a|, |b|, |c|)$. By induction, $|\sum_{k=1}^{N} a_k| \leq \max_{1 \leq k \leq N} |a_k|$ for any finite sum. Given $\varepsilon > 0$, choose $N$ so that $|x_n - x_{n+1}| < \varepsilon$ for all $n \geq N$. Then for $m > n \geq N$: \begin{align*} |x_m - x_n| \leq \max_{n \leq k \leq m-1} |x_{k+1} - x_k| < \varepsilon. \end{align*} So $(x_n)$ is Cauchy. [/guided] [/step] [step:Prove (2): in a complete non-archimedean field, consecutive decay implies convergence] If $K$ is additionally complete and $|x_n - x_{n+1}| \to 0$, then by part (1), $(x_n)$ is Cauchy in $K$. By the definition of completeness, every Cauchy sequence in $K$ converges. Therefore $(x_n)$ converges. [/step] [step:Prove (3): a series converges if and only if its terms tend to zero] Define the partial sums $S_n := \sum_{k=0}^{n} y_k$. The series $\sum_{n=0}^{\infty} y_n$ converges if and only if the sequence $(S_n)$ converges. **($\Leftarrow$):** Suppose $y_n \to 0$. Since $S_n - S_{n-1} = y_n \to 0$, the sequence $(S_n)$ satisfies $|S_n - S_{n+1}| = |y_{n+1}| \to 0$. By part (2) (using completeness), $(S_n)$ converges, so the series converges. **($\Rightarrow$):** Conversely, suppose the series converges, i.e., $(S_n)$ converges to some limit $S$. Then $y_n = S_n - S_{n-1} = (S_n - S) - (S_{n-1} - S) \to 0$ as $n \to \infty$. [guided] In archimedean analysis, the implication $y_n \to 0 \Rightarrow \sum y_n$ converges is famously false (consider the harmonic series $\sum 1/n$). The converse ($\sum y_n$ converges $\Rightarrow y_n \to 0$) holds in all metric spaces. In non-archimedean fields, both directions hold. The ($\Leftarrow$) direction is a direct consequence of parts (1) and (2): define $S_n = \sum_{k=0}^{n} y_k$. Then $S_{n+1} - S_n = y_{n+1} \to 0$, so by part (2) the sequence $(S_n)$ converges. The ($\Rightarrow$) direction is standard: if $S_n \to S$, then $y_n = S_n - S_{n-1} \to S - S = 0$. This makes convergence of series in non-archimedean fields remarkably simple to check: a series converges if and only if its general term tends to zero. No comparison tests, ratio tests, or root tests are needed — just verify that the terms go to zero. [/guided] [/step]