[proofplan]
We first prove the two-term case: assuming $|x| > |y|$, the ultrametric inequality gives $|x + y| \leq |x|$, and a reverse application (writing $x = (x + y) - y$) forces $|x| \leq |x + y|$, yielding equality. The general statement for convergent series with distinct non-zero absolute values follows by induction on partial sums, using the two-term result at each step to identify the dominant term.
[/proofplan]
[step:Establish $|x + y| \leq \max(|x|, |y|)$ from the ultrametric inequality]
The strong triangle inequality (non-archimedean property) gives
\begin{align*}
|x + y| \leq \max(|x|, |y|).
\end{align*}
Assuming $|x| > |y|$, this becomes $|x + y| \leq |x|$.
[/step]
[step:Prove the reverse inequality $|x| \leq |x + y|$ when $|x| > |y|$]
Write $x = (x + y) + (-y) = (x + y) - y$. Applying the ultrametric inequality:
\begin{align*}
|x| = |(x + y) - y| \leq \max(|x + y|, |{-y}|) = \max(|x + y|, |y|).
\end{align*}
Since $|x| > |y|$, the maximum on the right cannot be achieved by $|y|$ alone — that is, if $|x + y| \leq |y|$, we would have $|x| \leq |y|$, contradicting $|x| > |y|$. Therefore $|x + y| \geq |x|$.
[guided]
This is the crucial reverse direction. We know $|x + y| \leq |x|$ from the ultrametric inequality. To get equality, we need $|x + y| \geq |x|$.
Write $x = (x + y) - y$. The ultrametric inequality applied to this decomposition gives
\begin{align*}
|x| = |(x + y) + (-y)| \leq \max(|x + y|, |-y|) = \max(|x + y|, |y|),
\end{align*}
where we used $|-y| = |{-1}| \cdot |y| = |y|$ (since $|{-1}| = 1$ for any absolute value). Now we use the strict inequality $|x| > |y|$. If $|x + y| < |x|$, then $|x + y| \leq |y|$ (since $|x + y| < |x|$ and the only other candidate for the max is $|y|$), and thus $\max(|x+y|, |y|) = |y| < |x|$, contradicting $|x| \leq \max(|x+y|, |y|)$. So we must have $|x + y| \geq |x|$.
The intuition is that when two elements have different "sizes," their sum cannot cancel the larger one — the strong triangle inequality prevents the kind of cancellation that happens in archimedean settings (like $1 + (-1) = 0$ where $|1| = |-1| = 1$ but $|1 + (-1)| = 0$). Here, cancellation can only occur between elements of equal absolute value.
[/guided]
[/step]
[step:Combine to conclude $|x + y| = \max(|x|, |y|)$ when $|x| \neq |y|$]
From the previous two steps, when $|x| > |y|$:
\begin{align*}
|x| \leq |x + y| \leq |x| = \max(|x|, |y|).
\end{align*}
Therefore $|x + y| = |x| = \max(|x|, |y|)$. The case $|y| > |x|$ follows by symmetry (swap $x$ and $y$ and use $|x + y| = |y + x|$).
[/step]
[step:Deduce the general statement for convergent series with distinct absolute values by induction]
Let $x = \sum_{i=0}^{\infty} x_i$ be a convergent series in $K$ with the non-zero terms $|x_i|$ all distinct. Let $S_N = \sum_{i=0}^{N} x_i$ denote the partial sums. We show by induction on $N$ that $|S_N| = \max_{0 \leq i \leq N} |x_i|$.
**Base case.** For $N = 0$, $|S_0| = |x_0| = \max_{0 \leq i \leq 0} |x_i|$.
**Inductive step.** Suppose $|S_{N-1}| = \max_{0 \leq i \leq N-1} |x_i|$. Since the non-zero $|x_i|$ are all distinct, either $|x_N| = 0$ (in which case $S_N = S_{N-1}$ and the claim holds), or $|x_N| \neq |x_i|$ for all $0 \leq i \leq N - 1$, so $|x_N| \neq |S_{N-1}|$ (by the inductive hypothesis). The two-term result applied to $S_N = S_{N-1} + x_N$ gives
\begin{align*}
|S_N| = \max(|S_{N-1}|, |x_N|) = \max\!\left(\max_{0 \leq i \leq N-1} |x_i|,\, |x_N|\right) = \max_{0 \leq i \leq N} |x_i|.
\end{align*}
**Passage to the limit.** Since $x_i \to 0$ (as the series converges), the maximum $M := \max_{i \geq 0} |x_i|$ is attained at some finite index $j$. For all $N \geq j$, we have $|S_N| = M$. Since $S_N \to x$ and $|\cdot|$ is continuous, $|x| = \lim_{N \to \infty} |S_N| = M = \max_i |x_i|$.
[guided]
The general statement extends the two-term result to infinite series. The strategy is straightforward: use induction on partial sums, and then pass to the limit.
For the inductive step, the crucial point is that $|S_{N-1}| = \max_{0 \leq i \leq N-1} |x_i|$ by the inductive hypothesis. Since the non-zero $|x_i|$ are all distinct, the value $|S_{N-1}|$ is one of the $|x_i|$ for $i \leq N-1$, and $|x_N|$ (if non-zero) is different from all of them. Therefore $|S_{N-1}| \neq |x_N|$, and we can apply the two-term isoceles principle to $S_N = S_{N-1} + x_N$:
\begin{align*}
|S_N| = \max(|S_{N-1}|, |x_N|) = \max_{0 \leq i \leq N} |x_i|.
\end{align*}
For the passage to the limit: the series converges, so $x_i \to 0$. Since $|x_i| \to 0$ and the non-zero absolute values are distinct, only finitely many terms can have $|x_i| \geq \varepsilon$ for any $\varepsilon > 0$. In particular, $M = \max_{i \geq 0} |x_i|$ is attained at some finite index $j$. For all $N \geq j$, the partial sum satisfies $|S_N| = M$. Since $|S_N - x| = |{\sum_{i > N} x_i}| \to 0$ (by convergence) and $|\cdot|$ is continuous (being a metric), we conclude $|x| = \lim_{N \to \infty} |S_N| = M = \max_i |x_i|$.
[/guided]
[/step]
