[proofplan] The proof uses only the definition of convergence of a series through its partial sums. If the partial sums $s_N$ converge to a limit $S$, then consecutive partial sums $s_n$ and $s_{n-1}$ both become arbitrarily close to $S$. Since each term is the difference $a_n = s_n - s_{n-1}$, the triangle inequality forces $a_n$ to approach $0$. The divergence criterion is then the contrapositive of this implication. [/proofplan] [step:Express each term as a difference of consecutive partial sums] For each $n \geq 2$, the definitions of $s_n$ and $s_{n-1}$ give \begin{align*} s_n - s_{n-1} &= \sum_{k=1}^{n} a_k - \sum_{k=1}^{n-1} a_k \\ &= a_n. \end{align*} Thus \begin{align*} a_n = s_n - s_{n-1} \end{align*} for every $n \geq 2$. [/step] [step:Use convergence of the partial sums to bound each term] Assume that $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{K}$. By definition, there exists $S \in \mathbb{K}$ such that $s_N \to S$ as $N \to \infty$. Let $\varepsilon > 0$. Since $s_N \to S$, there exists $N_0 \in \mathbb{N}$ such that \begin{align*} |s_N - S| < \frac{\varepsilon}{2} \end{align*} for every $N \geq N_0$. If $n \geq N_0 + 1$, then both $n \geq N_0$ and $n-1 \geq N_0$. Using $a_n = s_n - s_{n-1}$ and the triangle inequality in $\mathbb{K}$, we obtain \begin{align*} |a_n| &= |s_n - s_{n-1}| \\ &= |(s_n - S) + (S - s_{n-1})| \\ &\leq |s_n - S| + |s_{n-1} - S| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align*} Therefore, for every $\varepsilon > 0$, there exists $N_0 + 1 \in \mathbb{N}$ such that $|a_n| < \varepsilon$ whenever $n \geq N_0 + 1$. Hence \begin{align*} \lim_{n \to \infty} a_n = 0. \end{align*} [/step] [step:Take the contrapositive to obtain the divergence criterion] We have proved the implication \begin{align*} \sum_{n=1}^{\infty} a_n \text{ converges in } \mathbb{K} \implies \lim_{n \to \infty} a_n = 0. \end{align*} Taking its contrapositive gives \begin{align*} \lim_{n \to \infty} a_n \neq 0 \implies \sum_{n=1}^{\infty} a_n \text{ does not converge in } \mathbb{K}. \end{align*} Here “$(a_n)$ does not converge to $0$” includes both the case where $(a_n)$ converges to a nonzero limit and the case where $(a_n)$ has no limit. This is exactly the asserted divergence criterion. [/step]