[proofplan] We rewrite the series as its sequence of partial sums. The displayed tail sum is exactly the difference of two partial sums, so the stated condition is precisely the Cauchy condition for the partial-sum sequence. Convergence of the series gives the condition by the triangle inequality, and the condition gives convergence because $\mathbb{R}$ and $\mathbb{C}$ are complete. [/proofplan] [step:Express tail sums as differences of partial sums] Define the partial-sum sequence $s: \mathbb{N} \to \mathbb{K}$ by \begin{align*} s_m := \sum_{n=1}^{m} a_n. \end{align*} By definition, the series $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{K}$ if and only if the sequence $(s_m)_{m=1}^{\infty}$ converges in $\mathbb{K}$. For integers $q > p \geq 1$, subtracting the two finite sums gives \begin{align*} s_q - s_p &= \sum_{n=1}^{q} a_n - \sum_{n=1}^{p} a_n \\ &= \sum_{n=p+1}^{q} a_n. \end{align*} Thus the tail expression in the theorem is exactly $|s_q - s_p|$. [/step] [step:Derive the tail estimate from convergence of the series] Assume that $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{K}$. Then there exists $S \in \mathbb{K}$ such that $s_m \to S$ as $m \to \infty$. Let $\varepsilon > 0$. By convergence of $(s_m)$ to $S$, there exists $N \in \mathbb{N}$ such that for every $m \geq N$, \begin{align*} |s_m - S| < \frac{\varepsilon}{2}. \end{align*} If $q > p \geq N$, then both $p$ and $q$ are at least $N$, so the triangle inequality gives \begin{align*} \left|\sum_{n=p+1}^{q} a_n\right| &= |s_q - s_p| \\ &= |(s_q - S) + (S - s_p)| \\ &\leq |s_q - S| + |s_p - S| \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon. \end{align*} This proves the stated Cauchy tail condition. [/step] [step:Use the tail estimate to prove convergence of the partial sums] Assume that for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that for all integers $q > p \geq N$, \begin{align*} \left|\sum_{n=p+1}^{q} a_n\right| < \varepsilon. \end{align*} We prove that $(s_m)_{m=1}^{\infty}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathbb{K}$. Let $\varepsilon > 0$, and choose $N \in \mathbb{N}$ from the assumed tail condition. Let $m, \ell \in \mathbb{N}$ satisfy $m \geq N$ and $\ell \geq N$. If $m = \ell$, then \begin{align*} |s_m - s_\ell| = 0 < \varepsilon. \end{align*} If $\ell > m$, then \begin{align*} |s_\ell - s_m| = \left|\sum_{n=m+1}^{\ell} a_n\right| < \varepsilon. \end{align*} If $m > \ell$, then \begin{align*} |s_m - s_\ell| = \left|\sum_{n=\ell+1}^{m} a_n\right| < \varepsilon. \end{align*} Therefore, for all $m, \ell \geq N$, \begin{align*} |s_m - s_\ell| < \varepsilon. \end{align*} Hence $(s_m)_{m=1}^{\infty}$ is Cauchy in $\mathbb{K}$. Since $\mathbb{K}$ is either $\mathbb{R}$ or $\mathbb{C}$, it is complete, so there exists $S \in \mathbb{K}$ such that $s_m \to S$. Therefore the series $\sum_{n=1}^{\infty} a_n$ converges in $\mathbb{K}$. [/step] [step:Conclude the equivalence] The previous two steps prove both implications: convergence of the series implies the stated Cauchy tail condition, and the stated Cauchy tail condition implies convergence of the sequence of partial sums. By the definition of convergence of a series, this proves the theorem. [/step]