# Proof with Cauchy criterion

Show that sequence $$x_n$$ is absolutely summable if and only if for every $$\epsilon>0$$ there is an $$N\in\mathbb{N}$$ such that for every finite subset $$F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$$, $$\displaystyle{\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$$.

$$\textbf{Proof:}$$ Let the sequence $$x_n$$ be absolutely summable so the series $$\displaystyle{\sum_{n=1}^\infty |x_n|}$$ is convergent. Therefore, by the Cauchy criterion, for every $$\epsilon > 0$$ there exists a natural number $$N$$ such that $$\bigl\lvert\lvert x_n \rvert +\lvert x_{n+1} \rvert + \dots + \lvert x_{n+p} \rvert \bigr\rvert \\ < \epsilon.$$ Thereby, implying $$\displaystyle{\sum_{i=n}^{n+p} |x_i| < \epsilon}$$ for all $$n\geq N$$ and $$p\geq 1.$$ This comes directly from the Cauchy Criterion of Sequence Convergence and has an alternate form $$\bigl\lvert \sum_{i=n}^m \lvert x_i \rvert \bigr\rvert \\ < \epsilon$$ for all $$n\geq N$$ and $$m>n$$.

Now, consider any finite subset $$F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$$ then let min$$F = n_1$$ and max$$F=m_1$$ where they exist since F is a finite subset of $$\mathbb{N}$$ and $$n_1\geq N$$ and $$m_1 \geq n_1$$. Therefore, as the Cauchy criterion we have, $$\bigl\lvert \sum_{i=n_1}^{m_1} \lvert x_i \rvert \bigr\rvert \\ < \epsilon$$ as $$n_1\geq N$$ implies $$\displaystyle{\sum_{i=n_1}^{m_1} |x_i| < \epsilon}.$$. Then, $$\displaystyle{\sum_{n\in F} |x_n| \leq \sum_{i=n_1}^{m_1} |x_i| < \epsilon}$$ implies $$\displaystyle{|\sum_{n\in F} x_n| \leq \sum_{n\in F} |x_n| < \epsilon}$$. Therefore, this show that there exists an $$N$$ such that for every finite subset $$F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$$, $$\displaystyle{\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$$. Clearly, we have a generalization of Cauchy Criterion and the condition came as Cauchy Criterion when $$F$$ has the form $$F=\{n, n+1, \dots, m\}$$.

You proved one direction: If $$\sum_{n=1}^\infty x_n$$ converges absolutely then for every $$\epsilon>0$$ there is an $$N\in\mathbb{N}$$ such that for every finite subset $$F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$$, $${\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$$.
For the opposite direction assume that $$\sum_{n=1}^\infty x_n$$ does not converge absolutely, i.e. that $$\sum_{n=1}^\infty |x_n|$$ diverges. From $$\sum_{n=1}^k |x_n| = \sum_{n=1}^k x_n^+ + \sum_{n=1}^k x_n^-$$ with $$x_n^+ = \max(x_n, 0)$$, $$x_n^- = \max(-x_n, 0)$$ it follows that at least one of the series $$\sum_{n=1}^\infty x_n^+$$, $$\sum_{n=1}^\infty x_n^-$$ diverges.
Without loss of generality assume that $$\sum_{n=1}^\infty x_n^+$$ diverges, i.e. $$\lim_{k \to \infty} \sum_{n=1}^k x_n^+ = + \infty \, .$$ It follows that for every $$N \in \Bbb N$$ $$\lim_{k \to \infty} \sum_{n=N}^k x_n^+ = + \infty$$ and that implies $$\bigg|\sum_{n\in \{ N, \ldots, m\}} x_n\bigg| = \sum_{n=N}^m x_n^+ > 1$$ for some $$m > N$$.
So we have shown that if $$\sum_{n=1}^\infty x_n$$ does not converge absolutely then for $$\epsilon = 1$$ there is no $$N \in \Bbb N$$ such that $${\bigg|\sum_{n\in F} x_n\bigg|<\epsilon}$$ for every finite subset $$F\subseteq\{n\in\mathbb{N}\colon n\ge N\}$$. That proves the other direction by contrapositive.