The proof proceeds in two stages: :$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to :$(2): \quad$ Showing that the candidate is indeed sought limit. That $\displaystyle \sum \set {v_i: i \mathop \in I}$ converges absolutely means that $\displaystyle \sum \set {\norm {v_i}: i \mathop \in I}$ converges. Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that: :$\displaystyle \sum_{i \mathop \in G} \norm {v_i} > \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n}$ for all finite $G$ with $F_n \subseteq G \subseteq I$ It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $\displaystyle F'_n = \bigcup_{m \mathop = 1}^n F_m$ if necessary. Define: : $\displaystyle v_n = \sum_{i \mathop \in F_n} v_i$ Next, it is to be shown that the sequence $\sequence {v_n}_{n \mathop \in \N}$ is Cauchy. So let $\epsilon > 0$, and let $N \in \N$ be such that $2^{-N} < \epsilon$. Then for $m \ge n \ge N$, have: {{begin-eqn}} {{eqn | l = \map d {v_m, v_n}       | r = \norm {\paren {\sum_{i \mathop \in F_m} v_i} - \paren {\sum_{i \mathop \in F_n} v_i} }       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | r = \norm {\sum_{i \mathop \in F_m \setminus F_n} v_i}       | c = $F_n \subseteq F_m$ }} {{eqn | o = \le       | r = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}       | c = Triangle Inequality for $\norm {\, \cdot \,}$ }} {{end-eqn}} Now to estimate this last quantity, observe: {{begin-eqn}} {{eqn | l = \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}       | o = <       | r = \sum_{i \mathop \in F_n} \norm {v_i} + \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}       | c = Defining property of $F_n$ }} {{eqn | r = \sum_{i \mathop \in F_m} \norm {v_i}       | c = Union with Relative Complement, $F_n \subseteq F_m$ }} {{eqn | o = \le       | r = \sum \set {\norm {v_i}: i \mathop \in I}       | c = Generalized Sum is Monotone }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop \in F_m \setminus F_n} \norm {v_i}       | o = <       | r = 2^{-n} }} {{end-eqn}} Finally, by the defining property of $N$, as $n \ge N: :2^{-n} < 2^{-N} < \epsilon$ Combining all of these estimates leads to the conclusion that: :$\map d {v_m, v_n} < \epsilon$ It follows that $\sequence {v_n}_{n \mathop \in \N}$ is a Cauchy sequence. As $V$ is a Banach space: :$\displaystyle \exists v \in V: \lim_{n \mathop \to \infty} v_n = v$ Having identified a candidate $v$ for the sum $\displaystyle \sum \set {v_i: i \in I}$ to converge to, it remains to verify that this is indeed the case. According to the definition of considered sum, the convergence is convergence of a net. Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $v$ to open balls centered at $v$. Now let $\epsilon > 0$. We want to find a finite $F \subseteq I$ such that: :$\map d {\displaystyle \sum_{i \mathop \in G} v_i, v} < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$. Now let $N \in \N$ such that: :$\forall n \ge N: \map d {v_n, v} < \dfrac \epsilon 2$ with the $v_n$ as above. By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well. Let us verify that the set $F_N$ defined above has sought properties. So let $G$ be finite with $F_N \subseteq G \subseteq I$. Then: {{begin-eqn}} {{eqn | l = \map d {\sum_{i \mathop \in G} v_i, v}       | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - v}       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | o = \le       | r = \norm {\paren {\sum_{i \mathop \in G} v_i} - \paren {\sum_{i \mathop \in F_N} v_i} } + \norm {\paren {\sum_{i \mathop \in F_N} v_i} - v}       | c = Triangle inequality for $\norm {\, \cdot \,}$ }} {{eqn | o = <       | r = \norm {\sum_{i \mathop \in G \setminus F_N} v_i} + \frac \epsilon 2       | c = $F_N \subseteq G$, defining property of $N$ }} {{eqn | o = \le       | r = \sum_{i \mathop \in G \setminus F_N} \norm {v_i} + \frac \epsilon 2       | c = Triangle inequality for $\norm {\, \cdot \,}$ }} {{end-eqn}} For the first of these terms, observe: {{begin-eqn}} {{eqn | l = \sum \set {\norm {v_i}: i \in I} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}       | o = <       | r = \sum_{i \mathop \in F_N} \norm {v_i} + \sum_{i \mathop \in G \setminus F_N} \norm {v_i}       | c = defining property of $F_N$ }} {{eqn | r = \sum_{i \mathop \in G} \norm {v_i}       | c = Union with Relative Complement, $F_N \subseteq G$ }} {{eqn | o = \le       | r = \sum \set {\norm {v_i}: i \in I}       | c = Generalized Sum is Monotone }} {{eqn | ll= \leadsto       | l = \sum_{i \mathop \in G \setminus F_N} \norm {v_i}       | o = <       | r = 2^{-N} }} {{end-eqn}} Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain: :$\displaystyle \map d {\sum_{i \mathop \in G} v_i, v} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$ By definition of convergence of a net, it follows that: :$\displaystyle \sum \set {v_i: i \in I} = v$ {{qed}}