Absolute Net Convergence Equivalent to Absolute Convergence/Absolute Net Convergence implies Absolute Convergence

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Theorem

Let $V$ be a Banach space.

Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.

Let $r \in \R_{\mathop \ge 0}$

Let the generalized sum $\ds \sum \set {v_n: n \in \N}$ be absolutely net convergent to $r$.

Then:

the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$.

Proof

Let $\epsilon \in \R_{\mathop > 0}$.

From Characterization of Convergent Net in Metric Space:

$(1) \quad \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\implies \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$

Let $N = \max \set{n : v_n \in F}$.

We have:

$F \subseteq \closedint 0 N$

Let $m \ge N$.

We have:

$\closedint 0 m \supseteq \closedint 0 N \supseteq F$

From $(1)$:

$\size{\ds \sum_{n \mathop \in \closedint 0 m} \norm{v_n} - r} < \epsilon$

By definition of summation over finite index:

$\size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$

Since $m \ge N$ was arbitrary, it follows that:

$\forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$

Since $\epsilon$ was arbitrary, it follows that the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$ by definition.

$\blacksquare$