Absolutely Convergent Generalized Sum Converges

Theorem
Let $V$ be a Banach space.

Let $\left\Vert{\cdot}\right\Vert$ denote the norm on $V$.

Let $d$ denote the corresponding induced metric.

Let $\left({v_i}\right)_{i \mathop \in I}$ be an indexed subset of $V$ such that the generalized sum $\displaystyle \sum_{i \mathop \in I} \left\{{v_i}\right\}$ converges absolutely.

Then the generalized sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges.

Proof
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 $\displaystyle \sum \left\{{v_i: i \mathop \in I}\right\}$ converges absolutely means that $\displaystyle \sum \left\{{\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\}$ converges.

Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that:


 * $\displaystyle \sum_{i \mathop \in G} \left\Vert{v_i}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\} - 2^{-n}$, 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 $\left({v_n}\right)_{n \mathop \in \N}$ 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:

Now to estimate this last quantity, observe:

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:
 * $d \left({v_m, v_n}\right) < \epsilon$

It follows that $\left({v_n}\right)_{n \in \N}$ is a Cauchy sequence.

As $V$ is a Banach space:
 * $\displaystyle \exists v \in V: \lim_{n \to \infty} v_n = v$

Having identified a candidate $v$ for the sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ 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:


 * $d \left({\displaystyle \sum_{i \mathop \in G} v_i, v}\right) < \epsilon$

for all finite $G$ with $F \subseteq G \subseteq I$.

Now let $N \in \N$ such that:
 * $\forall n \ge N: d \left({v_n, v}\right) < \dfrac \epsilon 2$

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:

For the first of these terms, observe:

Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain:


 * $\displaystyle d \left({\sum_{i \mathop \in G} v_i, v}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

By definition of convergence of a net, it follows that:


 * $\displaystyle \sum \left\{{v_i: i \in I}\right\} = v$

Also see

 * Absolutely Convergent Series is Convergent