I read on notes of a course on differential geometry the following statement but it doesn't seem so clear to me.
Let $I$ be a set and $b: I \rightarrow \mathbb{C}$ a map, where $b_i=b(i)$, namely we are given a sequence of complex numbers indexed by $I$. Now, suppose that $\sum_ib_i$ is absolutely convergent.
Therefore, the "series" converges to a certain complex number $b$ and this is characterized by the following property. For each $\epsilon >0$ and each finite subset $J \subset I$ there exists a finite subset $J \subset K \subset I$ such that $$|b-\sum_{i \in K }b_i|<\epsilon$$
I would like to know how to prove this characterization for the limit $b$.
Thank you
Let $c=\sup \sum_{j \in J}|b_j|$ taken on all finite subsets of $I$; by hypothesis $c$ is finite so for all $n \ge 1$ there is $K_n$ finite subset of $I$ st $0 \le c-\sum_{j \in K_n}|b_j| <\frac{1}{n}$ so in particular $\sum_{j \in J -K_n}|b_j| < \frac{1}{n}$ for all finite subsets $J$ of $I$ as otherwise $\sum_{j \in J \cup K_n}|b_j| >c$ contradicting the definition of $c$
But now if $a_n=\sum_{j \in K_n}b_j$ we have that $$|a_n-a_m| = |\sum_{j \in K_n -K_m}b_j-\sum_{j \in K_m -K_n}b_j| \le \sum_{j \in K_m -K_n}|b_j|+\sum_{j \in K_n -K_m}|b_j| \le \frac{1}{n}+\frac{1}{m}$$
This means that $a_n$ is Cauchy, hence it has a limit $b$ and for every $J$ finite, we have that $$|\sum_{j \in J \cup K_n}b_j-b| \le |\sum_{j \in K_n}b_j-b|+|\sum_{j \in J-K_n}b_j|\le |a_n-b|+\frac{1}{n}$$ which clearly gives the required result with $K=J \cup K_n$ and $n$ large enough so $|a_n-b|+\frac{1}{n} < \epsilon$