Countable subset of $(a,b)$. absolutely convergent series, convergent sequence Related :
Can absolute convergent series be expressed as sum of two series?
Let $E$ be a countable subset of $(a,b)$ in $\mathbb{R}$.
Let $x_n$ enumerate $E$.
Now, fix $m\in \mathbb{N}$.
Let $\{p_i\}$ be a sequence in $(x_m,b)$ such that $p_i \rightarrow x_m$.
Let $I_i = \{n\in \mathbb{N} | x_n \in (x_m,p_i)\}$
Let $\sum c_n$ be a absolutely convergent series.
Then, $\sum_{n\in I_i} c_n$ is well defined and it exists.
Here, how do I prove that $\lim_{i\to\infty} (\sum_{n\in I_i} c_n ) = 0$??
 A: Since $p_i\to x_m$, for each $k\in\mathbb N$ we have $k\notin I_i$ for almost all $i$.
This works also for finitely many $k$ at once, i.e. for almost all $i$ we have $I_i\cap \{1,\ldots,k\}=\emptyset$.
Thus for almost all $i$
$$\left|\sum_{n\in I_i}c_n\right|\le \sum_{n\in I_i}|c_n|\le \sum_{n>k}|c_n|.$$
By absolute convergenxce, the right hand side becomes $<\varepsilon$ for suitable $k$.
A: This is just a roundabout way of asking the following question:

Let $\{A_k:k\in\omega\}$ be a family of subsets of $\omega$ such that $A_{k+1}\subseteq A_k$ for all $k\in\omega$ and such that $\bigcap_{k\in\omega}A_k=\varnothing$. Show that if the series $\sum_{n\in\omega}c_n$ is absolutely convergent, then $$\lim_{n\to\infty}\sum_{k\in A_n}c_k=0\;.$$

For each $n\in\omega$ we have $$\left|\sum_{k\in A_n}c_k\right|\le\sum_{k\in A_n}|c_k|\le\sum_{k\ge\min A_n}|c_k|\;.$$ Clearly $\min A_n\to\infty$ as $n\to\infty$, so 
$$\lim_{n\to\infty}\sum_{k\ge\min A_n}|c_k|=0$$ and hence
$$\lim_{n\to\infty}\left|\sum_{k\in A_n}c_k\right|=0\;.$$
