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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 $\lim_{i\to\infty} (\sum_{n\in I_i} c_n ) = 0$??

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2 Answers 2

up vote 1 down vote accepted

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 $k$.

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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

$$\lim_{n\to\infty}\left|\sum_{k\in A_n}c_k\right|=0\;.$$

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Note that this is essentially the same as @Hagen’s answer, which I’ve only just seen. –  Brian M. Scott Oct 20 '12 at 19:03

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