Convergence on sum of sequence Let $\{ a_n \}^\infty_1 \subseteq \Bbb{R}$ be a sequence. If for any recursively enumerable set $S = \{k_1, k_2, ...\} \subseteq \Bbb{N}$, where $k_1 < k_2 < ...$ is an infinite ascending chain, $\sum^\infty_{i=1} a_{k_i}$ converges, can we conclude that $\sum^\infty_{i=1} a_{i}$ converges absolutely? Although we usually construct a conditionally convergent series using a particular pattern, I think there must be a counterexample with the help of a non-computable number. However I am not familiar with computability, so I ask for help.
 A: We'll construct a counterexample.
A maximal r.e. set is a set $M\subseteq\omega$ such that:
(1) $M$ is r.e.;
(2) $\omega\setminus M$ is infinite;
and (3) for every r.e. set $X\supseteq M,$ either $X\setminus M$ is finite or $X$ is cofinite.
($X$ is cofinite means that $\omega\setminus X$ is finite.)
Friedberg proved that maximal r.e. sets exist.
$$ $$
Let $M$ be a maximal r.e. set.  List $\omega\setminus M$ in increasing order: $u_1 \lt u_2 \lt u_3 \lt\dots.$  (This is not a recursive enumeration.)
Define $a_n$ for $n\lt\omega$ by setting:
$$a_n=\begin{cases}
\displaystyle\frac{(-1)^j}{j}, & \text{if }n=u_j;\cr
0, & \text{if }n\in M.
\end{cases}$$
Note that the sum $\sum_{n=0}^\infty a_n$ converges, but that series is not absolutely convergent.
Now let $S$ be any infinite r.e. set; list S in increasing order: $k_1\lt k_2 \lt k_3\dots.$  (This is not necessarily a recursive enumeration.)  We'll show that $\sum_{j=1}^\infty a_{k_j}$ converges.
Since $M\cup S$ is an r.e. set containing $M,$ we must have that either $S\setminus M = (M\cup S)\setminus M$ is finite or that $M\cup S$ is cofinite.  We handle each case separately:
$\bullet\;\;$ If $S\setminus M$ is finite, then the terms of $\sum_{j=1}^\infty a_{k_j}$ consist of infinitely many terms that are zero and only finitely many terms of the form $\frac{(-1)^j}j.$ So, in this case, the series converges.
$\bullet\;\;$ If $M\cup S$ is cofinite, then $S$ contains all but finitely many members of $\omega\setminus M$ (plus some members of $M$ as well).  So $\sum_{j=1}^\infty a_{k_j}$ consists of all but finitely many terms of $\sum_{j=1}^\infty \frac{(-1)^j}j$ (in the same order as in that series) interspersed with some zero terms, which again converges, completing the proof.
