Conditional convergence, partial sums, and rearrangement If a series $\sum a_{n}$ converges but not $\sum |a_{n}|$, then $\sum a_{n}$ can be rearranged so that the newly rearranged sum takes a value different from the original one.
This is a well-known fact and I do remember having learnt about it some long years ago.
But recently, I have been thinking about this. "There is a new rearrangement", but what sort of rearrangement is it? 
Suppose that $\sum_{n \geq 1}a_{n}$ converges and
$\mathbb{N} = A_{1} \cup A_{2} \cup \cdots$ with $A_{i}$ disjoint
from one another. 
Then what are the conditions on those subsets $A_{i}$'s so that
\[
\sum_{j \geq 1} \sum_{n \in A_{j}}a_{n} \not = \sum_{n \geq 1} a_{n}.
\]
The limit of its partial sums $\lim_{N \to \infty}\sum_{j \leq N} \sum_{n \in A_{j}}a_{n}$ seems to be the same regardless of what $A_{i}$'s are, 
as long as each $\sum_{n \in A_{j}}a_{n}$ converges absolutely and this way of summing up is concerned.
And I would like to think about this question in the context of
convergence of the product
\[
\prod_{p \leq N} (1 + a(p)p^{-s}) = \sum_{n \in A_{1} \cup \cdots \cup A_{N}}\frac{a(n)}{n^{s}}
\]
with $|a(n)| = 1$, $s$ some number $0 < s < 1$, and $p$ primes.
Here, the right series converges absolutely for $s > 1$. If the relation
\[
\lim_{N \to \infty}\prod_{p \leq N} (1 + a(p)p^{-s}) 
= \lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}}
\] 
is shown to be valid for  $a < \text{Re}(s) \leq 1$ for some $0 < a < 1$,
then we may have more restrictions on rearrangement of the sum
on the right (since by analytic continuation, rearrangements are restricted to those which assign the series the same original value).
The next question is, then, can we see the sum 
\[
\lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}}
\]
as a Dirichlet series? To put it in formulas, is it true that
\[
\lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}}
= \sum_{ n \geq 1} \frac{a(n)}{n^{s}} \quad ?
\]
Any objection, comments are welcome.
 A: 
The limit of its partial sums $\lim_{N \to \infty}\sum_{j \leq N} \sum_{n \in A_{j}}a_{n}$ seems to be the same regardless of what $A_{i}$'s are, 
  as long as each $\sum_{n \in A_{j}}a_{n}$ converges absolutely and this way of summing up is concerned.

That is not true. Consider a series $\sum a_n$ converges but not $\sum |a_n|$ and an arbitrary $M\in\mathbb R$. Since $\sum |a_n|$ diverges, the series over the positive $a_n$ diverges as the series over the negative $a_n$. Hence, you can pick positive $a_n$ until there sum exceeds $M$ and define the set of indices $A_1$. Next, you pick negative $a_n$ until you become less than $M$ and define the set of indices $A_2$ and so on. In the end, you will get
$$
\lim_{N\to\infty}\sum_{j=1}^N\sum_{n\in A_j}a_n = M.
$$
For further details, you should study the proof in https://en.wikipedia.org/wiki/Riemann_series_theorem .
Since the theorem states the existence of a rearrangement, you can't directly deduce conditions on $A_1,\ldots$ such that the limits matches.
