A theorem about absolutely convergent series. Please tell me a rigorous proof. I am reading a calculus book in Japanese.
In this book, there is the following theorem and its proof.
It is intuitively obvious that $|s-\sum_{i=1}^{m}s^{(i)}|\leq |a_{n+1}|+|a_{n+2}|+\dots$ holds, but I want a rigorous proof.
Please tell me a rigorous proof.


*

*Suppose that $M_1\cup M_2\cup\dots=\mathbb{N}$ and $M_i\cap M_j=\emptyset$ for any $i,j\in\mathbb{N}$ such that $i\neq j$.

*Suppose that $\sum_{n=1}^{\infty}a_n$ is an absolutely convergent series.

*Let $s^{(i)}:=\sum_{n\in M_i} a_n$.

Then, $$\sum_{i\in\mathbb{N}} s^{(i)}=\sum_{n=1}^{\infty}a_n.$$


Proof:
Let $s:=\sum_{n=1}^{\infty}a_n$.
For any $n\in\mathbb{N}$, there exists $m\in\mathbb{N}$ such that $\{1,2,\dots,n\}\subset M_1\cup M_2\cup\dots\cup M_m$.
Then, $$|s-\sum_{i=1}^{m}s^{(i)}|\leq |a_{n+1}|+|a_{n+2}|+\dots$$ holds.
For any positive real number $\epsilon$, there exists $n\in\mathbb{N}$ such that $$|a_{n+1}|+|a_{n+2}|+\dots < \epsilon.$$
So, $$\sum_{i\in\mathbb{N}} s^{(i)}=\sum_{n=1}^{\infty}a_n$$ holds.

 A: There is a rigorous and (to me) surprisingly simple proof that
closely follows the argument in the book, but it is hard to be sure
that it is what the author had in mind, because the quoted extract
does not define the meaning of $\sum_{i \in I}a_i$ when $I$ is an
infinite subset of $\mathbb{N} = \{1, 2, 3, \ldots\}$, and it would
be interesting to know if a definition was given earlier in the
book.
Such a sum may be defined in several ways (some of them quite
elaborate), but all give the same value.  I choose a definition that
is simple and convenient for this question.
For all $n \in \mathbb{N},$ define $[n] = \{1, 2, \ldots, n\}.$
Define $\sum_{i \in I}a_i$ to be the limit of the sequence of finite
partial sums $\sum_{i \in I \cap [n]}a_i,$ as $n$ tends to infinity,
if this limit exists.  We could restrict $n$ to take values in $I,$
but it is convenient to allow any $n \in \mathbb{N},$ because then
the result remains true even when $I$ is finite. (The sequence of
partial sums is then eventually constant.) This reduces the need to
consider special cases.
It may be reassuring to note that when $I$ is infinite, the
definition is equivalent to
$\sum_{i \in I}a_i = \sum_{r=1}^\infty a_{h(r)},$ where
$h \colon \mathbb{N} \to \mathbb{N}$ is the unique strictly
increasing function such that $h(\mathbb{N}) = I.$
Thus $\sum_{i \in I}a_i$ is the sum of a 'subseries' of
$\sum_{i=1}^\infty a_i$ (but I don't think this is a technical term,
unlike 'subsequence').
Let $I, J$ be disjoint subsets of $\mathbb{N}$ such that
$\sum_{i \in I}a_i, \sum_{i \in J}a_i$ exist.  Then the following
limit also exists:
\begin{align*}
\sum_{i \in I \cup J}a_i & =
\lim_{n \to \infty} \sum_{i \in (I \cup J) \cap [n]}a_i \\ & =
\lim_{n \to \infty} \sum_{i \in (I \cap [n]) \cup (J \cap [n])}a_i
\\ & =
\lim_{n \to \infty} \left( \sum_{i \in I \cap [n]}a_i +
\sum_{i \in J \cap [n]}a_i \right) \\ & =
\left( \lim_{n \to \infty} \sum_{i \in I \cap [n]}a_i \right) +
\left( \lim_{n \to \infty} \sum_{i \in J \cap [n]}a_i \right)
\\ & = \sum_{i \in I}a_i + \sum_{i \in J}a_i.
\end{align*}
It follows by induction on $m \in \mathbb{N}$ that if
$I_1, I_2, \ldots, I_m$ is a finite sequence of $m$ pairwise
disjoint subsets of $\mathbb{N}$ such that $\sum_{i \in I_k}a_i$
exists for $k = 1, 2, \ldots, m,$ then
\begin{equation}\label{4189981:eq:1}\tag{$1$}
\sum_{i \in I_1 \cup I_2 \cup \cdots \cup I_m} \!\! a_i =
\sum_{i \in I_1}a_i + \sum_{i \in I_2}a_i + \cdots +
\sum_{i \in I_m}a_i.
\end{equation}
Now consider the case of an infinite sequence
$(I_k)_{k \in \mathbb{N}}$ of pairwise disjoint subsets of
$\mathbb{N}.$  We begin to make use of the hypothesis that the
series $\sum_{i=1}^\infty a_i$ is absolutely convergent.  Because of
this hypothesis, we do not need to postulate additionally that
$\sum_{i \in I_k}a_i$ exists for all $k \in \mathbb{N}.$ In fact, it
is clear that for any subset $I \subseteq \mathbb{N},$ and
any $n \in \mathbb{N},$
$$
\sum_{i \in I \cap [n]}|a_i| \leqslant \sum_{i=1}^n|a_i|,
$$
therefore if $I$ is infinite, the 'subseries'
$\sum_{r=1}^\infty a_{h(r)}$ (where the function
$h \colon \mathbb{N} \to \mathbb{N}$ is as defined above) is
absolutely convergent, whence $\sum_{i \in I}a_i$ exists.
(We could have avoided some slight
awkwardness of expression here by defining a concept of 'absolute
convergence' especially for expressions of the form
$\sum_{i \in I}a_i,$ so that the function $h$ need not be mentioned,
and we need only observe that $\sum_{i \in I}|a_i|$ exists,
therefore $\sum_{i \in I}a_i$ exists.  Although that would have been
easy to do, it hardly seems worth even the small amount of trouble.)
It would be interesting to know if the author of the textbook quoted
in the question was taking all of that argument - or something
similar - for granted, as being 'obvious', or whether it was
stated explicitly earlier in the book.
Let $J = \bigcup_{k=1}^\infty I_k.$ (The question concerns the case
$J = \mathbb{N},$ but it is not much harder to consider the general
case.) For all $n \in \mathbb{N},$
$$
J \cap [n] = \bigcup_{k=1}^\infty (I_k \cap [n]).
$$
The finite set $J \cap [n]$ can only be a disjoint union of finitely
many distinct, non-empty sets, therefore there exists
$m \in \mathbb{N}$ such that
$$
J \cap [n] = \bigcup_{k=1}^m (I_k \cap [n]) =
(I_1 \cup I_2 \cup \cdots \cup I_m) \cap [n].
$$
Define
$$
K = \! \bigcup_{k=m+1}^\infty I_k \subseteq J \setminus [n].
$$ Then
$$
J = I_1 \cup I_2 \cup \cdots \cup I_m \cup K,
$$
and this is a union of pairwise disjoint sets, therefore, by
\eqref{4189981:eq:1},
$$
\sum_{i \in J}a_i =
\sum_{i \in I_1}a_i + \sum_{i \in I_2}a_i + \cdots +
\sum_{i \in I_m}a_i + \sum_{i \in K}a_i,
$$
whence
\begin{equation}\label{4189981:eq:2}\tag{$2$}
\left\lvert \sum_{i \in J}a_i -
\sum_{i \in I_1}a_i - \sum_{i \in I_2}a_i + \cdots -
\sum_{i \in I_m}a_i \right\rvert =
\left\lvert \sum_{i \in K}a_i \right\rvert
\leqslant \sum_{i \in J \setminus [n]} |a_i|.
\end{equation}
We pause briefly, to justify the last step in \eqref{4189981:eq:2}.
(Hopefully the rest is clear!)
Let $E, F$ be subsets of $\mathbb{N}$ (finite or infinite) such that
$E \subseteq F$ and $\sum_{i \in F}|a_i|$ exists. (Here, we
temporarily drop the hypothesis that $\sum_{i=1}^\infty a_i$ is
absolutely convergent: for now, it may be any series at all.) Then
for all $n \in \mathbb{N},$
$$
\sum_{i \in E \cap [n]}|a_i| \leqslant \sum_{i \in F \cap [n]}|a_i|.
$$
By hypothesis, the partial sums on the right are bounded above.
Therefore $\sum_{i \in E}|a_i|$ exists. Again using the rather
awkward identification of expressions like $\sum_{i \in E}a_i$ with
sums $\sum_{r=1}^\infty a_{h(r)}$ of 'subseries' of
$\sum_{i=1}^\infty a_i,$ we conclude that $\sum_{i \in E}a_i$ also
exists (is 'absolutely convergent' - perhaps after all it would
have been worth the effort of making this terminology usable
literally). This is in the case where $E$ (therefore also $F$) is
infinite; if $E$ is finite, then of course there is nothing to worry
about.  Finally, we have for all $n \in \mathbb{N},$
$$
\bigg\lvert \sum_{i \in E \cap [n]}a_i \bigg\rvert \leqslant
\sum_{i \in E \cap [n]}|a_i| \leqslant \sum_{i \in F \cap [n]}|a_i|,
$$
whence, taking limits as $n \to \infty,$
\begin{equation}\label{4189981:eq:3}\tag{$3$}
\bigg\lvert \sum_{i \in E}a_i \bigg\rvert \leqslant
\sum_{i \in F}|a_i|.
\end{equation}
Using \eqref{4189981:eq:3} with $E = K$ and $F = J \setminus [n],$
we have fully justified \eqref{4189981:eq:2}.
Because $\sum_{i=1}^\infty a_i$ is absolutely convergent (actually
it would have been enough to postulate that $\sum_{i \in J}|a_i|$
exists, but this is a minor point, and I shall carry on with the
more extravagant hypothesis), for every $\epsilon > 0,$ there exists
$n \in \mathbb{N}$ such that $\sum_{i=n+1}^\infty|a_i| < \epsilon.$
Using \eqref{4189981:eq:3} again (this awkwardness could have been
avoided, either by using the weaker hypothesis just mentioned, or
else by treating only the case $J = \mathbb{N}$), it follows that
$\sum_{i \in J \setminus [n]}|a_i| < \epsilon.$
By \eqref{4189981:eq:2}, therefore,
$$
\left\lvert \sum_{i \in J}a_i -
\sum_{i \in I_1}a_i - \sum_{i \in I_2}a_i + \cdots -
\sum_{i \in I_m}a_i \right\rvert < \epsilon.
$$
Since $\epsilon$ is arbitrary, we have proved
$$
\sum_{i \in J}a_i = \sum_{k=1}^\infty\sum_{i \in I_k}a_i.
$$
