I am reading "A First Course in Analysis vol.1" (in Japanese) by Sin Hitotumatu.
The following theorem is in this book.
Theorem 7.6:
Let $\mathbb{N}=N_1\cup N_2\cup\cdots$ and $N_i\cap N_j=\emptyset$ for any $i, j$ such that $i\neq j$.
We can write $N_i$ as $N_i=\{m_{i1},m_{i2},\dots\}$ for $i\in\{i\in\mathbb{N}\mid i=j+1,0\leq j<\#\{N_1,N_2,\dots\}\}$.
Let $\{a_n\}$ be a complex sequence.
Let $a_{ij}:=a_{m_{ij}}$.
(1)
Assume that a complex series $\sum a_n$ converges absolutely.
Let $s:=\sum a_n$.
Then, $\sum_{j=1}^{\# N_i} a_{ij}$ converges absolutely for $i\in\{i\in\mathbb{N}\mid i=j+1,0\leq j<\#\{N_1,N_2,\dots\}\}$.
Let $s_i:=\sum_{j=1}^{\# N_i} a_{ij}$ for $i\in\{i\in\mathbb{N}\mid i=j+1,0\leq j<\#\{N_1,N_2,\dots\}\}$.
Then, $\sum s_i$ converges absolutely and $s=\sum s_i$ holds.
(2)
Conversely, assume that $\sum_{j=1}^{\# N_i} a_{ij}$ converges absolutely for $i\in\{i\in\mathbb{N}\mid i=j+1,0\leq j<\#\{N_1,N_2,\dots\}\}$.
Let $s_i:=\sum_{j=1}^{\# N_i} |a_{ij}|$ for $i\in\{i\in\mathbb{N}\mid i=j+1,0\leq j<\#\{N_1,N_2,\dots\}\}$.
If $\sum s_i$ converges, then $\sum a_n$ converges absolutely.
In the proof of Theorem 7.6, the author wrote as follows:
If we can prove this theorem for a non-negative series $\sum a_n$, we can prove this theorem for a complex series $\sum a_n$ in a similar way to the proof of Theorem 7.5.
So we assume that $\sum a_n$ is a non-negative series and we prove this theorem for a non-negative series.
But I cannot understand why we need to prove this theorem only for a non-negative series $\sum a_n$.
Please tell me the reason.
The theorem 7.5 and its proof are the following:
Theorem 7.5:
A complex series which converges absolutely converges.
Proof:
Let $\sum_{i=1}^\infty a_n$ be a complex series.
Let $a_n=\alpha_n+i\beta_n$ ($\alpha_n,\beta_n\in\mathbb{R}$).
Then, obviously, $\sum_{n=1}^\infty a_n$ converges if and only if $\sum\alpha_n$ and $\sum\beta_n$ converge.
Since $|\alpha_n|\leq |a_n|$ and $|\beta_n|\leq |a_n|$, if $\sum a_n$ conveges absolutely, $\sum\alpha_n$ and $\sum\beta_n$ converge absolutely.
So, it is sufficient to prove if a real series $\sum a_n$ converges absolutely, then $\sum a_n$ converges.
Let $a_n^{+}:=\max(a_n,0)$ and $a_n^{-}:=\max(-a_n,0)$.
Then, $|a_n|=a_n^{+}+a_n^{-}$, $a_n=a_n^{+}-a_n^{-}$, $a_n^{+}\leq |a_n|$, and $a_n^{-}\leq |a_n|$ hold.
So, if $\sum a_n$ converges absolutely, $\sum a_n^{+}$ and $\sum a_n^{-}$ converge.
Since $a_n=a_n^{+}-a_n^{-}$, if $\sum a_n$ converges absolutely, $\sum a_n$ converges.
We use the following lemma to prove Theorem 7.6:
Thorem 7.2:
Let $\sum_{n=1}^\infty a_n$ be a non-negative series.
Let $A:=\{x\in\mathbb{R}\mid x=a_{n_1}+\dots+a_{n_m},\{n_1,\dots,n_m\}\subset\mathbb{N}\text{ and }\#\{n_1,\dots,n_m\}=m\}$.
$\sum a_n$ converges if and only if $A$ has an upper bound.
If $A$ has an upper bound, then $\sum a_n = \sup A$.
We omit the proof.
The proof of Theorem 7.6 is the following:
Proof of Theorem 7.6:
(1)
If we can prove this theorem for a non-negative series $\sum a_n$, we can prove this theorem for a complex series $\sum a_n$ in a similar way to the proof of Theorem 7.5.
So we assume that $\sum a_n$ is a non-negative series and we prove this theorem for a non-negative series.
Let $\sum a_n$ be a convergent non-negative series.
Let $A:=\{x\in\mathbb{R}\mid x=a_{n_1}+\dots+a_{n_m},\{n_1,\dots,n_m\}\subset\mathbb{N}\text{ and }\#\{n_1,\dots,n_m\}=m\}$.
Let $A_{N_i}:=\{x\in\mathbb{R}\mid x=a_{in_1}+\dots+a_{in_m},\{n_1,\dots,n_m\}\subset \{i\in\mathbb{N}\mid i=j+1, 0\leq j<\# N_i\}\text{ and }\#\{n_1,\dots,n_m\}=m\}$.
Since $\sum a_n$ converges, $A$ has an upper bound.
Since $A_i\subset A$, $A_i$ also has an upper bound.
So, $s_i:=\sum_{j=1}^{\# N_i} a_{ij}$ converges.
Let $s_{il}:=a_{i1}+\dots+a_{il}$.
Let $\epsilon$ be an arbitrary positive real number.
For sufficiently large integer $l_i$, $s_{il_i}>s_i-\frac{\epsilon}{2^i}$ holds.
For any positive integer $k$, $\sum_{i=1}^{k} s_i < \sum_{i=1}^{k} s_{il_i}+\sum_{i=1}^{k} \frac{\epsilon}{2^i}$.
Since $\sum_{i=1}^{k} s_{il_i}\in A$, $\sum_{i=1}^{k} s_{il_i}\leq\sup A=\sum a_n=s$.
And $\sum_{i=1}^{k} \frac{\epsilon}{2^i}<\epsilon$.
So, $\sum_{i=1}^{k} s_i < \sum_{i=1}^{k} s_{il_i}+\sum_{i=1}^{k} \frac{\epsilon}{2^i}<s+\epsilon$.
So, $\sum_{i=1}^{\infty} s_i$ exists and $\sum_{i=1}^{\infty} s_i\leq s+ \epsilon$.
Since $\epsilon$ is an arbitrary positive real number, $\sum_{i=1}^{\infty} s_i\leq s$.Let $x$ be an arbitrary element of $A$.
Then we can write $x$ as $x=a_{n_1}+\dots+a_{n_m}$, where $\{n_1,\dots,n_m\}\subset\mathbb{N}\text{ and }\#\{n_1,\dots,n_m\}=m$.
There exists a positive integer $p$ such that $\{n_1,\dots,n_m\}\subset N_1\cup\dots\cup N_p$.
There exist positive integers $l_1,\dots,l_p$ such that $x=a_{n_1}+\dots+a_{n_m}\leq s_{1l_1}+\dots+s_{pl_p}$.
Let $S:=\{x\in\mathbb{R}\mid x=s_{n_1}+\dots+s_{n_m},\{n_1,\dots,n_m\}\subset\mathbb{N}\text{ and }\#\{n_1,\dots,n_m\}=m\}$.
Since $s_{il_i}\in A_{N_i}$, $s_{il_i}\leq \sup A_{N_i}=s_i$.
So, $x=a_{n_1}+\dots+a_{n_m}\leq s_{1l_1}+\dots+s_{pl_p}\leq s_1+\dots+s_p$.
Since $s_1+\dots+s_p\in S$, $s_1+\dots+s_p\leq\sup S=\sum_{i=1}^{\infty} s_i$.
So, $x=a_{n_1}+\dots+a_{n_m}\leq s_{1l_1}+\dots+s_{pl_p}\leq s_1+\dots+s_p\leq\sum_{i=1}^{\infty} s_i$.
So, $\sup A=s\leq\sum_{i=1}^{\infty} s_i$.So, $\sum_{i=1}^{\infty} s_i=s$.
(2)
Assume that $s_i:=\sum_{j=1}^{\# N_i} a_{ij}$ converges.
Assume that $s:=\sum_{i=1}^{\#\{N_1,N_2,\cdots\}} s_i$ converges.
Let $x$ be an arbitrary element of $A$.
Then we can write $x$ as $x=a_{n_1}+\dots+a_{n_m}$, where $\{n_1,\dots,n_m\}\subset\mathbb{N}\text{ and }\#\{n_1,\dots,n_m\}=m$.
There exists a positive integer $p$ such that $x=a_{n_1}+\dots+a_{n_m}\leq\sum_{i=1}^{p} s_i\leq s$.
So, $\sum a_n$ converges.
We can prove that $\sum a_n = s$ in a similar way to (1).