Theorem 3.55 Rudin (rearrangement and convergence) 
If $\sum a_n$ is a series of complex numbers which converges absolutely, then every rearrangement of $\sum a_n$ converges, and they all converge to the same sum. 
Proof: Let $\sum a_n'$ be a rearrangement , with partial sums $s_n'$. Given $\epsilon > 0$ there exist an integer $N$ such that $m \geq m \geq N$ implies
  $$\sum\limits_{i=n}^m \vert a_i\vert \leq \epsilon. \tag{26}\label{26}$$
Now choose $p$ such that the integers $1, 2, 3, \ldots, N$ are all contained in the set $k_1, k_2, \ldots, k_p$.  [Here  $\{k_n\}, n= 1, 2, 3, \ldots$ is a sequence where every positive integer appears once and only once; $a_n' = a_{k_n}$ for each $n= 1, 2, 3, \ldots$. Then we say that $\sum a_n'$ is a rearrangement of $\sum a_n$.]  Then if $n > p$ , the numbers $a_1,...,a_N$ will cancel in the difference $s_n - s_n'$ , so that $\vert s_n - s_n' \vert \leq \epsilon$, by \eqref{26} Hence ${s_n'}$ converges to the same sum as $\{s_n\}$.

My question is how can I conclude this from \eqref{26};  $\vert s_n - s_n' \vert \leq \epsilon$
 A: The sum $s_n$ comprises $a_1,\, \dotsc,\, a_N$, and also $a_{N+1},\, \dotsc,\, a_n$. The sum $s_n'$ comprises $a_1,\, \dotsc,\, a_N$, and also the $a_k$ for $k$ in a finite set $F$ disjoint from $\{1,\, \dotsc,\, N\}$. Let $G = \{N+1,\, \dotsc,\, n\}$. Then
$$s_n - s_n' = \sum_{k\in G} a_k - \sum_{k\in F} a_k = \sum_{k\in G\setminus F} a _k - \sum_{k \in F\setminus G} a_k,$$
whence
$$\lvert s_n - s_n'\rvert \leqslant \sum_{k \in G\setminus F} \lvert a_k\rvert + \sum_{k \in F\setminus G} \lvert a_k\rvert = \sum_{k \in (G\setminus F) \cup (F\setminus G)} \lvert a_k\rvert \leqslant \varepsilon,$$
since $(G\setminus F) \cup (F\setminus G) \subset \{N+1,\, \dotsc,\, m\}$ for large enough $m$.
A: $$N \leq p < n.$$
$$s_n = a_1 + \cdots + a_N + a_{N+1} + \cdots + a_n.$$
$$s'_n = a_{k_1} + \cdots + a_{k_N} + \cdots + a_{k_p} + a_{k_{p+1}} + \cdots + a_{k_n}.$$
$$\{1, \cdots, N\} \subset \{1, \cdots, n\}.$$
$$\{1, \cdots, N\} \subset \{k_1, \cdots, k_p\} \subset \{k_1, \cdots, k_n\}.$$
$$\{1,\cdots,N\}\subset\{k_1,\cdots,k_n\}\cap\{1,\cdots,n\}.$$
$$T:=(\{k_1,\cdots,k_n\}\cup\{1,\cdots,n\})-(\{k_1,\cdots,k_n\}\cap\{1,\cdots,n\}).$$
$$T \cap \{1, \cdots, N\} = \emptyset.$$
$$\text{If } i \in T, \text{ then } i \geq N+1.$$
$$|s_n - s'_n| \leq \sum_{i \in T} |a_i| \leq \sum_{i=N+1}^{\max{T}} |a_i| \leq \epsilon.$$
