Proof explanation: Every absolutely convergent series is commutative. I found the following proof in Kuratowski's intro to calculus, and I was hoping someone could explain to me one of the steps. This is the proof:

$$
2.\ \textit{Every absolutely convergent series is commutative.}
$$
   In other words, if a series $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent and if $m_{1}, m_{2}, ...$ is a permutation of the sequence of positive integers, then 
  $$
\sum_{n=1}^{\infty}a_{m_{n}}=\sum_{n=1}^{\infty}a_{n}.\tag{22}
$$
  Let $\varepsilon>0$. The series $|a_{1}|+|a_{2}|+...$ being convergent, there exists a $k$ such that 
  $$
\sum_{i=k+1}^{\infty}|a_{i}|<\varepsilon. \tag{23}
$$
  Since the sequence $\{m_{n}\}$ contains all positive integers, a number $r$ exists such that the numbers $1,2,3,...,k$ all appear among the numbers $m_{1},m_{2},...,m_{r}$. But each positive integer appears in the sequence only once; therefore we have $m_{n}>k$ for each $n>r$. So by a given $n>r$, if we cancel the numbers $1,2,...,k$ in the system $m_{1},m_{2},...,m_{r},...,m_{n},$ then there remain in this system only numbers grater than $k$ (and different one from anothe). Thus, writing $s_{n}=a_{1}+...+a_{n}$ and $t_{n}=a_{m_{1}}+...+a_{m_{n}}$ and reducing in the difference $t_{n}-s_{n}$ the terms with equal indices, we obtain as remaining terms only terms with indices greater than $k$. This (and the formula for the absolute value of a sum) implies
  $$
|t_{n}-s_{n}|\le\sum_{i=k+1}^{\infty}|a_{i}|,\qquad\text{whence}\qquad|t_{n}-s_{n}|<\varepsilon\text{ by (23).}
$$
  Since the last inequality holds for each $n>r$, we obtain
  $$
\lim_{n=\infty}t_{n}=\lim_{n=\infty}s_{n}.
$$
  This is equivalent to formula (22).


My question arises when he defines $s_{n}$ and $t_{n}$. If both sums contain $n$ elements and you reduce the terms with equal indices, then you don't have any terms left, do you? I'm obviously misunderstanding, so any insights are welcome. Thanks in advance.
 A: Not an answer, just a few more details:
We need the following property of bijections of $\mathbb{Z}$: Suppose $m: \mathbb{N} \to \mathbb{N}$ is a bijection.
Pick some $N$. Then there is some $N'$ such that if $n \ge N'$ then $m_n \ge N$.
To see this, note that there are $k_1,...,k_{N-1}$ such that $m(k_1) = 1,...,
m(k_{N-1}) = N-1$, and if we let $N' = \max(N,k_1,...,k_{N-1})+1$, then if $n \ge N'$
we have $m_n \ge N$. 
In particular, note that by design $N' \ge N$ and that if $n \ge N'$ we have ${1,...,N-1} \subset \{ m_1,...,m_{N'-1} \}$.
So suppose $\sum_n |a_n|$ is summable. Then for any $\epsilon>0$ there is some $N$
such that $\sum_{n\ge N} |a_n| < \epsilon$. Then the above shows that there is some
$N'$ such that $\sum_{n\ge N'} |a_{m_n}| < \epsilon$.
Hence the limits $S = \sum_n a_n$, $S_m = \sum_n a_{m_n}$ exist, we just need to show that they are equal.
Let $\epsilon>0$ and choose $N,N'$ as above. Let $I=\{1,...,N-1\}$ and 
$I' = \{ m_1,...,m_{N'-1}\}$.
\begin{eqnarray}
|S-S_m| &\le& |S-\sum_{n \in I'} a_{n} | + |S_m-\sum_{n \in I'} a_{n} | \\
&\le& |S-\sum_{n \in I} a_{n} -\sum_{n \in I'\setminus I} a_{n}| + \sum_{n \ge N'} |a_{m_n} | \\
&\le& |S-\sum_{n \in I} a_{n}| + \sum_{n \in I'\setminus I} |a_{n}| + \sum_{n \ge N'} |a_{m_n} | \\
&\le& |S-\sum_{n \in I} a_{n}| + \sum_{n \ge N} |a_{n}| + \sum_{n \ge N'} |a_{m_n} | \\
&\le& \sum_{n \ge N} |a_{n}| + \sum_{n \ge N} |a_{n}| + \sum_{n \ge N'} |a_{m_n} | \\
&<& 3 \epsilon
\end{eqnarray}
