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.
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}