Proof that rearrangements of an absolutely convergent series have the same sum

Proof

Let $T = \sum_{n=0}^{\infty} \abs{a_n}$, and let $t_K = \sum_{k=0}^{K} \abs{a_k}$ and $s_K = \sum_{k=0}^{K} a_k$. The series $\sum a_n$ converges, with sum $A$, since absolutely convergent series converge. Let $B_N = \sum_{n=0}^{N} a_{\sigma(n)}$ be the partial sums of the rearrangement.

1. By the criterion for nonnegative series, $t_K \le T$ for every $K$.
2. Let $N$ be a natural number and let $K = \max \set{\sigma(0), \dots, \sigma(N)}$. Since $\sigma$ is injective, the indices $\sigma(0), \dots, \sigma(N)$ are distinct elements of $\set{0, \dots, K}$.
3. So $\sum_{n=0}^{N} \abs{a_{\sigma(n)}} \le t_K \le T$, because the terms are nonnegative.
4. The partial sums of $\sum \abs{a_{\sigma(n)}}$ are bounded above by $T$. By the same criterion, the rearrangement converges absolutely.
5. Let $\varepsilon > 0$. Since $t_M \to T$ and $s_M \to A$, there is $M$ with $T - t_M < \varepsilon$ and $\abs{s_M - A} < \varepsilon$, by the definition of the limit.
6. Since $\sigma$ is surjective, each of $0, 1, \dots, M$ equals $\sigma(n)$ for some $n$. Let $N_0$ be the largest of these $M + 1$ values of $n$.
7. Let $N \ge N_0$. Then $\set{0, \dots, M} \subseteq \sigma(\set{0, \dots, N})$. Let $D = \sigma(\set{0, \dots, N}) \setminus \set{0, \dots, M}$, a finite set of indices larger than $M$.
8. Since $\sigma$ is injective, $B_N = \sum_{k \in \sigma(\set{0, \dots, N})} a_k = s_M + \sum_{k \in D} a_k$.
9. Let $K$ be an integer at least $M$ and every element of $D$. By the triangle inequality and step 1, $\abs{\sum_{k \in D} a_k} \le \sum_{k=M+1}^{K} \abs{a_k} = t_K - t_M \le T - t_M < \varepsilon$.
10. By steps 5, 8 and 9, $\abs{B_N - A} \le \abs{s_M - A} + \abs{\sum_{k \in D} a_k} < 2\varepsilon$ for all $N \ge N_0$.
11. So $B_N \to A$. The rearrangement converges absolutely, and its sum is $A$.