[proofplan] We show that any rearrangement of an absolutely convergent series converges to the same sum. The key idea is that absolute convergence makes the tail $\sum_{j=M+1}^{\infty} |a_j|$ arbitrarily small, and for $N$ large enough, the partial sum of the rearrangement includes all terms up to index $M$, so the difference from $S$ is controlled by the tail of the absolute series. [/proofplan] [step:Show that the rearranged partial sums eventually capture all initial terms] Let $S = \sum_{j=1}^{\infty} a_j$ and $T_N = \sum_{j=1}^{N} a_{\sigma(j)}$. Let $\epsilon > 0$ be given. Since $\sum_{j=1}^{\infty} |a_j|$ converges, there exists $M \in \mathbb{N}$ such that \begin{align*} \sum_{j=M+1}^{\infty} |a_j| < \epsilon. \end{align*} Define $K = \max\{\sigma^{-1}(1), \sigma^{-1}(2), \ldots, \sigma^{-1}(M)\}$. For all $N \geq K$, the set $\{\sigma(1), \sigma(2), \ldots, \sigma(N)\}$ contains $\{1, 2, \ldots, M\}$, so $T_N$ includes all terms $a_1, a_2, \ldots, a_M$. [guided] Since $\sigma: \mathbb{N} \to \mathbb{N}$ is a bijection, every natural number appears exactly once in the sequence $\sigma(1), \sigma(2), \sigma(3), \ldots$ The index $\sigma^{-1}(m)$ tells us where the original term $a_m$ appears in the rearrangement. Taking $K = \max\{\sigma^{-1}(1), \ldots, \sigma^{-1}(M)\}$ ensures that by position $K$, all of $a_1, \ldots, a_M$ have appeared in the rearranged partial sum. This is where absolute convergence is used: it guarantees the tail $\sum_{j=M+1}^{\infty} |a_j|$ can be made arbitrarily small, which controls the terms that have not yet been matched between the two partial sums. [/guided] [/step] [step:Bound $|T_N - S|$ using the tail of the absolute series] For $N \geq K$, the partial sum $T_N$ contains all terms $a_1, \ldots, a_M$ plus some terms $a_j$ with $j > M$. Therefore \begin{align*} T_N - S &= T_N - \sum_{j=1}^{\infty} a_j \\ &= \sum_{\substack{j \leq N \\ \sigma(j) > M}} a_{\sigma(j)} - \sum_{j=M+1}^{\infty} a_j. \end{align*} Taking absolute values and using the triangle inequality: \begin{align*} |T_N - S| &\leq \sum_{\substack{j \leq N \\ \sigma(j) > M}} |a_{\sigma(j)}| + \left|\sum_{j=M+1}^{\infty} a_j - \sum_{\substack{j \leq N \\ \sigma(j) > M}} a_{\sigma(j)}\right|. \end{align*} More directly, since $T_N$ and $S$ agree on the terms $a_1, \ldots, a_M$, the difference $|T_N - S|$ involves only terms $a_j$ with $j > M$. Each such term satisfies $|a_j| \leq \sum_{j=M+1}^{\infty} |a_j| < \epsilon$. Therefore \begin{align*} |T_N - S| \leq \sum_{j=M+1}^{\infty} |a_j| < \epsilon. \end{align*} Since $\epsilon > 0$ is arbitrary, $T_N \to S$ as $N \to \infty$. [/step]