Let $\seq{x_n} \subset \complex$ be such that $\exists \sum_{n = 1}^{\infty}x_n$, then $\sum_{n = 1}^{\infty}x_n = \sum_{n = 1}^{\infty}x_{\sigma(n)}$ for any bijection $\sigma: \nat \to \nat$, if and only if $\sum_{n = 1}^{\infty}\abs{x_n} < \infty$. *Proof*. Suppose that $\seq{x_n}$ is absolutely convergent, then since $\abs{\re{x_n}} \le \abs{x_n}$ and $\abs{\im{x_n}} \le \abs{x_n}$, both $\seq{\re{x_n}}$ and $\seq{\im{x_n}}$ are absolutely convergent. By the rearrangement theorem for real-valued sequences, $ \begin{align*} \sum_{n = 1}^{\infty}x_{n} &= \sum_{n = 1}^{\infty}\re{x_n} + i\sum_{n = 1}^{\infty}\im{x_n} \\ &= \sum_{n = 1}^{\infty}\re{x_{\sigma (n)}} + i\sum_{n = 1}^{\infty}\im{x_{\sigma(n)}} \\ &= \sum_{n = 1}^{\infty}x_{\sigma(n)} \end{align*} $ Now suppose that $\seq{x_n}$ is not absolutely convergent. Since $ \sum_{n \in \nat}\abs{x_n} \le \sum_{n \in \nat}\abs{\re{x_n}} + \sum_{n \in \nat}\abs{\im{x_n}} $ At least one of $\seq{\re{x_n}}$ and $\seq{\im{x_n}}$ is not absolutely convergent. Assume without loss of generality that $\seq{\re{x_n}}$ is not absolutely convergent, then there exists a rearrangement $\sigma$ such that $\sum_{n = 1}^{\infty}\re{x_n} \ne \sum_{n = 1}^{\infty}\re{x_{\sigma(n)}}$. Therefore $\sum_{n = 1}^{\infty}x_n \ne \sum_{n = 1}^{\infty}x_{\sigma(n)}$.