Theorem 10.2.4 (Riemann’s Rearrangement Theorem). Let $\seq{x_n}\subset \real$ and $N = P \sqcup N$ such that $x_{n} \ge 0$ for all $n \in P$ and $x_{n} \le 0$ for all $n \in N$, then
If $\sum_{n \in P}x_{n} = \infty$ and $\sum_{n \in N}x_{n} = -\infty$, then there exists bijections $\sigma, \tau: \natp \to \natp$ such that
\[\sum_{n = 1}^{\infty} x_{\sigma(n)}= \infty \quad \sum_{n = 1}^{\infty} x_{\tau(n)}= -\infty\]If $\sum_{n \in P}x_{n} = \infty$, $\sum_{n \in N}x_{n} = -\infty$, and $\sum_{n =1}^{\infty} x_{n}$ converges, then for any $x \in \ol{\real}$, there exists a bijection $\sigma: \natp \to \natp$ such that $\sum_{n = 1}^{\infty} x_{\sigma(n)}= x$.
If $\sum_{n \in P}x_{n} = \infty$ but $\sum_{n \in N}x_{n} > -\infty$, then $\sum_{n = 1}^{\infty} x_{n}$ converges to $\infty$ unconditionally.
If $\sum_{n \in N}x_{n} = -\infty$ but $\sum_{n \in P}x_{n} < \infty$, then $\sum_{n = 1}^{\infty} x_{n}$ converges to $-\infty$ unconditionally.
If $\sum_{n \in \natp}|x_{n}| < \infty$, then $\sum_{n = 1}^{\infty} x_{n}$ converges unconditionally.
In other words, a series in $\real$ converges unconditionally if and only if its positive parts or its negative parts are finite.
Proof. By inserting zeroes, assume without loss of generality that $P$ and $N$ are both infinite. Let $\seq{p_k}= P$ be an enumeration of $P$ and $\seq{n_k}= N$ be an enumeration of $N$.
(1): By taking $\sum_{n \in P}-x_{n}$, assume without loss of generality that $\sum_{n \in P}x_{n} = \infty$.
Let $k_{0} = 0$. Let $n \in \natp$ and suppose inductively that $k_{n}$ has been constructed such that
Since $\sum_{n \in \natp}p_{n} = \infty$, there exists $k_{n+1}\in \natp$ such that $\sum_{k = k_{n}+1}^{k_{n+1}}x_{p_k}\ge x_{n_{n+1}}+ 1$.
Therefore there exists $\bracs{k_n}_{1}^{K} \subset \natp$ such that
for all $1 \le n < K$. In which case, if $\sigma: \natp \to \natp$ is defined by
Let $n \in \natp$, then for any $K \ge k_{n}+ n$, $\sum_{k = 1}^{K} x_{\sigma(k)}\ge n$, so $\sum_{k = 1}^{\infty} x_{\sigma(k)}= \infty$.
(2): Assume without loss of generality that $x > 0$. Define $m_{0} = 0$ and $n_{0} = 0$. Let $n \in \natz$. If $n \ge 1$, suppose inductively that $m_{n}, n_{n} \in \natp$ has been constructed such that
Since $\sum_{k \in \natp}x_{p_k}= \infty$, there exists $m_{n+1}> m_{n}$ such that
As $\sum_{k \in \natp}x_{n_k}= -\infty$, there exists $n_{n+1}> n_{n}$ such that
Let $\sigma: \natp \to \natp$ be defined by
then for any $K \in [m_{n} + n_{n}, m_{n+1}+ n_{n+1}]$,
Since $x_{n} \to 0$ as $n \to \infty$, $\sum_{k = 1}^{K} x_{\sigma(k)}\to x$ as $K \to \infty$.
(3): Let $\sigma: \natp \to \natp$ be a bijection and $\alpha > 0$, then there exists $K \in \natp$ such that $\sum_{k = 1}^{K} x_{p_k}> \alpha - \sum_{k \in \natp}x_{n_k}$. Since $\sigma$ is a bijection, there exists $K' \in \natp$ such that $\sigma([1, K']) \supset \bracs{p_k|1 \le k \le K}$. In which case, for any $k \ge K'$,
(4): By applying (3) to $\sum_{n = 1}^{\infty} -x_{n}$.
(5): Let $\sigma: \natp \to \natp$ be a bijection, then for any $N \in \natp$, there exists $K \in \natp$ such that $\sigma([1, K]) \supset [1, N]$. In which case,
so