> [!theoremb] Theorem > > Let $\seq{x_n}$ be a real-valued [[Sequence|sequence]], and let $\seq{p_n}, \seq{q_n} \subset \seq{x_n}$ be the non-negative and negative terms of $\seq{x_n}$ (filling in $0$s if the sequences are finite), then > - **Part 1:** If the [[Series|series]] $\sum_{n \in \nat}p_n = \infty$, $\sum_{n \in \nat}q_n = -\infty$ and then for any $c = \pm \infty$, there exists a rearrangement $\seq{x_n'}$ of the sum such that $\sum_{n = 1}^{\infty}x_n' = c$. > - **Part 2:** If both series diverge, but $\sum_{n = 1}^{\infty}x_n$ converges, then for any $c \in \ol\real$, there exists a rearrangement $\seq{x_n'}$ such that $\sum_{n = 1}^{\infty}x_n' = c$. > - **Part 3:** If only one series diverge, then the sum is equal to $\pm \infty$, based on the divergent part, regardless of rearrangements. > - **Part 4:** If both series converge, then the sum converges absolutely and to the same value, regardless of rearrangements. > > In other words, the sum evaluates to the same value regardless of arrangements if and only if its non-negative or negative parts are finite. It evaluates to a finite value if and only if both are finite, in which case the sum converges absolutely. > > ### Part 1 > > Let $\seq{x_n}$ be a real-valued sequence and $\seq{p_n}, \seq{q_n} \subset \seq{x_n}$ be its non-negative and negative terms. If $\sum_{n \in \nat}p_n = \infty$ and $\sum_{n \in \nat}q_n = -\infty$, then there exists a rearrangement of $\seq{x_n}$ such that $\sum_{n = 1}^{\infty}x_n = \infty$ or $\sum_{n = 1}^{\infty}x_n = -\infty$. > > *Proof*. Let $m_0 = 0$. Fix $n \in \nat_0$, then since $\sum_{n \in \nat}p_n = \infty$, there exists $m_{n + 1} \in \nat$ such that $\sum_{n = m_{n} + 1}^{m_{n + 1}} > q_n + 1$. Rearrange the sequence as > $ > \bracs{x_n'} = (q_1, x_1, \cdots, x_{m_{1}}, q_2, x_{m_1 + 1}, \cdots, x_{m_{2}}, \cdots) > $ > then this is a rearrangement and $\sum_{n = 1}^{\infty}x'_n = \infty$. > > For the negative case, simply swap the roles of $p_n$ and $q_n$ in the above process. > > > ### Part 2 > > Let $\seq{x_n}$ be a real-valued [[Sequence|sequence]] such that the [[Series|series]] $\sum x_n$ [[Limit|converges]], but not absolutely. Let $c \in \real$, then there exists an rearrangement $\sum x_n'$ with partial sums $s'_n$ such that > $ > \limv{n}s'_n = c > $ > > *Proof*. Let $\seq{p_n} \subset \seq{x_n}$ be the non-negative terms of the sum, and $\seq{q_n} \subset \seq{x_n}$ be the negative terms of the sum. Since the sum is not absolutely convergent, $\sum_{n \in \nat}p_n = \infty$ or $\sum_{n \in \nat}q_n = -\infty$. As the series is convergent, both must be true. > > Let $c_0 = 0$, $m_0 = 0$, $n_0 = 0$. Fix $n \in \nat_0$. Since $\sum_{k = 1}^{\infty}p_k = \infty$, there exists $m_{n + 1} \in \nat$ such that > $ > c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k > c > $ > Choose the smallest such $m_{n+1}$ such that > $ > c_n + \sum_{k = m_n + 1}^{m_{n + 1} - 1}p_k \le c < c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k > $ > As $\sum_{k = 1}^{\infty}q_n = -\infty$, there exists $n_{n + 1}$ such that > $ > c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k + \sum_{k = n_{n} + 1}^{n_{n + 1}}q_k < c > $ > Choose the smallest $n_{n + 1}$ such that > $ > c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k + \sum_{k = n_{n} + 1}^{n_{n + 1}}q_k < c \le c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k + \sum_{k = n_{n} + 1}^{n_{n + 1} - 1}q_k > $ > Finally, let > $ > c_{n + 1} = c_n + \sum_{k = m_n + 1}^{m_{n + 1}}p_k + \sum_{k = n_{n} + 1}^{n_{n + 1}}q_k = \sum_{k = 1}^{m_{n+1}}p_k + \sum_{k = 1}^{n_{n+1}}q_k > $ > Then $|c_{n + 1} - c| < q_{n_{n+1}}$. As $\limv{n}q_{n} = 0$, $\limv{n}c_n = c$. > > Now arrange the sequence as > $ > \seq{x_n'} = (p_1, \cdots, p_{m_1}, q_1, \cdots, q_{n_1}, p_{m_1 + 1}, \cdots) > $ > Then since each step of this process (except the first) changes the partial sum from strictly greater than $c$ to strictly less than $c$ (or the other way around), the segment added to the partial sum is non-empty, meaning that $\seq{m_n}, \seq{n_n}$ are both strictly increasing. Therefore for any $n$, $p_n, q_n$ is included in the partial sum represented by $c_n$. As each $p_n$ and $q_n$ is only used once, this constitutes as a rearrangement. > > Moreover, since the partial sums reverse in direction immediately after exceeding $c$, $\limv{n}s_n' = c$ as well. > > > ### Part 3 > > Let $\seq{x_n}$ be a real-valued sequence and $\seq{p_{m_k}}, \seq{q_{n_k}} \subset \seq{x_n}$. > > If $\seq{q_{n_k}}$ is convergent or finite (there are only finitely many of them) and $\sum_{k \in \nat}p_{m_k} = \infty$, then $\sum_{n = 1}^{\infty}x_n = \infty$. > > If $\seq{p_{m_k}}$ is convergent or finite, and $\sum_{k \in \nat}q_{n_k} = \infty$, then $\sum_{n = 1}^{\infty}x_n = -\infty$. > > *Proof*. Let $S = \sum_{k \in \nat}q_{n_k}$ and suppose that $\sum_{k \in \nat}p_{m_k} = \infty$, then for all $\alpha \in \real$, > $ > \forall \alpha \in \real, \exists K \in \nat: \sum_{j = 1}^{k}p_{m_j} > \alpha - S \quad \forall k \ge K > $ > Let $k \ge K$ and $n \ge n_k$, then > $ > \sum_{j = 1}^{n}x_n \ge \sum_{k = 1}^{k}p_{n_k} + \sum_{n_k \le n}q_{n_k} \ge \sum_{k = 1}^{k}p_{n_k} + S > \alpha > $ > and taking $N = n_K$ yields > $ > \sum_{j = 1}^{N}x_n > \alpha \quad \forall n \ge N > $ > therefore $\sum_{n = 1}^{\infty}x_n = \infty$. > > For the negative case, swap the role of $\seq{p_{m_k}}$ and $\seq{q_{n_k}}$. > > > ### Part 4 > > Let $\seq{x_n} \subset \real$ be a sequence such that $\sum_{n = 1}^{\infty}x_n$ converges absolutely, then every rearrangement of $\sum_{n = 1}^{\infty}x_n$ converges to the same sum. Moreover, if $\bracs{p_n} \subset \bracs{x_n}$ are the non-negative terms and $\bracs{q_n} \subset \bracs{x_n}$ are the negative terms, > $ > \sum_{n = 1}^{\infty}x_n = \sum_{n \in \nat}p_n + \sum_{n \in \nat}q_n > $ > where > $ > \sum_{n \in \nat}p_n = \sup\bracs{\sum_{i \in I}p_i: I \subset \nat, |I| < |\nat|} > $ > and > $ > \sum_{n \in \nat}q_n = \inf\bracs{\sum_{i \in I}p_i: I \subset \nat, |I| < |\nat|} > $ > *Proof.* Since the sum converges absolutely, > $ > \sum_{n = 1}^{\infty}|x_{\sigma(n)}| = \sum_{n \in \nat}|x_{n}| > $ > for any rearrangement $\sigma: \nat \to \nat$ of the indices, meaning that any rearranged series also converges. Let $\varepsilon > 0$, then > $ > \exists N \in \nat: \sum_{n > N}|x_n| < \varepsilon/4 > $ > Let $N' = \max(\sigma^{-1}([1, N]))$, then $\sigma([1, N']) \supseteq [1, N]$ and $\sigma(n) > N$ for all $n > N'$. This means that > $ > \begin{align*} > \abs{\sum_{i = 1}^{N'}x_n - \sum_{i = 1}^{N'}x_{\sigma(n)}} &= \abs{\sum_{i = N+1}^{N'}x_n - \sum_{n \in [1, N'] \setminus \sigma^{-1}([1, N])}x_{\sigma(n)}} \\ > &\le \abs{\sum_{i = N + 1}^{N'}x_n} + \abs{\sum_{n \in [1, N'] \setminus \sigma^{-1}([1, N])}x_{\sigma(n)}} \\ > &\le \sum_{i = N + 1}^{N'}|x_n| + \sum_{n \in [1, N'] \setminus \sigma^{-1}([1, N])}\abs{x_{\sigma(n)}} \\ > &\le 2\sum_{n > N}|x_n| > \end{align*} > $ > which allows > $ > \begin{align*} > \abs{\sum_{i = 1}^{\infty}x_n - \sum_{i = 1}^{\infty}x_{\sigma(n)}} &\le > \abs{ > \sum_{i = N' + 1}^{\infty}x_n - \sum_{i = N'+1}^{\infty}x_{\sigma(n)}} \\ > &+ \abs{\sum_{i = 1}^{N'}x_n - \sum_{i = 1}^{N'}x_{\sigma(n)}} \\ > &\le 4\sum_{n > N}|x_n| < \varepsilon > \end{align*} > $ > Therefore the sequence of partial sums for $\sum x_{\sigma(n)}$ and $\sum x_{n}$ converge to the same value. > > Now, let $\sigma: \nat \to \nat$ be a rearrangement of $\seq{x_n}$ such that it is non-negative on odd indices, and negative on even indices (setting their value to $0$ once the non-negative/negative parts run out), then > $ > \begin{align*} > \sum_{n = 1}^{\infty}x_{\sigma(n)} &= \sum_{n = 1}^{\infty}x_{\sigma(2n - 1)} + \sum_{n = 1}^{\infty}x_{\sigma(2n)} \\ > &= \sum_{n \in \nat}x_{\sigma(2n - 1)} - \sum_{n \in \nat}(-x_{\sigma(2n)}) \\ > &= \sum_{n \in \nat}p_n + \sum_{n \in \nat}q_n > \end{align*} > $ >