[proofplan] We separate the given conditionally convergent series into its positive, negative, and zero terms. The positive subseries must diverge to $\infty$, the negative subseries must diverge to $-\infty$, and all terms tend to $0$. To obtain a prescribed limit $L$, we alternately add unused positive terms until the partial sum exceeds $L$ and unused negative terms until it falls below $L$; the overshoots vanish because the terms tend to $0$. The divergent and oscillatory rearrangements are built by using long enough blocks of positive and negative terms to force the partial sums above or below prescribed levels. [/proofplan]

[step:Separate the positive, negative, and zero terms] Define the index sets \begin{align*} P &:= \{n \in \mathbb{N} : a_n > 0\},\\ N &:= \{n \in \mathbb{N} : a_n < 0\},\\ Z &:= \{n \in \mathbb{N} : a_n = 0\}. \end{align*} Let $(p_k)_{k \in \mathbb{N}}$ be the increasing enumeration of $P$, and let $(q_k)_{k \in \mathbb{N}}$ be the increasing enumeration of $N$. If $Z$ is infinite, let $(z_k)_{k \in \mathbb{N}}$ be its increasing enumeration; if $Z$ is finite, list its elements once as $z_1,\dots,z_m$. [claim:The positive and negative subseries are infinite and diverge in opposite directions] The sets $P$ and $N$ are infinite, and \begin{align*} \sum_{k=1}^{\infty} a_{p_k} = \infty, \qquad \sum_{k=1}^{\infty} a_{q_k} = -\infty. \end{align*} [/claim] [proof] Since $\sum_{n=1}^{\infty} a_n$ converges, its terms satisfy $a_n \to 0$ as $n \to \infty$. Define the positive and negative parts \begin{align*} a_n^+ &:= \max\{a_n,0\},\\ a_n^- &:= \max\{-a_n,0\}. \end{align*} Then $a_n = a_n^+ - a_n^-$ and $|a_n| = a_n^+ + a_n^-$ for every $n \in \mathbb{N}$. Suppose first that $\sum_{n=1}^{\infty} a_n^+$ converges. Since $\sum_{n=1}^{\infty} a_n$ converges, the identity $a_n^- = a_n^+ - a_n$ implies that $\sum_{n=1}^{\infty} a_n^-$ also converges. Hence \begin{align*} \sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} a_n^+ + \sum_{n=1}^{\infty} a_n^- \end{align*} would converge, contradicting the hypothesis that $\sum_{n=1}^{\infty} |a_n|$ diverges. Thus $\sum_{n=1}^{\infty} a_n^+$ diverges. Because each $a_n^+ \geq 0$, its partial sums are increasing, so the divergence is to $\infty$. It remains to prove that $\sum_{n=1}^{\infty} a_n^-$ diverges to $\infty$. Suppose instead that $\sum_{n=1}^{\infty} a_n^-$ converges. Since $\sum_{n=1}^{\infty} a_n$ converges and $a_n^+ = a_n + a_n^-$ for every $n \in \mathbb{N}$, the series $\sum_{n=1}^{\infty} a_n^+$ would also converge. Then \begin{align*} \sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} a_n^+ + \sum_{n=1}^{\infty} a_n^- \end{align*} would converge, contradicting the conditional convergence hypothesis. Thus $\sum_{n=1}^{\infty} a_n^-$ diverges. Because each $a_n^- \geq 0$, its partial sums are increasing, so the divergence is to $\infty$. Therefore the positive terms have total sum $\infty$, and the negative terms have total sum $-\infty$: \begin{align*} \sum_{k=1}^{\infty} a_{p_k} = \infty, \qquad \sum_{k=1}^{\infty} a_{q_k} = -\infty. \end{align*} If $P$ were finite, then only finitely many $a_n^+$ would be nonzero, contradicting $\sum_{n=1}^{\infty} a_n^+ = \infty$. If $N$ were finite, then only finitely many $a_n^-$ would be nonzero, contradicting $\sum_{n=1}^{\infty} a_n^- = \infty$. Hence both $P$ and $N$ are infinite. [/proof] [/step]

[step:Construct a rearrangement converging to the prescribed real number]Fix $L \in \mathbb{R}$. We define a sequence $(b_m)_{m \in \mathbb{N}}$ consisting of all terms of the original sequence exactly once. Start with no positive or negative terms used. At the beginning of stage $j$, insert the zero term $a_{z_j}$ if such a zero term exists and has not yet been inserted. Then append unused positive terms in their increasing-index order until the current partial sum first becomes strictly larger than $L$. Next append unused negative terms in their increasing-index order until the current partial sum first becomes strictly smaller than $L$. This process is well-defined at every stage. Indeed, after finitely many positive and negative terms have been used, the remaining positive terms still have divergent sum $\infty$, and the remaining negative terms still have divergent sum $-\infty$. Hence each crossing of $L$ occurs after finitely many additional terms. Let $(s_m)_{m \in \mathbb{N}}$ be the partial sums of the constructed series: \begin{align*} s_m := \sum_{i=1}^{m} b_i. \end{align*} We prove that $s_m \to L$. During a positive block, the partial sums increase from a value below $L$ to a value above $L$. If the last term of that positive block is $a_{p_k}$, then the excess above $L$ is at most $a_{p_k}$, because the preceding partial sum was at most $L$. During a negative block, the partial sums decrease from a value above $L$ to a value below $L$. If the last term of that negative block is $a_{q_k}$, then the deficit below $L$ has absolute value at most $|a_{q_k}|$, because the preceding partial sum was at least $L$. Since every positive and negative block uses new indices and $a_n \to 0$, the final terms in these blocks tend to $0$. Fix $\varepsilon > 0$. Choose a stage after which every positive crossing term has size less than $\varepsilon$ and every negative crossing term has absolute value less than $\varepsilon$. After that stage, each positive-block endpoint lies in $(L,L+\varepsilon)$ and each negative-block endpoint lies in $(L-\varepsilon,L)$. Inside a positive block the partial sums increase from the preceding undershoot to the next overshoot, and inside a negative block the partial sums decrease from the preceding overshoot to the next undershoot. Hence every sufficiently late partial sum lies between $L-\varepsilon$ and $L+\varepsilon$. Thus \begin{align*} \lim_{m \to \infty} s_m = L. \end{align*} The construction uses each positive term exactly once, each negative term exactly once, and each zero term exactly once. Therefore there exists a bijection $\pi: \mathbb{N} \to \mathbb{N}$ such that $b_m = a_{\pi(m)}$ for every $m \in \mathbb{N}$, and the rearranged series satisfies \begin{align*} \sum_{m=1}^{\infty} a_{\pi(m)} = L. \end{align*}[/step]

[guided]The construction is the standard overshoot-and-correct procedure. Since the positive terms have infinite total mass, we can always push the partial sum above $L$ using only unused positive terms. Since the negative terms have total mass $-\infty$, we can always pull it back below $L$ using only unused negative terms. Formally, begin with no positive or negative terms used. At stage $j$, first insert the zero term $a_{z_j}$ if such a zero term exists and has not yet been inserted. Zero terms do not change any partial sum, but this convention ensures that the final rearrangement is a bijection from $\mathbb{N}$ onto all original indices. Then add unused positive terms in the order $a_{p_1},a_{p_2},\dots$ until the partial sum is strictly greater than $L$. This takes finitely many steps because the unused tail of the positive subseries still has sum $\infty$. After that, add unused negative terms in the order $a_{q_1},a_{q_2},\dots$ until the partial sum is strictly less than $L$. This also takes finitely many steps because the unused tail of the negative subseries still has sum $-\infty$. Let \begin{align*} s_m := \sum_{i=1}^{m} b_i \end{align*} be the partial sum after $m$ terms of the newly constructed sequence. We now check convergence. Consider a positive block, and suppose its last added term is $a_{p_k}$. The partial sum immediately before this last term was at most $L$, because the block stops at the first crossing above $L$. Therefore the final overshoot satisfies \begin{align*} 0 < s_m - L \leq a_{p_k}. \end{align*} Similarly, consider a negative block, and suppose its last added term is $a_{q_k}$. The partial sum immediately before this last term was at least $L$, because the block stops at the first crossing below $L$. Therefore \begin{align*} 0 < L - s_m \leq |a_{q_k}|. \end{align*} The crucial point is that these error bounds go to $0$. Since $\sum a_n$ converges, we have $a_n \to 0$. The indices $p_k$ and $q_k$ used at the ends of later blocks tend to infinity, so $a_{p_k} \to 0$ and $a_{q_k} \to 0$. Hence, for every $\varepsilon > 0$, all sufficiently late overshoots and undershoots have magnitude less than $\varepsilon$, and every partial sum inside a block lies between the two neighboring crossings. Thus all sufficiently late partial sums lie in $(L-\varepsilon,L+\varepsilon)$. Therefore $s_m \to L$. Since the construction exhausts every positive index, every negative index, and every zero index exactly once, the sequence $(b_m)$ is a rearrangement of the original sequence. Equivalently, there is a bijection $\pi:\mathbb{N}\to\mathbb{N}$ with $b_m=a_{\pi(m)}$, and \begin{align*} \sum_{m=1}^{\infty} a_{\pi(m)} = L. \end{align*}[/guided]

[step:Force a rearrangement whose partial sums diverge to $\infty$]We construct a sequence $(c_m)_{m \in \mathbb{N}}$ containing each $a_n$ exactly once. At stage $j$, first append the next unused negative term, if one remains in the prescribed enumeration. Then append unused positive terms until the current partial sum is greater than $j$. Insert the next unused zero term, if one remains. This stage always terminates because the unused positive terms have divergent sum $\infty$. Let \begin{align*} t_m := \sum_{i=1}^{m} c_i \end{align*} be the corresponding partial sums. After stage $j$, the partial sum is greater than $j$. The only possible downward movements between the end of stage $j$ and the end of a later stage come from finitely many individual negative terms. Since $a_{q_k} \to 0$, for every fixed $M \in \mathbb{R}$ there exists $j_0 \in \mathbb{N}$ such that $j_0 > M+1$ and $|a_{q_k}| < 1$ for all negative terms used after stage $j_0$. Hence, after stage $j_0$, even immediately after inserting one negative term, the partial sum remains greater than $M$. Thus $t_m \to \infty$. The construction uses every negative term and every zero term exactly once by design. It also uses every positive term: if some positive term $a_{p_r}$ were never used, then, because positive terms are appended in increasing-index order, no positive term $a_{p_k}$ with $k \geq r$ would be used. The total positive contribution after that point would be bounded, so the partial sums could not exceed every level $j$, contradicting the construction. Hence every index is used exactly once, and there exists a bijection $\pi_+:\mathbb{N}\to\mathbb{N}$ such that the partial sums of $\sum_{m=1}^{\infty} a_{\pi_+(m)}$ diverge to $\infty$.[/step]

[guided]The idea is to make the positive blocks dominate the occasional negative insertions. At stage $j$, we first append the next unused negative term if there is one. This guarantees that negative terms are not postponed forever. Then we append unused positive terms, in the order $a_{p_1},a_{p_2},\dots$, until the partial sum is greater than $j$. This stopping time is finite because deleting finitely many already-used positive terms from a subseries with total sum $\infty$ still leaves a tail with total sum $\infty$. Let \begin{align*} t_m := \sum_{i=1}^{m} c_i \end{align*} be the partial sum after $m$ terms of the constructed sequence. After the end of stage $j$, the partial sum is greater than $j$. The only terms that can decrease the partial sum between the end of one stage and a later positive-block endpoint are the single negative terms inserted at the beginnings of later stages. Since the original series converges, $a_n \to 0$, and therefore $a_{q_k} \to 0$ along the negative subsequence. Given $M \in \mathbb{R}$, choose $j_0 \in \mathbb{N}$ such that $j_0 > M+1$ and every negative term used after stage $j_0$ has absolute value less than $1$. Then after stage $j_0$, even immediately after inserting a negative term, the partial sum is greater than $M$. Thus $t_m \to \infty$. We also check that this is a rearrangement, not merely a subsequence with extra terms. Every negative term is inserted when its turn comes, and every zero term is inserted when its turn comes. Every positive term is used as well: if $a_{p_r}$ were never used, then the increasing-index rule would prevent all $a_{p_k}$ with $k \geq r$ from being used. Only finitely many positive terms would then ever contribute, and the partial sums could not be forced above all levels $j$. This contradicts the construction. Therefore the sequence $(c_m)$ contains every original term exactly once, so it equals $(a_{\pi_+(m)})_{m \in \mathbb{N}}$ for a bijection $\pi_+:\mathbb{N}\to\mathbb{N}$, and its partial sums diverge to $\infty$.[/guided]

[step:Force a rearrangement whose partial sums diverge to $-\infty$]The construction for divergence to $-\infty$ is symmetric. At stage $j$, append the next unused positive term, if one remains in the prescribed enumeration. Then append unused negative terms until the current partial sum is less than $-j$. Insert the next unused zero term, if one remains. The unused negative terms always allow this stage to terminate because their total sum is $-\infty$. If $(u_m)_{m \in \mathbb{N}}$ denotes the partial sums of the resulting rearranged series, then after stage $j$ the partial sum is less than $-j$. The only possible upward movements between the end of stage $j$ and the end of a later stage come from individual positive terms, and these terms tend to $0$. Therefore, for every fixed $M \in \mathbb{R}$, all sufficiently late partial sums are less than $M$. Thus $u_m \to -\infty$. The construction uses every positive term and every zero term exactly once by design. It also uses every negative term: if some negative term $a_{q_r}$ were never used, then, because negative terms are appended in increasing-index order, no negative term $a_{q_k}$ with $k \geq r$ would be used. The total negative contribution after that point would be bounded below, so the partial sums could not be forced below every level $-j$, contradicting the construction. Hence the construction exhausts all indices exactly once, so there exists a bijection $\pi_-:\mathbb{N}\to\mathbb{N}$ whose rearranged partial sums diverge to $-\infty$.[/step]

[guided]This construction is the downward analogue of the previous one, but we spell out the estimates because the conclusion is about all late partial sums, not only the endpoints of stages. At stage $j$, append the next unused positive term if one remains, then append unused negative terms in the order $a_{q_1},a_{q_2},\dots$ until the partial sum is less than $-j$. The stage terminates after finitely many negative terms because removing finitely many already-used negative terms from a subseries with total sum $-\infty$ leaves a tail whose total sum is still $-\infty$. Let $(u_m)_{m \in \mathbb{N}}$ be the partial sums of the resulting series. After stage $j$, the partial sum is less than $-j$. The only terms that can increase a partial sum between the end of one stage and a later negative-block endpoint are the single positive terms inserted at the beginnings of later stages. Since $a_n \to 0$, the positive subsequence satisfies $a_{p_k} \to 0$. Given $M \in \mathbb{R}$, choose $j_0 \in \mathbb{N}$ such that $-j_0 + 1 < M$ and every positive term used after stage $j_0$ has size less than $1$. Then every partial sum after stage $j_0$ is less than $M$, including the partial sums immediately after a positive insertion. Hence $u_m \to -\infty$. Finally, the construction exhausts the original terms. Positive and zero terms are inserted one at a time in their prescribed order. If a negative term $a_{q_r}$ were never inserted, then the increasing-index rule for the negative list would prevent every $a_{q_k}$ with $k \geq r$ from being inserted. The remaining negative contribution would then be bounded below, and the stages could not force the partial sums below all levels $-j$. This contradiction shows that every negative term is used. Thus the resulting sequence is $(a_{\pi_-(m)})_{m \in \mathbb{N}}$ for a bijection $\pi_-:\mathbb{N}\to\mathbb{N}$, and its partial sums diverge to $-\infty$.[/guided]

[step:Force a rearrangement whose partial sums have no extended real limit]We construct a sequence $(d_m)_{m \in \mathbb{N}}$ containing each $a_n$ exactly once and whose partial sums cross above $1$ and below $-1$ infinitely many times. At stage $j$, append unused positive terms until the current partial sum is greater than $1$. Then append unused negative terms until the current partial sum is less than $-1$. Insert the next unused zero term, if one remains. Each positive block terminates because the unused positive terms have sum $\infty$, and each negative block terminates because the unused negative terms have sum $-\infty$. Let \begin{align*} v_m := \sum_{i=1}^{m} d_i \end{align*} be the partial sums of this rearrangement. Infinitely many of the partial sums are greater than $1$, and infinitely many are less than $-1$. Hence $(v_m)_{m \in \mathbb{N}}$ cannot converge to a real number. It also cannot diverge to $\infty$, because it is less than $-1$ infinitely often, and it cannot diverge to $-\infty$, because it is greater than $1$ infinitely often. The construction uses all zero terms exactly once by design. It uses all positive terms: if some $a_{p_r}$ were never used, then no later positive term $a_{p_k}$ with $k \geq r$ would be used, so after finitely many positive insertions the construction could no longer force crossings above $1$. It uses all negative terms by the analogous ordered-tail argument: if some $a_{q_r}$ were never used, then no later negative term would be used, so after finitely many negative insertions the construction could no longer force crossings below $-1$. Thus every original term is used exactly once, and there exists a bijection $\pi_0:\mathbb{N}\to\mathbb{N}$ such that the partial sums of $\sum_{m=1}^{\infty} a_{\pi_0(m)}$ fail to have a limit in $\mathbb{R}\cup\{-\infty,\infty\}$.[/step]

[guided]Here the goal is not convergence to a finite target or divergence to one side. We deliberately force persistent oscillation. At stage $j$, append unused positive terms in the order $a_{p_1},a_{p_2},\dots$ until the current partial sum is greater than $1$, then append unused negative terms in the order $a_{q_1},a_{q_2},\dots$ until the current partial sum is less than $-1$. Each positive block terminates because every unused tail of the positive subseries has sum $\infty$, and each negative block terminates because every unused tail of the negative subseries has sum $-\infty$. Let \begin{align*} v_m := \sum_{i=1}^{m} d_i \end{align*} be the partial sums. At the end of every positive block, some partial sum is greater than $1$. At the end of every negative block, some later partial sum is less than $-1$. Therefore the sequence $(v_m)_{m \in \mathbb{N}}$ has a subsequence lying in $(1,\infty)$ and another subsequence lying in $(-\infty,-1)$. A sequence with two such subsequences cannot converge to a real number. It also cannot diverge to $\infty$, because values below $-1$ occur infinitely often, and it cannot diverge to $-\infty$, because values above $1$ occur infinitely often. We must still verify that the sequence is a rearrangement of the original series. Zero terms are inserted one at a time when available. If a positive term $a_{p_r}$ were never used, the increasing-index rule would prevent all later positive terms from being used, leaving only finitely many positive insertions; then the construction could not continue to force crossings above $1$. This contradicts the prescribed stages. The same ordered-tail argument applies to the negative terms: if $a_{q_r}$ were never used, then all later negative terms would also be unused, and the construction could not continue to force crossings below $-1$. Hence every original index occurs exactly once, so there is a bijection $\pi_0:\mathbb{N}\to\mathbb{N}$ with $d_m=a_{\pi_0(m)}$, and the corresponding partial sums have no limit in $\mathbb{R}\cup\{-\infty,\infty\}$.[/guided]

[step:Conclude all asserted rearrangements exist] The second step proves that for every prescribed $L \in \mathbb{R}$ there is a bijection $\pi:\mathbb{N}\to\mathbb{N}$ for which the rearranged series converges to $L$. The next three steps construct bijections $\pi_+$, $\pi_-$, and $\pi_0$ whose partial sums diverge to $\infty$, diverge to $-\infty$, and have no extended real limit, respectively. These are exactly the assertions of the theorem. [/step]