The Riemann series theorem: a conditionally convergent real series has, for every $c \in \mathbb{R}$, a rearrangement with sum $c$, and rearrangements diverging to $+\infty$, to $-\infty$, and oscillating with any prescribed $\liminf \le \limsup$ in $\overline{\mathbb{R}}$

Since $\sum a_k$ converges, $a_k \to 0$.

For every $K \in \mathbb{N}$ there is $k \ge K$ with $k \in P$: otherwise $a_k < 0$ for every $k \ge K$, so $a_k^{+} = 0$ for every $k \ge K$, so the partial sums of $\sum a_k^{+}$ are constant from $K$ on and hence bounded, contradicting [L1]. The same argument with $a_k^{-}$ shows that for every $K$ there is $k \ge K$ with $k \in N$.

In particular $P$ and $N$ are nonempty, and for every $k$ the sets $\{m \in P : m > k\}$ and $\{m \in N : m > k\}$ are nonempty; so by [L3] each has a least element.

Define $p : \mathbb{N} \to \mathbb{N}$ by $p_0 := \min P$ and $p_{i+1} := \min\{m \in P : m > p_i\}$, and $q : \mathbb{N} \to \mathbb{N}$ by $q_0 := \min N$ and $q_{l+1} := \min\{m \in N : m > q_l\}$; both are legitimate applications of the recursion theorem, the "next element" operations being total functions $\mathbb{N} \to \mathbb{N}$ by step 2.1. Both $p$ and $q$ take values in $P$, respectively $N$, and are strictly increasing.

An induction gives $p_i \ge i$ and $q_l \ge l$ for every index, since $p_0 \ge 0$ and $p_{i+1} > p_i \ge i$ forces $p_{i+1} \ge i+1$.

An induction on $i$ gives $P \cap \{k : k < p_i\} = \{p_{i'} : i' < i\}$: at $i = 0$ both sides are empty because $p_0$ is the least element of $P$; and passing from $i$ to $i+1$ adds exactly $p_i$, since $p_{i+1}$ is the least element of $P$ strictly greater than $p_i$, so no element of $P$ lies strictly between them. The same holds for $q$ and $N$.

Fix real sequences $(u_j)$ and $(v_j)$ with $u_j \le v_j$ and $u_j \le v_{j+1}$ for every $j$. Put $A := \mathbb{N} \times \mathbb{N} \times \mathbb{N} \times \mathbb{R} \times \{0,1\}$, whose elements are written $(i, l, j, s, m)$, and define $\mathrm{out} : A \to \mathbb{N}$ and $f : A \to A$ by: if $m = 0$ and $s \le v_j$, then $\mathrm{out} := p_i$ and $f := (i+1, l, j, s + a_{p_i}, 0)$; if $m = 0$ and $s > v_j$, then $\mathrm{out} := q_l$ and $f := (i, l+1, j, s + a_{q_l}, 1)$; if $m = 1$ and $s \ge u_j$, then $\mathrm{out} := q_l$ and $f := (i, l+1, j, s + a_{q_l}, 1)$; if $m = 1$ and $s < u_j$, then $\mathrm{out} := p_i$ and $f := (i+1, l, j+1, s + a_{p_i}, 0)$. The four cases are exhaustive and mutually exclusive, the order on $\mathbb{R}$ being total, so $f$ and $\mathrm{out}$ are functions.

Every element of $P$ is some $p_i$, and every element of $N$ is some $q_l$: given $k \in P$, the set $\{i : p_i > k\}$ is nonempty by step 4.1, so it has a least element $i_0$; $i_0 \ne 0$ since $p_0 = \min P \le k$, and $p_{i_0 - 1} \le k < p_{i_0}$, so $k \in P \cap \{m : m < p_{i_0}\} = \{p_{i'} : i' < i_0\}$ by step 4.2. Together with step 3.1 this says that $p$ is a bijection onto $P$ and $q$ a bijection onto $N$; both are injective because they are strictly increasing.

An induction on $i$ gives $\sum_{i' < i} a_{p_{i'}} = \sum_{k < p_i} a_k^{+}$: at $i = 0$ every $k < p_0$ lies in $N$, so $a_k^{+} = 0$ and both sides are $0$; and splitting $\sum_{k<p_{i+1}} a_k^{+}$ at $p_i$ and at $p_i + 1$ isolates the single term $a_{p_i}^{+} = a_{p_i}$, all remaining indices $k$ with $p_i < k < p_{i+1}$ lying in $N$ by step 4.2 and contributing $0$. The same argument gives $\sum_{l' < l} a_{q_{l'}} = -\sum_{k<q_l} a_k^{-}$.

By the recursion theorem let $g : \mathbb{N} \to A$ satisfy $g(0) = (0,0,0,0,0)$ and $g(n+1) = f(g(n))$, write $g(n) = (i_n, l_n, j_n, s_n, m_n)$, and define $\sigma(n) := \mathrm{out}(g(n))$.

For general $\alpha \le \beta$ choose real sequences with $u_j \le v_j$ and $u_j \le v_{j+1}$ as follows: if $\alpha, \beta$ are real, $u_j := \alpha$ and $v_j := \beta$; if $\alpha = -\infty$ and $\beta$ is real, $u_j := \beta - (j+1)$ and $v_j := \beta$; if $\alpha$ is real and $\beta = +\infty$, $u_j := \alpha$ and $v_j := \alpha + (j+1)$; if $\alpha = \beta = +\infty$, $u_j := j$ and $v_j := j+1$; if $\alpha = \beta = -\infty$, $u_j := -(j+2)$ and $v_j := -(j+1)$; and if $\alpha = -\infty$, $\beta = +\infty$, $u_j := -(j+1)$ and $v_j := j+1$. In every case $(u_j)$ tends to $\alpha$ and $(v_j)$ to $\beta$ in $\overline{\mathbb{R}}$, and both conditions of step 4.3 hold.

Hence $\sum_{i'<i} a_{p_{i'}} \to +\infty$ as $i \to \infty$ and $\sum_{l'<l} a_{q_{l'}} \to -\infty$ as $l \to \infty$: the left-hand sides are the values of the partial sums of $\sum a_k^{+}$, respectively of $-\sum a_k^{-}$, at the strictly increasing indices $p_i$, respectively $q_l$, and by step 4.1 those indices are at least $i$, respectively $l$.

An induction on $n$ gives $i_n + l_n = n$ and $s_n = \sum_{k<n} a_{\sigma(k)}$: both hold at $n = 0$, and each transition increases exactly one of $i, l$ by one and adds to $s$ exactly the term $a_{\sigma(n)}$ indexed by the emitted natural. So $s_n = T_n$, the $n$-th partial sum of the rearranged series.

Consequently, for every $i_0 \in \mathbb{N}$ and every real $M$ there is $i > i_0$ with $\sum_{i'=i_0}^{i-1} a_{p_{i'}} > M$, and for every $l_0$ and every real $M$ there is $l > l_0$ with $\sum_{l' = l_0}^{l-1} a_{q_{l'}} < M$; this is step 6.1 together with splitting of finite sums, the omitted initial block being a fixed real.

An induction on $n$ gives that $\sigma(n) = p_{i_n}$ at every step that increments $i$, and $\sigma(n) = q_{l_n}$ at every step that increments $l$; since $(i_n)$ and $(l_n)$ are nondecreasing and increase by one exactly at those steps, distinct steps of the first kind carry distinct values of $i_n$ and distinct steps of the second kind distinct values of $l_n$. As $p$ and $q$ are injective with disjoint ranges $P$ and $N$, the map $\sigma$ is injective.

There are infinitely many steps of each kind: if from some step $n_0$ on no step increments $l$, then $m_n$ is eventually constantly $0$, because a step with $m = 1$ that does not increment $l$ sets $m$ to $0$ and a step with $m = 0$ that does not increment $l$ leaves $m$ at $0$; then $j_n$ is eventually constant, say $j$, and every subsequent step satisfies $s_n \le v_{j}$, while by step 7.1 the values $s_n$, which from $n_0$ on increase by the successive terms $a_{p_i}$, exceed $v_j$ for some $n$. Symmetrically, if from some step on no step increments $i$, then $m_n$ is eventually constantly $1$, $j_n$ is eventually constant $j$, every subsequent step satisfies $s_n \ge u_j$, and step 7.1 makes $s_n$ fall below $u_j$.

Hence $i_n \to \infty$ and $l_n \to \infty$, so every $p_i$ and every $q_l$ occurs as some $\sigma(n)$; since $P \cup N = \mathbb{N}$ and $p, q$ enumerate $P$ and $N$, the map $\sigma$ is surjective, and with step 7.2 it is a bijection of $\mathbb{N}$.

Likewise $j_n \to \infty$: if $j_n$ were eventually constant $j$, then from some step on no round is completed, so no step has $m = 1$ and $s < u_j$; by the argument of step 8.1 the mode is then eventually constant, and either it is $0$ forever, whence $s_n \le v_j$ always while $s_n$ increases past $v_j$, or it is $1$ forever, whence $s_n \ge u_j$ always while $s_n$ falls below $u_j$.

For each $j \ge 1$ let $\beta_j$ be the step at which the mode of round $j$ changes from $0$ to $1$, that is the unique $n$ with $j_n = j$, $m_n = 0$ and $s_n > v_j$, and let $\alpha_j$ be the step at which round $j$ is completed, the unique $n$ with $j_n = j$, $m_n = 1$ and $s_n < u_j$; both exist by step 8.1 and step 9.2, and $\alpha_{j-1} < \beta_j < \alpha_j$.

The step $\beta_j$ is preceded, within round $j$, either by a step that added a term $a_{p_i} \ge 0$ to a value $s \le v_j$, or by the completing step $\alpha_{j-1}$ of the previous round, which added a term $a_{p_i} \ge 0$ to a value $s < u_{j-1} \le v_j$. In both situations $v_j < T_{\beta_j} \le v_j + a_{p_i}$ for the index $i$ used at the immediately preceding step.

Likewise the step $\alpha_j$ is preceded within round $j$ by a step that added a term $a_{q_l} < 0$ to a value $s \ge u_j$, that step being either an earlier descent step or the switch $\beta_j$ itself, at which $s > v_j \ge u_j$; so $u_j - |a_{q_l}| \le T_{\alpha_j} < u_j$ for the index $l$ used at that step.

For $\alpha_{j-1} \le n \le \beta_j$ the partial sums increase, every step of the climb adding a term $a_{p_i} \ge 0$; for $\beta_j \le n \le \alpha_j$ they decrease, every step of the descent adding a term $a_{q_l} < 0$. Hence for every $n$ with $\alpha_{j-1} \le n \le \alpha_j$ one has $\min\{T_{\alpha_{j-1}}, T_{\alpha_j}\} \le T_n \le T_{\beta_j}$.

Put $\delta_j := \max\{a_{p_{i(j)}},\, |a_{q_{l(j)}}|\}$ for the two indices appearing in step 11.1 and step 11.2. As $j \to \infty$ those indices tend to infinity, by step 8.1 and step 9.2, so $p_{i(j)} \to \infty$ and $q_{l(j)} \to \infty$ by step 4.1, and $\delta_j \to 0$ by step 1.1. Thus $v_j < T_{\beta_j} \le v_j + \delta_j$ and $u_j - \delta_j \le T_{\alpha_j} < u_j$ for every $j \ge 1$.

Fix $n$ and let $J$ be least with $\alpha_{J-1} \ge n$, which exists because the $\alpha_j$ are strictly increasing. By step 11.3 every $m \ge \alpha_{J-1}$ satisfies $T_m \le \sup\{T_{\beta_j} : j \ge J\}$, and only the finitely many indices $m$ with $n \le m < \alpha_{J-1}$ are unaccounted for; each of those lies in a round of index at most $J-1$ and so is at most $\max\{T_{\beta_j} : 1 \le j \le J-1\}$ together with $T_n$ itself. Hence $\sup\{T_m : m \ge n\}$ is finite or $+\infty$ according as $\sup\{T_{\beta_j} : j \ge J\}$ is, and taking the infimum over $n$, which drives $J$ to infinity, gives $\limsup_n T_n = \limsup_j T_{\beta_j}$.

Take $u_j = v_j = c$ for all $j$, which satisfies the two conditions of step 4.3. Then $c < T_{\beta_j} \le c + \delta_j$ and $c - \delta_j \le T_{\alpha_j} < c$, so by step 11.3 every $n$ with $\alpha_{j-1} \le n \le \alpha_j$ has $|T_n - c| \le \max\{\delta_{j-1}, \delta_j\}$. Given a real $\varepsilon > 0$, choose $J \ge 2$ with $\delta_j < \varepsilon$ for all $j \ge J-1$; then $|T_n - c| < \varepsilon$ for all $n \ge \alpha_{J-1}$, so $T_n \to c$ and the rearranged series converges with sum $c$. This is claim 1.

Take $v_j = j+1$ and $u_j = j$, which satisfy the two conditions. Then $T_{\alpha_j} \ge u_j - \delta_j = j - \delta_j$, so by step 11.3 every $n$ with $\alpha_{j-1} \le n \le \alpha_j$ has $T_n \ge \min\{j-1-\delta_{j-1},\, j - \delta_j\}$, a quantity that exceeds any prescribed real for all large $j$; hence $T_n \to +\infty$. Taking instead $v_j = -(j+1)$ and $u_j = -(j+2)$, which also satisfy the two conditions, gives $T_n \le T_{\beta_j} \le v_j + \delta_j = -(j+1) + \delta_j$ on the same ranges, hence $T_n \to -\infty$. This is claim 2.

By step 12.1 the subsequence $(T_{\beta_j})_{j \ge 1}$ tends to $\beta$ and $(T_{\alpha_j})_{j\ge1}$ tends to $\alpha$, in $\overline{\mathbb{R}}$: when the target sequence is real-valued and convergent the two-sided bound of step 12.1 with $\delta_j \to 0$ gives it, and when the target sequence diverges the one-sided bound does.

By step 13.3 and [L11], $\limsup_j T_{\beta_j} = \beta$; so $\limsup_n T_n = \beta$. The same argument applied to infima, with $\alpha_j$ in place of $\beta_j$ and the lower bound of step 11.3 in place of the upper one, gives $\liminf_n T_n = \liminf_j T_{\alpha_j} = \alpha$.

The bijection $\sigma$ of step 5.3, built from the targets chosen in step 5.4, is therefore a rearrangement of $\sum a_k$ whose partial sums have limit inferior $\alpha$ and limit superior $\beta$; claims 1 and 2 are the special cases computed directly in step 13.1 and step 13.2, and claim 3 is the case $\alpha < \beta$.