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 $+\mathrm{\infty }$, to $-\mathrm{\infty }$, and oscillating with any prescribed $\mathrm{lim\hspace{0.17em}inf}\le \mathrm{lim\hspace{0.17em}sup}$ in $\bar{\mathbb{R}}$

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 $P\cap \{k:k<p_i\}=\{p_{i^{\mathrm{\prime }}}:i^{\mathrm{\prime }}<i\}$: 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 $\mathrm{o}\mathrm{u}\mathrm{t}:A\to \mathbb{N}$ and $f:A\to A$ by: if $m=0$ and $s\le v_j$, then $\mathrm{o}\mathrm{u}\mathrm{t}:=p_i$ and $f:=(i+1,l,j,s+a_{p_i},0)$; if $m=0$ and $s>v_j$, then $\mathrm{o}\mathrm{u}\mathrm{t}:=q_l$ and $f:=(i,l+1,j,s+a_{q_l},1)$; if $m=1$ and $s\ge u_j$, then $\mathrm{o}\mathrm{u}\mathrm{t}:=q_l$ and $f:=(i,l+1,j,s+a_{q_l},1)$; if $m=1$ and $s<u_j$, then $\mathrm{o}\mathrm{u}\mathrm{t}:=p_i$ 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 $\mathrm{o}\mathrm{u}\mathrm{t}$ 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 $i_0$; $i_0\mathrm{\ne }0$ since $p_0=\min P\le k$, and $p_{i_0-1}\le k<p_{i_0}$, so $k\in P\cap \{m:m<p_{i_0}\}=\{p_{i^{\mathrm{\prime }}}:i^{\mathrm{\prime }}<i_0\}$ 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 ${\sum }_{i^{\mathrm{\prime }}<i}{a}_{p_{i^{\mathrm{\prime }}}}={\sum }_{k<p_i}{a}_k^{+}$: 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 ${\sum }_{l^{\mathrm{\prime }}<l}{a}_{q_{l^{\mathrm{\prime }}}}=-{\sum }_{k<q_l}{a}_k^{-}$.

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 $\sigma (n):=\mathrm{o}\mathrm{u}\mathrm{t}(g(n))$.

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 $\alpha =-\mathrm{\infty }$ and $\beta$ is real, $u_j:=\beta -(j+1)$ and $v_j:=\beta$; if $\alpha$ is real and $\beta =+\mathrm{\infty }$, $u_j:=\alpha$ and $v_j:=\alpha +(j+1)$; if $\alpha =\beta =+\mathrm{\infty }$, $u_j:=j$ and $v_j:=j+1$; if $\alpha =\beta =-\mathrm{\infty }$, $u_j:=-(j+2)$ and $v_j:=-(j+1)$; and if $\alpha =-\mathrm{\infty }$, $\beta =+\mathrm{\infty }$, $u_j:=-(j+1)$ and $v_j:=j+1$. In every case $(u_j)$ tends to $\alpha$ and $(v_j)$ to $\beta$ in $\bar{\mathbb{R}}$, and both conditions of step 4.3 hold.

Hence ${\sum }_{i^{\mathrm{\prime }}<i}{a}_{p_{i^{\mathrm{\prime }}}}\to +\mathrm{\infty }$ as $i\to \mathrm{\infty }$ and ${\sum }_{l^{\mathrm{\prime }}<l}{a}_{q_{l^{\mathrm{\prime }}}}\to -\mathrm{\infty }$ as $l\to \mathrm{\infty }$: 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 ${\sum }_{i^{\mathrm{\prime }}=i_0}^{i-1}{a}_{p_{i^{\mathrm{\prime }}}}>M$, and for every $l_0$ and every real $M$ there is $l>l_0$ with ${\sum }_{l^{\mathrm{\prime }}=l_0}^{l-1}{a}_{q_{l^{\mathrm{\prime }}}}<M$; 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 $i_n\to \mathrm{\infty }$ and $l_n\to \mathrm{\infty }$, 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 $j_n\to \mathrm{\infty }$: 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 $u_j-\mathrm{\mid }{a}_{q_l}\mathrm{\mid }\le T_{{\alpha }_j}<u_j$ 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 ${\delta }_j:=\max \{a_{p_{i(j)}},\mathrm{\mid }{a}_{q_{l(j)}}\mathrm{\mid }\}$ for the two indices appearing in step 11.1 and step 11.2. As $j\to \mathrm{\infty }$ those indices tend to infinity, by step 8.1 and step 9.2, so $p_{i(j)}\to \mathrm{\infty }$ and $q_{l(j)}\to \mathrm{\infty }$ 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 $+\mathrm{\infty }$ according as $\sup \{T_{{\beta }_j}:j\ge J\}$ is, and taking the infimum over $n$, which drives $J$ to infinity, gives ${\mathrm{lim\hspace{0.17em}sup}}_n{T}_n={\mathrm{lim\hspace{0.17em}sup}}_j{T}_{{\beta }_j}$.

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 $\mathrm{\mid }{T}_n-c\mathrm{\mid }\le \max \{{\delta }_{j-1},{\delta }_j\}$. Given a real $\epsilon >0$, choose $J\ge 2$ with ${\delta }_j<\epsilon$ for all $j\ge J-1$; then $\mathrm{\mid }{T}_n-c\mathrm{\mid }<\epsilon$ 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 $T_n\to +\mathrm{\infty }$. 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 $T_n\to -\mathrm{\infty }$. This is claim 2.

By step 12.1 the subsequence $(T_{{\beta }_j}{)}_{j\ge 1}$ tends to $\beta$ and $(T_{{\alpha }_j}{)}_{j\ge 1}$ tends to $\alpha$, in $\bar{\mathbb{R}}$: 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], ${\mathrm{lim\hspace{0.17em}sup}}_j{T}_{{\beta }_j}=\beta$; so ${\mathrm{lim\hspace{0.17em}sup}}_n{T}_n=\beta$. 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 ${\mathrm{lim\hspace{0.17em}inf}}_n{T}_n={\mathrm{lim\hspace{0.17em}inf}}_j{T}_{{\alpha }_j}=\alpha$.

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 $\alpha <\beta$.