Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum

Finite domination. For every $n \in \mathbb{N}$ the following holds: for every sequence $(c_k)$ of nonnegative reals, every $Q \in \mathbb{N}$ and every injective map $\tau$ from $\{k : k < n\}$ into $\{k : k < Q\}$, one has $\sum_{k<n} c_{\tau(k)} \le \sum_{k<Q} c_k$. This is proved by induction on $n$, the sequence, $Q$ and $\tau$ being universally quantified inside the induction statement. At $n = 0$ the left side is the empty sum $0$ and the right side is nonnegative. Assume the statement at $n$, and let $\tau$ be injective from $\{k : k < n+1\}$ into $\{k : k < Q\}$; put $p := \tau(n)$, so $p < Q$, and let $(c'_k)$ agree with $(c_k)$ except that $c'_p := 0$, again a nonnegative sequence. The restriction of $\tau$ to $\{k : k < n\}$ is injective into $\{k : k < Q\}$ and never takes the value $p$, so $c'_{\tau(k)} = c_{\tau(k)}$ for $k < n$, and the induction hypothesis gives $\sum_{k<n} c_{\tau(k)} = \sum_{k<n} c'_{\tau(k)} \le \sum_{k<Q} c'_k$. Splitting the sum $\sum_{k<Q}$ at $p$ and at $p+1$ shows $\sum_{k<Q} c'_k = \sum_{k<Q} c_k - c_p$, so adding $c_p$ to both sides gives $\sum_{k<n+1} c_{\tau(k)} \le \sum_{k<Q} c_k$.

Bounding index. For every injective $\rho : \mathbb{N} \to \mathbb{N}$ and every $n \in \mathbb{N}$ there is $Q \in \mathbb{N}$ with $\rho(k) < Q$ for all $k < n$: at $n = 0$ take $Q = 0$, and if $Q$ works for $n$ then the greater of $Q$ and $\rho(n)+1$ works for $n+1$, the order on $\mathbb{N}$ being total.

Since $\sigma$ is a bijection, for every $j \in \mathbb{N}$ there is exactly one $k$ with $\sigma(k) = j$; write $\sigma^{-1}(j)$ for that $k$. Then $\sigma^{-1}$ is a bijection of $\mathbb{N}$, with $\sigma(\sigma^{-1}(j)) = j$ for every $j$.

By [L7] both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge; write $U$ and $V$ for their sums. Since $a_k = a_k^{+} - a_k^{-}$, linearity gives $\sum_{k=0}^{\infty} a_k = U - V$.

The positive and negative parts are defined pointwise from the value of the term, so the positive part of $a_{\sigma(k)}$ is $a_{\sigma(k)}^{+}$ and its negative part is $a_{\sigma(k)}^{-}$; both are nonnegative sequences in the index $k$.

The nonnegative case, one inequality. Let $(c_k)$ be a sequence of nonnegative reals with $\sum c_k$ convergent of sum $M$, and let $\rho$ be a bijection of $\mathbb{N}$. For each $n$ pick $Q$ as in step 1.2; then $\rho$ restricted to $\{k : k<n\}$ is injective into $\{k : k<Q\}$, so $\sum_{k<n} c_{\rho(k)} \le \sum_{k<Q} c_k \le M$ by step 1.1 and [L3]. The terms $c_{\rho(k)}$ are nonnegative, so the partial sums of $\sum c_{\rho(k)}$ are bounded above by $M$; hence that series converges, and since each partial sum is at most $M$ its sum is at most $M$.

The nonnegative case, equality. With $(c_k)$, $M$ and $\rho$ as in step 2.1, write $M'$ for the sum of $\sum c_{\rho(k)}$, so $M' \le M$. The sequence $(c_{\rho(k)})_k$ is nonnegative with convergent series of sum $M'$, and its rearrangement along the bijection $\rho^{-1}$ is $j \mapsto c_{\rho(\rho^{-1}(j))} = c_j$; so step 2.1, applied to that sequence and that bijection, gives $M \le M'$. Hence $M' = M$.

Applying step 3.1 to the nonnegative sequence $(|a_k|)$, whose series converges by hypothesis, and to $\sigma$: the series $\sum |a_{\sigma(k)}|$ converges with the same sum as $\sum |a_k|$, which is claim 1.

Applying step 3.1 to $(a_k^{+})$ and to $(a_k^{-})$, each with the bijection $\sigma$: the series $\sum a_{\sigma(k)}^{+}$ and $\sum a_{\sigma(k)}^{-}$ converge, with sums $U$ and $V$ respectively.

Since $a_{\sigma(k)} = a_{\sigma(k)}^{+} - a_{\sigma(k)}^{-}$ for every $k$, linearity gives that $\sum a_{\sigma(k)}$ converges with sum $U - V$, which by step 1.4 equals $\sum_{k=0}^{\infty} a_k$; this is claim 2.

The same conclusion is available from claim 1 alone: $\sum |a_{\sigma(k)}|$ converges, so $\sum a_{\sigma(k)}$ converges; step 5.1 is what identifies its sum.

Claims 1 and 2 hold for an arbitrary bijection $\sigma$, so $\sum a_k$ converges and every rearrangement of it converges to the same sum, that is, $\sum a_k$ converges unconditionally.