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 $(c_k^{\mathrm{\prime }})$ agree with $(c_k)$ except that $c_p^{\mathrm{\prime }}:=0$, 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 $c_{\tau (k)}^{\mathrm{\prime }}=c_{\tau (k)}$ for $k<n$, and the induction hypothesis gives ${\sum }_{k<n}{c}_{\tau (k)}={\sum }_{k<n}{c}_{\tau (k)}^{\mathrm{\prime }}\le {\sum }_{k<Q}{c}_k^{\mathrm{\prime }}$. Splitting the sum ${\sum }_{k<Q}$ at $p$ and at $p+1$ shows ${\sum }_{k<Q}{c}_k^{\mathrm{\prime }}={\sum }_{k<Q}{c}_k-c_p$, 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 ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k=U-V$.

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 $M^{\mathrm{\prime }}$ for the sum of $\sum c_{\rho (k)}$, so $M^{\mathrm{\prime }}\le M$. The sequence $(c_{\rho (k)}{)}_k$ is nonnegative with convergent series of sum $M^{\mathrm{\prime }}$, 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 $M\le M^{\mathrm{\prime }}$. Hence $M^{\mathrm{\prime }}=M$.

Applying step 3.1 to the nonnegative sequence $(\mathrm{\mid }{a}_k\mathrm{\mid })$, whose series converges by hypothesis, and to $\sigma$: the series $\sum \mathrm{\mid }{a}_{\sigma (k)}\mathrm{\mid }$ converges with the same sum as $\sum \mathrm{\mid }{a}_k\mathrm{\mid }$, 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 ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k$; this is claim 2.

The same conclusion is available from claim 1 alone: $\sum \mathrm{\mid }{a}_{\sigma (k)}\mathrm{\mid }$ 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.