Rearranging Absolutely Convergent Series

Why Absolute Convergence Controls Rearrangement

A series is ordinarily summed in the order of its indices: first $a_1$, then $a_2$, and so on. A rearrangement changes that order without deleting or repeating any term. For a series with terms of both signs, changing the order can have important consequences. Absolute convergence provides a strong form of control: the total size of terms in a sufficiently late part of the original series is small, and every sufficiently late part of any rearrangement can be made to use only those small terms.

The previous tutorial, Series of Nonnegative Terms, established the Finite-Subset Characterization: for a convergent series of nonnegative terms, every finite-subset sum is bounded by the sum of the series. Applied to $|a_n|$, this lets us control finite collections of terms selected from a tail, even when those terms are not consecutive in their original order.

Definition: A permutation of the positive integers is a bijection $\pi:\mathbb{N}\to\mathbb{N}$. Given a series $\sum_{n=1}^{\infty}a_n$, the series $\sum_{k=1}^{\infty}a_{\pi(k)}$ is called a rearrangement of the original series. The bijection condition means that every original term appears exactly once.

The central question here is whether a rearrangement of an absolutely convergent series converges. We will prove that it does, and that it is itself absolutely convergent. The proof will not yet identify its sum with the sum of the original series; that requires a further comparison of partial sums.

Transferring an Original Tail Bound

Let $\sum_{n=1}^{\infty}|a_n|$ converge. Its remainders tend to zero, so for any $\varepsilon>0$ we can choose $N$ such that the original absolute tail after $N$ is less than $\varepsilon$:

\sum_{n=N+1}^{\infty}|a_n|<\varepsilon.

Only finitely many original indices are at most $N$. Since $\pi$ is a bijection, each of those indices occurs at some finite position in the rearrangement. Choose a position beyond all of them. After that position, every term in the rearrangement comes from an original index greater than $N$. The following lemma makes the resulting estimate precise.

Lemma (Tail Transfer for a Rearrangement): Suppose $\sum_{n=1}^{\infty}|a_n|$ converges and $\pi$ is a permutation of the positive integers. For every $\varepsilon>0$, there is an integer $K$ such that, for all integers $q\geq p>K$, $$ \sum_{k=p}^{q}|a_{\pi(k)}|<\varepsilon. $$

Proof. Choose $N$ so that $\sum_{n=N+1}^{\infty}|a_n|<\varepsilon$. The set of positions of the first $N$ original indices,

\{\pi^{-1}(1),\pi^{-1}(2),\ldots,\pi^{-1}(N)\},

is finite and nonempty. Let $K$ be its maximum. If $k>K$, then $\pi(k)>N$: otherwise $\pi(k)$ would be one of the indices $1,\ldots,N$, whose position is at most $K$. Thus, whenever $q\geq p>K$, all indices $\pi(p),\ldots,\pi(q)$ exceed $N$. They are distinct because $\pi$ is a bijection. By the Finite-Subset Characterization for the convergent nonnegative series $\sum |a_n|$,

0\leq\sum_{k=p}^{q}|a_{\pi(k)}| \leq\sum_{n=N+1}^{\infty}|a_n| <\varepsilon.

This proves the claim. $\square$

The choice of $K$ depends on the positions of the first $N$ original indices. It need not be close to $N$: a permutation can move one of those indices very far out. What matters is that there are only finitely many such indices, so all of them eventually occur.

Every Rearrangement Converges Absolutely

The Tail Transfer Lemma gives the Cauchy condition for the series of reordered absolute values. Once that series converges, the earlier theorem Absolute Convergence Implies Convergence applies to the rearranged series itself.

Theorem (Absolute Convergence Is Preserved by Rearrangement): If $\sum_{n=1}^{\infty}|a_n|$ converges and $\pi$ is a permutation of the positive integers, then $$ \sum_{k=1}^{\infty}|a_{\pi(k)}| $$ converges. Consequently, the rearranged series $\sum_{k=1}^{\infty}a_{\pi(k)}$ converges absolutely.

Proof. Fix $\varepsilon>0$. By the Tail Transfer Lemma, there is an integer $K$ such that, whenever $q\geq p>K$,

\left|\sum_{k=p}^{q}|a_{\pi(k)}|\right| =\sum_{k=p}^{q}|a_{\pi(k)}| <\varepsilon.

The equality holds because every term in this sum is nonnegative. The Cauchy Criterion for Series therefore proves that $\sum_{k=1}^{\infty}|a_{\pi(k)}|$ converges. By definition, this means that $\sum_{k=1}^{\infty}a_{\pi(k)}$ converges absolutely. The theorem Absolute Convergence Implies Convergence also gives its convergence as a series of real terms. $\square$

The argument relies on two separate facts. Absolute convergence makes the original tail small. Bijectivity ensures that the rearranged series eventually leaves behind every index in any fixed finite initial segment. Neither fact alone would give the conclusion.

Worked Applications

Worked Example: Swapping Adjacent Terms of a Geometric Series

Consider $a_n=2^{-n}$, and let $\pi$ swap each adjacent pair:

\pi(2j-1)=2j,\qquad \pi(2j)=2j-1

for every positive integer $j$. Each positive integer belongs to exactly one pair, and each pair is swapped, so $\pi$ is a permutation. The rearranged terms begin

a_{\pi(1)},a_{\pi(2)},a_{\pi(3)},a_{\pi(4)},\ldots =\frac14,\frac12,\frac1{16},\frac18,\ldots.

The original absolute series is $\sum_{n=1}^{\infty}2^{-n}$, which converges. To see the tail transfer explicitly, choose $N$ so that

\sum_{n=N+1}^{\infty}2^{-n}=2^{-N}<\varepsilon.

Every original index $n\leq N$ appears in the rearrangement by position $N+1$: an odd index $n$ appears at position $n+1$, while an even index appears at position $n-1$. Thus, after position $K=N+1$, all reordered indices exceed $N$. For $q\geq p>K$,

\sum_{k=p}^{q}|a_{\pi(k)}| \leq\sum_{n=N+1}^{\infty}2^{-n} =2^{-N}<\varepsilon.

This verifies directly that the reordered absolute tails become arbitrarily small.

Worked Example: A Sparse Set of Swaps in a P-Series

Let $a_n=1/n^2$. Define $\pi$ by swapping the pairs $(2^j,2^j+1)$ for each positive integer $j$, and fixing every index not in one of these pairs. The pairs are disjoint: the pair beginning at $2^j$ ends at $2^j+1$, which is less than $2^{j+1}$, the beginning of the next pair. Hence this rule defines a permutation.

The $p$-Series Convergence Criterion shows that $\sum 1/n^2$ converges. For a fixed $N$, every index $n\leq N$ is either fixed or moved by one position. Its position in the rearrangement is therefore at most $N+1$. Consequently, after position $K=N+1$, the rearranged terms have original indices greater than $N$, and

\sum_{k=p}^{q}|a_{\pi(k)}| \leq\sum_{n=N+1}^{\infty}\frac{1}{n^2} \qquad(q\geq p>N+1).

For $N\geq1$, the tail on the right can also be bounded explicitly. For $n\geq N+1$,

\frac{1}{n^2}\leq\frac{1}{n(n-1)} =\frac{1}{n-1}-\frac{1}{n}.

Thus the comparison series telescopes, giving

\sum_{n=N+1}^{\infty}\frac{1}{n^2} \leq\sum_{n=N+1}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right) =\frac{1}{N}.

Choosing $N$ large enough that $1/N<\varepsilon$ gives a concrete tail bound for every sufficiently late block of this rearrangement.

Worked Example: Why Omitting Terms Is Not Rearranging

For $a_n=2^{-n}$, keep only the even-indexed terms. The resulting series is

\sum_{k=1}^{\infty}a_{2k} =\sum_{k=1}^{\infty}4^{-k}.

This series converges, but it is not a rearrangement of $\sum 2^{-n}$: it omits every odd-indexed term. Its partial sums are

\sum_{k=1}^{M}4^{-k} =\frac{1}{3}(1-4^{-M}),

so its sum is $1/3$. The original geometric series has sum $1$. This example shows why the definition requires a bijection: a one-to-one selection of terms that omits some indices is a subseries, not a rearrangement. The theorem above concerns permutations and does not claim that arbitrary subseries preserve the original sum.

What the Tail Estimate Does—and Does Not—Say

The Tail Transfer Lemma controls blocks of reordered absolute values. In particular, it establishes convergence of the rearranged absolute series and hence convergence of the rearranged signed series. It does not, by itself, show that the rearranged series has the same sum as the original one. The Cauchy Criterion proves that each series has a limit, but a separate argument is needed to compare those limits.

A common mistake is to assume that a late block in the rearrangement corresponds to a consecutive block in the original order. It generally does not. The proof avoids that assumption: it first waits until every original index from $1$ through $N$ has appeared, and then bounds the remaining finite selection by the full original absolute tail. The finite-subset control from the previous tutorial is what makes this possible.

Takeaway: In an absolutely convergent series, choose a small original absolute tail and then wait until the rearrangement has placed every index before that tail. All subsequent rearranged blocks are bounded by the chosen tail, so every rearrangement converges absolutely.

Check Your Understanding

Use the definition of a permutation and the tail estimate to answer the following questions.

Why must a rearrangement use a bijection rather than merely an injective map?
Given $N$, how is a position $K$ chosen so that every rearranged term after $K$ has original index greater than $N$?
Where does the Finite-Subset Characterization enter the Tail Transfer Lemma?
Why does the Cauchy Criterion apply to the rearranged series of absolute values?
What additional issue remains after proving that a rearrangement converges absolutely?