Proof of Rearrangement Invariance

From Rearranged Convergence to the Same Sum

In Rearranging Absolutely Convergent Series, we proved that every rearrangement of an absolutely convergent series is itself absolutely convergent. That establishes that the reordered series has a sum, but it does not yet show that this sum equals the original one. To compare the sums, we need to control the rearranged partial sums themselves.

The key idea is to choose a finite initial segment of the original series whose absolute tail is small. A permutation eventually places every index in that initial segment somewhere in the rearrangement. After that point, the rearranged partial sum contains all those original terms; any additional terms come from the small absolute tail. This lets us compare the two sums without assuming that a rearranged block is consecutive in the original order.

Definition: Let $\sum_{n=1}^{\infty}a_n$ converge to $S$, and let $\pi$ be a permutation of the positive integers. The series $\sum_{k=1}^{\infty}a_{\pi(k)}$ is a rearrangement of the original series. Rearrangement invariance means that this rearranged series converges to the same sum $S$.

Capturing a Finite Set of Indices

A permutation may move an original index far from its usual position, but each individual index still appears at a finite position: the position of $n$ is $\pi^{-1}(n)$. For any fixed finite collection of original indices, there is therefore a single position by which all of them have appeared.

Lemma (Eventual Capture of a Finite Initial Segment): Let $\pi$ be a permutation of the positive integers. For every positive integer $N$, there is an integer $K$ such that $$ \{1,2,\ldots,N\}\subseteq\{\pi(1),\pi(2),\ldots,\pi(K)\}. $$

Proof. Each of the indices $1,\ldots,N$ occurs at the finite position $\pi^{-1}(n)$. Set

K=\max\{\pi^{-1}(1),\pi^{-1}(2),\ldots,\pi^{-1}(N)\}.

This maximum exists because it is the maximum of a finite, nonempty set of positive integers. For every $n\leq N$, we have $\pi^{-1}(n)\leq K$, so $n=\pi(\pi^{-1}(n))$ is among $\pi(1),\ldots,\pi(K)$. Thus all indices from $1$ through $N$ have appeared by position $K$. $\square$

There may be other indices among the first $K$ positions as well. The lemma does not require the rearrangement to place the original terms in their usual order. It only guarantees that no index in the chosen finite initial segment is left out.

The Sum Is Invariant

We now compare partial sums directly. Write

S=\sum_{n=1}^{\infty}a_n, \qquad T_M=\sum_{k=1}^{M}a_{\pi(k)}.

For a chosen $N$, once $M$ is at least the position $K$ from the lemma, all terms $a_1,\ldots,a_N$ occur in $T_M$. Every other term in that partial sum has an original index greater than $N$. Thus the difference between $T_M$ and the sum of the first $N$ original terms is a finite sum selected from the original absolute tail. The Finite-Subset Characterization for convergent nonnegative series, established in Series of Nonnegative Terms, bounds that selection by the full absolute tail.

Theorem (Rearrangement Invariance for Absolutely Convergent Series): Suppose $\sum_{n=1}^{\infty}|a_n|$ converges to a finite value, and let $\pi$ be a permutation of the positive integers. Then $$ \sum_{k=1}^{\infty}a_{\pi(k)}=\sum_{n=1}^{\infty}a_n. $$

Proof. Let $S=\sum_{n=1}^{\infty}a_n$, and fix $\varepsilon>0$. Since the series $\sum |a_n|$ converges, its tails tend to zero. Choose $N\geq1$ such that

\sum_{n=N+1}^{\infty}|a_n|<\frac{\varepsilon}{2}.

By the Eventual Capture Lemma, there is a $K$ such that every original index from $1$ through $N$ occurs among $\pi(1),\ldots,\pi(K)$. Take any $M\geq K$, and let

A_M=\{\pi(k):1\leq k\leq M,\ \pi(k)>N\}.

Because $\pi$ is a permutation, its values at distinct positions are distinct. Hence $A_M$ is a finite subset of $\{N+1,N+2,\ldots\}$. The first $M$ rearranged terms include all terms with indices at most $N$, together with precisely the terms whose indices belong to $A_M$. Therefore

T_M=\sum_{n=1}^{N}a_n+\sum_{n\in A_M}a_n.

Meanwhile, convergence of the original series gives

S=\sum_{n=1}^{N}a_n+\sum_{n=N+1}^{\infty}a_n.

Subtracting these expressions and applying the triangle inequality yields

|T_M-S| \leq \left|\sum_{n\in A_M}a_n\right| + \left|\sum_{n=N+1}^{\infty}a_n\right|.

The Finite-Subset Characterization applied to the nonnegative series $\sum_{n=N+1}^{\infty}|a_n|$ gives

\left|\sum_{n\in A_M}a_n\right| \leq\sum_{n\in A_M}|a_n| \leq\sum_{n=N+1}^{\infty}|a_n| <\frac{\varepsilon}{2}.

Also, by the triangle inequality for the convergent series,

\left|\sum_{n=N+1}^{\infty}a_n\right| \leq\sum_{n=N+1}^{\infty}|a_n| <\frac{\varepsilon}{2}.

Combining the bounds shows that $|T_M-S|<\varepsilon$ for every $M\geq K$. Thus $T_M\to S$. By definition, the rearranged series converges to $S$, as claimed. $\square$

The proof uses two separate controls. Absolute convergence makes the original absolute tail small, and bijectivity ensures that the rearrangement eventually includes every index in the chosen finite initial segment. The finite-subset bound then controls any extra terms drawn from that tail. No assumption about the order in which those extra terms occur is needed.

Worked Applications

Worked Example: Cycling Terms in a Geometric Series

Let $a_n=(-1)^{n-1}4^{-n}$. Its absolute series is geometric:

\sum_{n=1}^{\infty}|a_n| =\sum_{n=1}^{\infty}4^{-n} =\frac{1/4}{1-1/4} =\frac{1}{3}.

Define a permutation by cycling each consecutive block of three indices:

\pi(3j-2)=3j-1,\qquad \pi(3j-1)=3j,\qquad \pi(3j)=3j-2

for every $j\geq1$. Within each block, this rule uses each of the three indices exactly once, and the blocks partition the positive integers. It therefore defines a permutation. The first six reordered indices are $2,3,1,5,6,4$, so the first six terms are

-\frac{1}{16},\quad \frac{1}{64},\quad \frac{1}{4},\quad \frac{1}{1024},\quad -\frac{1}{4096},\quad -\frac{1}{256}.

The original series has sum

\sum_{n=1}^{\infty}(-1)^{n-1}4^{-n} =\sum_{n=1}^{\infty}\left(-\frac14\right)^{n-1}\frac14 =\frac{1/4}{1-(-1/4)} =\frac{1}{5}.

The series is absolutely convergent, so the theorem proves that the series in the cycled order also sums to $1/5$. This conclusion does not depend on the individual partial sums of the rearrangement matching the original partial sums at corresponding positions.

Worked Example: Reversing Dyadic Blocks of a P-Series

Consider $a_n=1/n^2$. The $p$-Series Convergence Criterion gives convergence of $\sum 1/n^2$, so its sum is invariant under every permutation. Partition the positive integers into the blocks

B_j=\{2^j,2^j+1,\ldots,2^{j+1}-1\},\qquad j\geq0.

Reverse the order within each block. Explicitly, for $2^j\leq k<2^{j+1}$, set

\pi(k)=3\cdot2^j-1-k.

At the first index of the block this gives $\pi(2^j)=2^{j+1}-1$, and at the last index it gives $\pi(2^{j+1}-1)=2^j$. The formula reverses each finite block, so it maps each block bijectively to itself; as the blocks partition the positive integers, $\pi$ is a permutation. The reordered indices begin

1,\quad 3,2,\quad 7,6,5,4,\quad 15,14,13,12,11,10,9,8,\ldots.

For example, the first seven terms of the rearrangement sum to

1+\frac{1}{3^2}+\frac{1}{2^2} +\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}+\frac{1}{4^2}.

Although these partial sums are not the original partial sums, the theorem shows that they converge to the same sum as $\sum 1/n^2$. Reversing a block changes the order locally, while the proof of invariance handles the full infinite rearrangement by controlling its absolute tail.

Worked Example: Moving Terms a Long Distance

Let $a_n=1/(n(n+1))$. The identity

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

shows by telescoping that the partial sum through $N$ is

\sum_{n=1}^{N}\frac{1}{n(n+1)} =1-\frac{1}{N+1}.

Consequently, the series converges to $1$. Its terms are positive, so it is absolutely convergent as well. For each $j\geq1$, swap the two indices $2^{2j}$ and $2^{2j+1}$, and fix every other index. These pairs are disjoint: the pair for $j$ ends at $2^{2j+1}$, while the next pair begins at $2^{2j+2}$. Thus these swaps define a permutation.

The first such swap exchanges indices $4$ and $8$. In the rearrangement, the term $a_8=1/(8\cdot9)=1/72$ appears at position $4$, while $a_4=1/(4\cdot5)=1/20$ appears at position $8$. Later swaps move terms across still larger distances. Nevertheless, rearrangement invariance proves that the reordered series converges to $1$. The size of the displacement of individual terms is not itself the control; the small absolute tail is.

Why Absolute Convergence Is Essential Here

The proof compares a rearranged partial sum with the original sum by bounding selected terms from the tail using their absolute values. This is precisely where absolute convergence matters. If the absolute tail does not become small, the estimates in the proof do not provide control over the terms gathered into a rearranged partial sum. The argument therefore cannot be applied to a conditionally convergent series.

A related pitfall is to say that, after a finite initial segment has appeared, every later partial sum contains only terms from the tail, and conclude that the error is at most one tail. In general, the difference between a rearranged partial sum and the original sum has two contributions: terms selected from the tail and the original tail itself. The proof above bounds each contribution by the absolute tail, giving a bound of twice that tail. This is already enough, because the tail can be chosen smaller than $\varepsilon/2$.

The theorem completes the result from Rearranging Absolutely Convergent Series: every rearrangement of an absolutely convergent series converges absolutely and has the original sum. The next topic, the Riemann Rearrangement Theorem, concerns what can happen when a convergent series is not absolutely convergent.

Takeaway: Choose an original finite initial segment with a small absolute tail, then wait until the rearrangement has included every index in that segment. The remaining discrepancy is bounded by the selected part of the absolute tail and the original absolute tail, so the rearranged partial sums approach the original sum.

Check Your Understanding

Use the finite-segment capture argument and the rearrangement-invariance proof to answer the following questions.

Why is the maximum of the positions $\pi^{-1}(1),\ldots,\pi^{-1}(N)$ finite?
Once every index from $1$ through $N$ has appeared, what kinds of terms may still occur in a rearranged partial sum?
Where does the Finite-Subset Characterization enter the proof of rearrangement invariance?
Why is it enough to choose the original absolute tail smaller than $\varepsilon/2$?
Which step of the proof would no longer be justified for a conditionally convergent series?