Why Rearranging Terms Requires a Proof
A series is the limit of its partial sums, so changing the order of its terms changes the sequence of partial sums. For a finite sum, order does not affect the result. For an infinite series, however, a rearrangement can produce different partial sums along the way. We need an argument that controls the effect of those changes.
For an absolutely convergent real series, that control comes from the tails of the series of absolute values. Those tails can be made as small as desired. This lets us show that, once a rearranged partial sum includes a sufficiently long initial block of the original series, the difference from the original sum is small.
Definitions
Let \(\sum_{n=1}^{\infty} a_n\) be a real series, with partial sums \(S_N=\sum_{n=1}^{N}a_n\). The series converges absolutely if the series of absolute values \(\sum_{n=1}^{\infty}|a_n|\) converges.
A rearrangement of the series is obtained by choosing a bijection \(\pi:\mathbb{N}\to\mathbb{N}\) and listing the terms in the order \(a_{\pi(1)},a_{\pi(2)},a_{\pi(3)},\ldots\). The requirement that \(\pi\) is a bijection means that every original term appears exactly once: no index is omitted or repeated.
The Tail-Bound Proof
Write \(R_M=\sum_{n=M+1}^{\infty}|a_n|\) for the tail of the absolute-value series after index \(M\). Absolute convergence implies \(R_M\to 0\) as \(M\to\infty\). We will compare a rearranged partial sum with the original sum by separating the terms up to \(M\) from the terms after \(M\).
Given \(\varepsilon>0\), choose \(M\) large enough that \(R_M<\varepsilon/2\). This is possible because the absolute-value series converges.
Each of the finitely many indices \(1,\ldots,M\) occurs somewhere in the rearrangement. Let \(K\) be the largest of their positions. Every rearranged partial sum with \(k\ge K\) therefore includes all terms \(a_1,\ldots,a_M\).
After the common initial block, the original sum contains all terms with indices greater than \(M\), while the rearranged partial sum contains a finite selection of them. The selected terms appear in both sums and cancel, so only the terms not yet selected remain, and their total absolute value is at most \(R_M\).
Here is the precise estimate. For \(k\ge K\), let \(E\) be the set of indices greater than \(M\) that occur among \(\pi(1),\ldots,\pi(k)\). The rearranged partial sum \(T_k=\sum_{j=1}^{k}a_{\pi(j)}\) consists of the common initial block and the terms indexed by \(E\). The original sum \(S\) consists of that same block and the entire tail. Since the tail is absolutely convergent, it can be split into the terms indexed by \(E\) and those indexed by its complement among indices greater than \(M\). The terms indexed by \(E\) occur in both \(S\) and \(T_k\) and cancel in the difference. Consequently,
Taking absolute values and applying the triangle inequality gives
The unselected indices are a subset of the indices greater than \(M\), which explains why their absolute-value sum is at most \(R_M\). Thus, for every \(\varepsilon>0\), all \(k\ge K\) satisfy \(|S-T_k|<\varepsilon\). This is exactly the definition that \(T_k\to S\). The rearranged series therefore converges to the original sum.
Worked Examples
Worked Example: Swapping Adjacent Terms of a Geometric Series
Consider \(\sum_{n=1}^{\infty}2^{-n}\), and rearrange it by swapping each adjacent pair. The index order is \(2,1,4,3,6,5,\ldots\), so the rearranged series begins
The original geometric series has first term \(1/2\) and common ratio \(1/2\), so its sum is \((1/2)/(1-1/2)=1\). For the first four rearranged terms, the indices used are \(2,1,4,3\). Their sum is
The omitted indices are \(5,6,7,\ldots\), not just the odd indices from \(5\) onward. Their terms sum to
Thus \(T_4+1/16=15/16+1/16=1\), as required. More generally, the first \(2m\) rearranged terms contain exactly the original indices \(1,\ldots,2m\), so their sum is \(1-2^{-2m}\), which tends to \(1\). This agrees with the theorem.
Worked Example: Rearranging an Alternating Geometric Series
Let \(a_n=(-1)^{n+1}2^{-n}\). The series of absolute values is geometric, with sum \(\sum_{n=1}^{\infty}2^{-n}=1\); hence the original series converges absolutely. Pairing its consecutive terms gives
Now use the permutation \(\pi(2j-1)=2j\) and \(\pi(2j)=2j-1\), which swaps each adjacent pair. The rearranged terms occur as \(-1/4,+1/2,-1/16,+1/8,\ldots\). Each complete pair has sum
Therefore the first \(2m\) rearranged terms sum to \(\sum_{j=1}^{m}4^{-j}=\frac13(1-4^{-m})\). This tends to \(1/3\). The next, odd-positioned partial sum adds the term \(a_{2m+2}=-2^{-(2m+2)}=-1/(4\cdot4^m)\), which tends to zero, so those partial sums also tend to \(1/3\). As a direct check, the first two rearranged terms sum to \(-1/4+1/2=1/4\), the first original pair. The limit agrees with the original sum, as the theorem requires.
Worked Example: Applying the Bound to a Specific Rearranged Partial Sum
Take \(a_n=3^{-n}\), whose sum is \((1/3)/(1-1/3)=1/2\). Consider the rearrangement beginning with the indices \(5,2,7,1,4,3,6,8,9,\ldots\); after these first seven indices, it lists \(8,9,\ldots\) in order. This is a bijection of the positive integers.
The first six rearranged terms use indices \(5,2,7,1,4,3\), so
The omitted indices are \(6,8,9,10,\ldots\). Their terms sum to
Indeed, \(1/2-T_6=2187/4374-2180/4374=7/4374\). The exact difference matches the omitted terms. The general proof also gives a bound using the first three original terms: the absolute tail after \(M=3\) is \(\sum_{n=4}^{\infty}3^{-n}=1/54\). The first six rearranged terms include indices \(1,2,3\), so the theorem's estimate gives \(|1/2-T_6|\le 2(1/54)=1/27\). The computed difference \(7/4374\) is less than \(1/27\), confirming the bound.
A Common Pitfall: Losing Track of an Index
When checking a rearranged partial sum, it is easy to omit an index accidentally and then report the wrong tail. For the adjacent-swap permutation above, the first four indices are \(2,1,4,3\). A mistaken calculation might claim that the omitted indices are \(5,7,8,\ldots\), and then write
This is incorrect because index \(6\) is omitted as well. The correct omitted indices are \(5,6,7,\ldots\), and their sum is
This also checks against the difference between the full sum and the partial sum: \(1-15/16=1/16\). A reliable check is to list the indices actually used before identifying the omitted ones.
What the Result Does—and Does Not—Say
The theorem is powerful because it applies to every permutation, even one that places very large indices early and leaves some small indices until much later. The proof does not require the rearrangement to preserve consecutive blocks or to follow any regular pattern. It requires only that every index appears exactly once and that the original series converges absolutely.
The tail bound is also why the hypothesis matters. Mere convergence of \(\sum a_n\) does not provide a small bound on \(\sum |a_n|\). Without that absolute-value control, the argument above cannot establish invariance. Whether a series without absolute convergence can change its sum under rearrangement is a separate question; it is not a consequence of this theorem.
Check Your Understanding
Use the definitions and tail-bound argument to answer these questions.
- What property of the index map makes a rearrangement include every original term exactly once?
- Why can one choose \(M\) so that \(\sum_{n=M+1}^{\infty}|a_n|<\varepsilon/2\)?
- For a fixed \(M\), why does some rearranged partial sum contain all terms \(a_1,\ldots,a_M\)?
- In the proof, why does the difference \(S-T_k\) involve only the unselected tail terms, and why are they controlled by \(R_M\)?
- For the adjacent-swap rearrangement of \(\sum 2^{-n}\), list the omitted indices after the first four terms and verify their sum.
