Series rearrangement, grouping and Cauchy products

Learning goals and definitions

Separate a finite reordering from an infinite rearrangement, prove preservation under absolute convergence, and construct a target-sum rearrangement of the alternating harmonic series. We also examine grouping and a product whose terms fail to tend to zero. Prerequisites are the Cauchy criterion, absolute and alternating convergence, and basic finite-sum algebra.

A rearrangement of $\sum_{n\ge1}a_n$ is $\sum_{n\ge1}a_{\sigma(n)}$ for a permutation $\sigma:\mathbb N_{\ge1}\to\mathbb N_{\ge1}$. Every original index must occur exactly once at a finite new position. “Take all positive terms first, then all negative terms” is not such a permutation when both collections are infinite: no finite position finishes the first collection and begins the second.

Grouping instead replaces successive finite blocks by their sums while preserving their original order. It does not reorder individual indices, although it examines only the block-endpoint subsequence of partial sums. These distinctions matter when passing from finite identities to infinite limits.

1. Absolute convergence makes every permutation safe

Theorem

If $\sum a_n=s$ and $\sum|a_n|<\infty$, then every rearrangement converges to the same $s$, and it also converges absolutely.

Proof. Given $\varepsilon>0$, choose $N$ with $\sum_{n>N}|a_n|<\varepsilon/2$. Since the permutation eventually includes each of the finitely many indices $1,\ldots,N$, there is $M_0$ such that all these indices appear within its first $M_0$ places. For $M\ge M_0$, subtract the original first $N$ terms from the rearranged first $M$ terms. What remains is a finite selection of original tail terms, whose absolute sum is below $\varepsilon/2$. The original remainder $s-\sum_{n=1}^Na_n$ has magnitude below $\varepsilon/2$ as well. The triangle inequality therefore makes the rearranged partial sum within $\varepsilon$ of $s$. Applying the same reasoning to the nonnegative absolutely summable terms shows the rearranged absolute sum equals the original one. Nothing in this proof assumes that a permutation has a uniformly bounded displacement.

Absolute convergence is a sufficient condition for every permutation. It is not a necessary condition for every particular permutation to preserve a sum; even a conditionally convergent series is unchanged by swapping only its first two terms.

2. Conditional harmonic terms can be steered to any real target

The alternating harmonic series has positive terms $p_j=1/(2j-1)$ and negative magnitudes $q_j=1/(2j)$. Both $\sum p_j$ and $\sum q_j$ diverge to infinity: the second is half the harmonic series and $p_j\ge q_j$ makes the first diverge as well. Both individual sequences tend to zero.

Fix any real target $T$. Starting from zero, take unused positive terms in their original positive order until the running sum is strictly above $T$; if already above it, begin with the negative stage. Then take unused negative terms in their original negative order until the sum is strictly below $T$. Continue alternating these stages. If the sum exactly equals $T$ during a stage, continue that stage until it strictly crosses.

Each stage ends after finitely many terms because the remaining positive or negative total is unbounded. After the initial stage, every stage uses at least one new term, so the counts used from both queues tend to infinity. Thus every original term eventually appears, exactly once: the construction defines a genuine permutation.

At an upward crossing, the overshoot above $T$ is at most the last added positive term, since the preceding partial sum was at or below $T$. At a downward crossing, the undershoot is at most the last subtracted negative magnitude. Those last terms tend to zero because both queues advance indefinitely. Hence all crossing sums tend to $T$. Every intermediate partial sum lies between its adjacent crossing sums, since the terms within one stage all have the same sign. Therefore the entire rearranged partial-sum sequence tends to $T$.

This is a complete target-sum construction for this conditional series, not merely a claim that “infinity minus infinity is arbitrary.” Every term has a finite assigned position, and the convergence of all partial sums, not just the crossings, has been proved.

3. Grouping is safe in only one direction automatically

If a series already converges to $s$, grouping consecutive finite blocks gives partial sums that are a subsequence of the original partial sums, so the grouped series also converges to $s$. The converse fails: $1-1+1-1+\cdots$ grouped into successive pairs has every block zero, while the original partial sums alternate between one and zero and do not converge.

A grouped calculation may therefore confirm an already established convergent value, but it cannot establish original convergence without controlling the partial sums inside each block. The target-sum construction above supplied that missing control by trapping them between crossing sums. Simply writing parenthesized infinite expressions would not have done so.

4. Absolutely convergent Cauchy products

For series indexed from zero, the Cauchy product has coefficients $c_n=\sum_{k=0}^na_kb_{n-k}$. If both original series converge absolutely to $A,B$, then the product series converges absolutely to $AB$.

Proof of absolute convergence. Let $\alpha=\sum|a_k|$ and $\beta=\sum|b_k|$. Every finite triangular sum of $|a_kb_j|$ is bounded by a finite rectangular product, hence by $\alpha\beta$. The partial sums of $\sum|c_n|$ are nondecreasing and bounded by those triangular absolute sums, so they converge.

Proof of its value. Let $A_m=\sum_{k=0}^ma_k$, $B_m=\sum_{j=0}^mb_j$, and $S_N=\sum_{k+j\le N}a_kb_j$. When $N\ge2m$, the triangle includes the square $0\le k,j\le m$. Every additional term has $k>m$ or $j>m$, so

$$|S_N-A_mB_m|\le\left(\sum_{k>m}|a_k|\right)\beta+\alpha\left(\sum_{j>m}|b_j|\right).$$

Choose $m$ large enough to make both tails small and $A_mB_m$ close to $AB$; the estimate then holds for all $N\ge2m$. Thus $S_N\to AB$. This uses explicit finite squares, triangles and tail estimates instead of an unjustified interchange of two infinite sums.

5. A conditional product can fail the term test

Let $a_n=b_n=(-1)^n/\sqrt{n+1}$ for $n\ge0$. Each original series converges by the alternating test but not absolutely, since $1/\sqrt{n+1}\ge1/(n+1)$ and the harmonic series diverges. Its Cauchy-product coefficient is

$$c_n=(-1)^n\sum_{k=0}^n\frac1{\sqrt{(k+1)(n-k+1)}}.$$

For every $k$, the two positive factors inside the square root sum to $n+2$, so their product is at most $((n+2)/2)^2$. Each reciprocal is therefore at least $2/(n+2)$, giving $|c_n|\ge2(n+1)/(n+2)$. These coefficients do not tend to zero, so the product series diverges. Finite distributive algebra remains true, but the particular infinite triangular limit cannot be justified by convergence of its two conditional factors alone.

Graded proof and error-diagnosis tasks

1. Absolute tail argument. In the rearrangement theorem, why can one threshold ensure that every original index up to $N$ has appeared? Explain why infinitely many indices cannot be handled by the same finite maximum argument.
2. Algorithm validity. Prove that the target-sum procedure uses every positive and every negative harmonic term, and explain why “all positives first” fails the permutation definition.
3. Grouping. Give an explicit grouped series that converges while its ungrouped series diverges. Identify which implication remains universally valid for already convergent series.
4. Product diagnosis. Verify the lower bound on $|c_n|$ in the square-root example and state the exact convergence test it violates.
5. Finite changes. Prove that altering the order of finitely many terms of any convergent series preserves its sum, even without absolute convergence.
6. Challenge. Let $\sigma$ be a permutation with $|\sigma(n)-n|\le K$ for one fixed integer $K$. Prove it preserves the sum of every convergent real series, using only that its terms tend to zero and comparing prefix sets.

Full solutions

1. Each index $j\le N$ has one finite preimage position $\sigma^{-1}(j)$. The maximum of these finitely many positions is finite and works for all of them. There is no finite maximum of all natural positions, so the argument handles a finite core while a small absolute tail handles the infinitely many remaining terms. Omitting that tail estimate would leave the proof incomplete.
2. Each stage terminates by divergence of the appropriate remaining nonnegative queue. After an upward crossing, a negative term is required to cross downward, and conversely, so infinitely many stages use infinitely many terms of both queues. Since each queue is always consumed in its own order, any fixed term is eventually reached. An ordering with all infinitely many positives before the first negative gives no finite new index for any negative term, so it is not a permutation of the original indices.
3. Pair the consecutive terms of $1-1+1-1+\cdots$ into $(1-1)+(1-1)+\cdots$. The grouped partial sums are zero, but the original partial sums alternate between one and zero. For any already convergent series, grouping consecutive finite blocks preserves its sum because the block endpoints form a subsequence of a convergent sequence.
4. The product $(k+1)(n-k+1)$ is at most $((n+2)/2)^2$ by the square $(k+1-(n-k+1))^2\ge0$. Taking positive square roots and reciprocals makes each summand at least $2/(n+2)$. There are $n+1$ terms, giving the bound. Since this lower bound tends to two rather than zero, the necessary zero-term condition for convergence fails.
5. Choose a finite index threshold containing every changed index. Every sufficiently long partial sum then contains exactly the same finitely reordered terms and the same unchanged tail as the original partial sum, hence is literally equal to it. Their limits agree. This is finite commutativity plus eventual equality, not an arbitrary infinite rearrangement theorem.
6. The first $N$ permuted indices lie in $\{1,\ldots,N+K\}$. Every index at most $N-K$ must already be present, because if $\sigma(j)=i\le N-K$, then $j\le i+K\le N$. Thus the symmetric difference of the original and permuted prefix sets lies in the band from $N-K+1$ through $N+K$, with at most $2K$ entries. The difference of partial sums has magnitude at most $2K$ times the largest $|a_i|$ in that band, tending to zero since $a_i\to0$. The original sums converge, so the permuted ones have the same limit. For $K=0$ the prefixes are identical directly.

Summary and terms

A rearrangement is a permutation of indices; grouping combines consecutive finite blocks. Absolute convergence controls every reordered tail without relying on cancellation. Conditional convergence can allow a target-sum construction, but that construction must still place every term at a finite position. A Cauchy product sums finite diagonal convolutions and needs justified limiting control, not only a formal distributive expansion.