Rearranging the Alternating Harmonic Series

Why Rearrange a Conditionally Convergent Series?

The alternating harmonic series has positive and negative terms:

In its displayed order, this series converges. But it does not converge absolutely: the sum of the magnitudes of its terms is the harmonic series, which diverges. A conditionally convergent series can behave differently when its terms are reordered. Here we will construct, for any prescribed real number \(L\), a rearrangement whose partial sums converge to \(L\).

The construction uses a simple idea. Add positive terms until the partial sum is strictly above \(L\); then add negative terms until it is strictly below \(L\). Repeat, always using the next unused term of each sign. Since the terms become small, each crossing eventually has a small overshoot. Since the positive and negative terms each have infinite total magnitude, every crossing can be reached.

The Terms Available for the Construction

Write the positive terms as \(p_n=1/(2n-1)\) and the magnitudes of the negative terms as \(q_n=1/(2n)\), where \(n\in\mathbb{N}\). Thus the series consists of \(p_1,-q_1,p_2,-q_2,\ldots\). The positive terms, in their original relative order, are \(1,1/3,1/5,\ldots\); the negative terms are \(-1/2,-1/4,-1/6,\ldots\).

Both sign lists have divergent total magnitude. Indeed, \(1/(2n-1)\geq 1/(2n)\), so the sum of the positive terms diverges by comparison with half the harmonic series. Also, \(1/(2n)=\tfrac12(1/n)\), so the sum of the negative-term magnitudes diverges. Removing a finite number of terms from either list does not change this divergence.

A second fact is that individual terms become arbitrarily small: \(p_n\to0\) and \(q_n\to0\). These two properties play different roles. Divergence ensures that a crossing takes place after finitely many terms; the terms tending to zero ensures that the crossing overshoot tends to zero.

The Crossing Rule

Start with partial sum \(S=0\). If \(L\geq0\), add positive terms in order until the partial sum is strictly greater than \(L\). Then add negative terms in order until it is strictly less than \(L\). If \(L<0\), begin by adding negative terms until the sum is strictly less than \(L\), then add positive terms until it is strictly greater than \(L\). In either case, continue alternating these two kinds of blocks.

The word strictly matters. A block does not stop merely because its sum equals \(L\); it continues until it passes to the other side. Each block takes the next unused term of its sign, so the procedure neither skips nor repeats terms.

Worked Example: Target \(L=\frac14\)

Worked Example: Target \(L=0\)

Why Every Crossing Happens

Suppose a positive block has begun with partial sum at or below \(L\). If it never passed above \(L\), then adding all the remaining positive terms would keep every resulting sum at or below \(L\). But the sum of those unused positive terms diverges, so their cumulative addition eventually exceeds the finite gap to \(L\). This contradiction shows that the block crosses after finitely many terms.

The negative case is the same argument with signs reversed. Starting above \(L\), subtracting the magnitudes of the unused negative terms must eventually take the partial sum below \(L\), because those magnitudes have divergent sum. Thus each block is finite. After the initial block, every block starts on the side opposite the direction in which its terms move, so this reasoning applies at every stage.

Why the Rearranged Series Converges to \(L\)

To prove this, consider a positive block that starts with sum \(A\leq L\) and ends with sum \(B>L\). Let \(p\) be the final positive term added in that block. Immediately before adding \(p\), the sum was at or below \(L\). Therefore the crossing error satisfies

The inequality is not necessarily strict: if the sum immediately before \(p\) equals \(L\), then \(B-L=p\). Similarly, if a negative block starts at \(C\geq L\) and ends at \(D<L\), and its last term has magnitude \(q\), then

Each sign's terms are used in their original order. As the construction continues, the index of the last term in each later block tends to infinity: each block uses at least one new term of its sign, and there are infinitely many alternating blocks. Since \(p_n\to0\) and \(q_n\to0\), the last-term bounds show that the distances from the successive crossing sums to \(L\) tend to zero.

It remains to check partial sums between crossings, not just the crossing sums themselves. During a positive block, partial sums increase monotonically from the previous negative crossing to the next positive crossing. Every such partial sum lies between those two endpoint sums. During a negative block, partial sums decrease monotonically between the corresponding positive and negative crossing sums. If both endpoints are within \(\varepsilon\) of \(L\), every number between them is also within \(\varepsilon\) of \(L\). Since the crossing errors tend to zero, all sufficiently late partial sums are within any chosen \(\varepsilon>0\) of \(L\). This proves convergence to \(L\).

Finally, the construction really does use every term. Each block is finite and uses consecutive unused terms from its sign list. There are infinitely many blocks of each sign, and each uses at least one term. Hence the number of positive terms used and the number of negative terms used both increase without bound. Every term in both lists is eventually included, exactly once, so the result is a rearrangement.

Worked Example: Target \(L=1\)

A Common Mistake: Stopping at Equality

A tempting version of the rule says to stop as soon as a partial sum reaches or passes the target. That fails to guarantee that the block has crossed to the opposite side. For instance, in the \(L=1/4\) example,

This is equality, not a crossing below \(1/4\). Stopping here would leave the next block starting at the target, and the claimed strict crossing bound would not describe that block. The correction is to continue until the sum is strictly on the other side, as in \(1/4-1/6=1/12<1/4\). Then the correct general bound is “at most the final term's magnitude,” not “strictly less”: in this very step the error is \(1/6\), exactly the final term's magnitude.

Why the Result Matters

The original alternating harmonic series has a definite sum in its displayed order, but its terms are not absolutely summable. The construction shows how strongly order can matter in this conditional case: the same terms can be arranged to converge to any real target. The essential ingredients are the unbounded total available in each sign and the fact that individual terms shrink to zero. Neither feature alone would establish the result: divergent sign totals make each crossing possible, while shrinking terms make the successive overshoots vanish.