Why Conditional Series Depend on Order

Why the Order Can Matter

A convergent series is defined by the limit of its partial sums in a specified order. If the series is conditionally convergent, changing that order can change the limit. The reason is that the positive terms have an infinite total, and so do the magnitudes of the negative terms. By choosing how much of each to take at successive stages, we can control the partial sums.

The result is striking: for a conditionally convergent real series, there is a rearrangement that converges to any real number we choose. The proof gives a method for constructing one, rather than merely asserting that such a rearrangement exists.

Definition: A real series \(\sum_{n=1}^{\infty} a_n\) is conditionally convergent if it converges to a finite real number, but its series of absolute values \(\sum_{n=1}^{\infty}|a_n|\) diverges.

Positive and Negative Terms Have Infinite Totals

For a convergent series, \(a_n\to 0\). Separate its nonzero terms into the positive terms and the negative terms, keeping the order in which each occurs. Write the positive terms as \(p_1,p_2,\ldots\), where \(p_j>0\), and write the negative terms as \(-q_1,-q_2,\ldots\), where \(q_j>0\). Zero terms contribute nothing and can be handled separately.

Both lists must be infinite, and their sums must each diverge. To see why, let \(P_N\) be the sum of the positive terms among \(a_1,\ldots,a_N\), and \(Q_N\) the sum of the magnitudes of the negative terms among them. The \(N\)th partial sum is \(P_N-Q_N\), while the sum of the absolute values through \(N\) is \(P_N+Q_N\).

Key fact: If \(\sum a_n\) is conditionally convergent, then \(\sum_j p_j=+\infty\) and \(\sum_j q_j=+\infty\).

Here is the argument. Since the original series converges, \(P_N-Q_N\) has a finite limit. If \(P_N\) were bounded, its nondecreasing partial sums would converge to a finite value. Then \(Q_N=P_N-(P_N-Q_N)\) would also be bounded and would converge to a finite value. That would make \(P_N+Q_N\) bounded, contradicting divergence of \(\sum |a_n|\). Thus \(P_N\) is unbounded. The same reasoning with \(P_N=Q_N+(P_N-Q_N)\) shows that \(Q_N\) is unbounded. Because both are nondecreasing, each tends to \(+\infty\).

Worked Example: A Concrete Conditionally Convergent Series

Worked Example: Verify the supply of positive and negative terms

For \(k\ge1\), define \(a_{2k-1}=1/k\) and \(a_{2k}=-(1/k+1/k^2)\). We first check convergence in the stated order. Each pair sums to

a_{2k-1}+a_{2k} =\frac1k-\left(\frac1k+\frac1{k^2}\right) =-\frac1{k^2}.

Therefore the even partial sums are \(-\sum_{k=1}^{m}1/k^2\), which converge to a finite limit because the \(p\)-series with exponent \(2\) converges. The odd partial sums differ from the preceding even partial sums by \(1/(m+1)\), and \(1/(m+1)\to0\); they have the same limit. So the original series converges.

Its positive terms sum to \(\sum_{k=1}^{\infty}1/k\), which diverges. The magnitudes of its negative terms sum to \(\sum_{k=1}^{\infty}(1/k+1/k^2)\), which diverges since it is at least the divergent harmonic series. Thus this series is conditionally convergent, with both signs available in unlimited total amount.

The Riemann Rearrangement Theorem

Theorem: If a real series is conditionally convergent, then for every \(L\in\mathbb R\), its terms can be rearranged so that the rearranged series converges to \(L\).

The construction alternates blocks of positive and negative terms. Add positive terms, in their original relative order, until the running sum is greater than \(L\). Then add negative terms, in their original relative order, until the running sum is less than \(L\). Repeat these two stages. If \(L\ge0\), begin with a positive block; if \(L<0\), begin with a negative block. This ensures that the first block is not skipped.

Choose the first sign.
If \(L\ge0\), add positive terms until the sum exceeds \(L\). If \(L<0\), add negative terms until it falls below \(L\).

Switch signs at each crossing.
After a positive block, add negative terms until the sum is less than \(L\). After a negative block, add positive terms until it is greater than \(L\).

Continue in the original order within each sign.
This uses every positive and negative term in a definite order while controlling each crossing of \(L\).

Why the Construction Works

Each block ends after finitely many terms. For example, from any starting sum, adding the positive terms eventually raises the sum above \(L\), because their total is infinite. The negative terms have infinite total magnitude, so a negative block likewise eventually takes the sum below \(L\). Every block contains at least one term. Since there are infinitely many alternating blocks, each successive term in either sign-list is eventually used. Thus the construction includes every nonzero term exactly once.

Now consider the size of the overshoot. Suppose a positive block begins at \(S\le L\) and ends at \(S'>L\). Immediately before its final term \(p_j\) is added, the running sum is at most \(L\). Consequently, \(S'-L\le p_j\). For a negative block beginning at \(T\ge L\) and ending at \(T'<L\), the same reasoning gives \(L-T'\le q_j\), where \(q_j\) is the magnitude of its final term.

The terms of the original convergent series tend to zero. As the blocks progress, their final terms come later and later in the positive and negative lists, so the overshoots tend to zero. The block endpoints therefore approach \(L\) from alternating sides. Within each block, the partial sums move monotonically between its starting value and its endpoint. Their distance from \(L\) cannot exceed the larger of the distances at those two ends. Since those endpoint distances tend to zero, all partial sums of the rearranged series tend to \(L\). This proves the theorem.

If the series contains zero terms, they do not change any partial sum. A finite number can be placed at the beginning; if there are infinitely many, insert one between successive blocks. This includes all zero terms without affecting the convergence argument.

Worked Example: Rearrange to Converge to Zero

Worked Example: Follow the crossing rule

Use the series from the earlier example, whose positive terms begin \(1,1/2,1/3,1/4,\ldots\) and whose negative terms begin \(-2,-3/4,-4/9,\ldots\). For target \(L=0\), first add positive terms until the sum is positive. Then add negative terms until it is negative, and repeat.

The first positive block is just \(1\), giving \(1>0\). The first negative term gives \(1-2=-1<0\). Starting from \(-1\), the next positive terms give

-1+\frac12=-\frac12,\qquad -\frac12+\frac13=-\frac16,\qquad -\frac16+\frac14=\frac1{12}>0.

So the positive block ends after \(1/4\). The next negative term is \(-3/4\), and

\frac1{12}-\frac34 =\frac1{12}-\frac9{12} =-\frac8{12} =-\frac23<0.

The crossing rule has been followed in both blocks. This finite calculation illustrates the construction; the theorem’s shrinking-overshoot argument, not a few computed terms, establishes convergence to zero.

Worked Example: Rearrange to Converge to a Negative Target

Worked Example: Target \(L=-2\)

Because the initial sum \(0\) is already greater than \(-2\), begin with negative terms and continue until the sum is below \(-2\). The first negative term gives \(0-2=-2\), which is not yet below the target. The next gives

-2-\frac34=-\frac{11}{4}<-2.

Now add positive terms until the sum exceeds \(-2\). The next positive term is \(1\), so

-\frac{11}{4}+1=-\frac74>-2.

The next negative term has magnitude \(4/9\), giving

-\frac74-\frac49 =-\frac{63}{36}-\frac{16}{36} =-\frac{79}{36}<-2.

The next positive term is \(1/2\), so the sum becomes

-\frac{79}{36}+\frac12 =-\frac{79}{36}+\frac{18}{36} =-\frac{61}{36}>-2.

These stages continue, with overshoots tending to zero. The theorem guarantees that the complete rearrangement converges to \(-2\).

A Common Mistake: Switching Too Soon

The instruction is to continue a block until the sum crosses the target, not merely to add one term of the chosen sign. In the zero-target example, after the negative term \(-2\), the sum is \(-1\). Adding just the next positive term gives \(-1+1/2=-1/2\), which is still below zero. Stopping there does not complete the positive block: the required condition is a sum greater than zero. Continuing gives \(-1/6\) after \(1/3\), and \(1/12\) after \(1/4\), which finally crosses zero.

There is also an important contrast with absolute convergence. The earlier result on rearrangement invariance states that an absolutely convergent series keeps its sum under rearrangement. Conditional convergence is precisely where the positive and negative totals can each be infinite, allowing the block-selection method to shift the limit.

Check Your Understanding

Use the definitions and crossing argument to answer these questions.

What two properties distinguish conditional convergence from absolute convergence?
Why must the positive terms of a conditionally convergent real series have an infinite sum?
In the rearrangement construction, why does every positive block end after finitely many terms?
Why does the overshoot at the end of a positive block tend to zero?
For a target \(L<0\), which sign should the construction use first, and why?