Conditional Convergence

Convergence by Cancellation

Absolute convergence guarantees convergence without requiring any cancellation: the total magnitude of the terms is finite. Conditional convergence describes a different situation. The signed terms have a finite sum, but their magnitudes do not. In that case, cancellation is essential, and the order of the terms can matter.

Definition (Conditional Convergence): A series $\sum_{n=1}^{\infty}a_n$ is conditionally convergent if $\sum_{n=1}^{\infty}a_n$ converges but $\sum_{n=1}^{\infty}|a_n|$ diverges.

The theorem Absolute Convergence Implies Convergence establishes one direction: absolute convergence implies convergence. Conditional convergence is precisely the convergent case in which the converse fails. The distinction is not merely about the signs alternating regularly. It is about how positive and negative terms together allow the signed partial sums to settle while the total amount of positive and negative mass remains infinite.

Both Signs Must Contribute Without Bound

For a conditionally convergent series, neither the positive terms nor the negative terms can have a finite total magnitude. If, for example, the positive terms had finite sum, convergence of the signed series would force the negative terms to have finite total magnitude as well. That would make the series absolutely convergent, contrary to the definition.

Theorem (Positive and Negative Parts of a Conditionally Convergent Series): Suppose $\sum_{n=1}^{\infty}a_n$ converges but $\sum_{n=1}^{\infty}|a_n|$ diverges. Then $$ \sum_{n=1}^{\infty}\max(a_n,0)=\infty \qquad\text{and}\qquad \sum_{n=1}^{\infty}\max(-a_n,0)=\infty. $$

Proof. Define $$ P_N=\sum_{n=1}^{N}\max(a_n,0), \qquad Q_N=\sum_{n=1}^{N}\max(-a_n,0). $$ For each $n$, $a_n=\max(a_n,0)-\max(-a_n,0)$ and $|a_n|=\max(a_n,0)+\max(-a_n,0)$. Thus the signed partial sum is $P_N-Q_N$, and the partial sum of the magnitudes is $P_N+Q_N$. Both $P_N$ and $Q_N$ are nondecreasing sequences.

Suppose, for contradiction, that the positive-part series has finite sum. Then $P_N$ converges to a finite limit. Since the signed series converges, $P_N-Q_N$ converges to a finite limit as well. Therefore $$ Q_N=P_N-(P_N-Q_N) $$ converges to a finite limit. It follows that $P_N+Q_N$ converges to a finite limit, so $\sum |a_n|$ converges, a contradiction. Hence the positive-part series diverges.

Now suppose instead that the negative-part series has finite sum. Then $Q_N$ converges, and the convergence of $P_N-Q_N$ implies that $$ P_N=(P_N-Q_N)+Q_N $$ converges to a finite limit. Again $P_N+Q_N$ would converge, contradicting divergence of $\sum |a_n|$. The negative-part series must also diverge. $\square$

In particular, a conditionally convergent series has infinitely many positive and infinitely many negative terms, and the magnitudes of each sign have an infinite total. Also, the Necessary Condition for Series Convergence says $a_n\to0$. These two facts—unbounded total magnitude in each sign, but individual terms tending to zero—are exactly what make the rearrangement result possible.

Worked Examples

Worked Example: The Alternating Harmonic Series

Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} =1-\frac12+\frac13-\frac14+\cdots. $$ The terms have the form $(-1)^{n-1}b_n$, where $b_n=1/n$. The sequence $b_n$ is nonnegative, nonincreasing, and tends to zero. The Alternating Series Test therefore shows that the signed series converges.

Its series of magnitudes is the harmonic series $$ \sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{n}\right| =\sum_{n=1}^{\infty}\frac1n, $$ which diverges by the p-Series Convergence Criterion with $p=1$. The original series is therefore conditionally convergent.

The positive terms are $1/(2k-1)$, and the negative terms have magnitudes $1/(2k)$, for positive integers $k$. Each negative magnitude satisfies $$ \frac{1}{2k}=\frac12\cdot\frac1k. $$ Thus the negative magnitudes have infinite sum. Since $2k-1\leq 2k$, we also have $1/(2k-1)\geq1/(2k)$, so the positive terms have infinite sum as well. This verifies directly, in this example, the two-sided divergence in the theorem.

Worked Example: A Different Conditionally Convergent Series

Consider $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}. $$ Set $b_n=n/(n^2+1)$. These terms are positive and tend to zero, since $$ 0<\frac{n}{n^2+1}\leq\frac1n $$ and $1/n\to0$. To check that $b_n$ is decreasing, compare consecutive terms. For $n\geq1$, all denominators are positive, and $$ \frac{n}{n^2+1}>\frac{n+1}{(n+1)^2+1} $$ is equivalent to $$ n\big((n+1)^2+1\big)>(n+1)(n^2+1). $$ The left side minus the right side is $n^2+n-1$, which is positive for every $n\geq1$. Hence $b_n$ is decreasing. The Alternating Series Test gives convergence of the signed series.

For the series of magnitudes, use $n^2+1\leq2n^2$, valid for $n\geq1$. It gives $$ \frac{n}{n^2+1}\geq\frac{1}{2n}. $$ The series $\sum 1/(2n)$ diverges, since it is one-half of the harmonic series. By comparison, $\sum n/(n^2+1)$ diverges. The displayed signed series is therefore conditionally convergent.

Rearranging a Conditionally Convergent Series

A rearrangement is a series formed by listing the same terms in a different order, without omitting or repeating any term. For absolutely convergent series, changing the order preserves the sum. Conditional convergence behaves differently: the infinite supply of positive and negative magnitude allows the terms to be ordered so that the partial sums approach any chosen real number.

Theorem (Riemann Rearrangement Theorem): If $\sum_{n=1}^{\infty}a_n$ is conditionally convergent, then for every real number $x$, there is a rearrangement of its terms that converges to $x$.

Proof. By the theorem on positive and negative parts, the sum of the positive terms is infinite, and the sum of the magnitudes of the negative terms is infinite. By the Necessary Condition for Series Convergence, the original terms tend to zero. Consequently, the positive terms, listed in their original order, tend to zero; the magnitudes of the negative terms, also listed in their original order, tend to zero.

First suppose $x<0$. Starting from the sum $0$, add unused negative terms in their original order until the partial sum is at most $x$. This requires only finitely many terms: the total magnitude of the negative terms is infinite, so their successive additions eventually bring the sum below $x$. If $x\geq0$, no such initial step is needed, since the starting sum $0$ is already at most $x$.

Now repeatedly carry out these two stages:

Add unused positive terms, in their original order, until the partial sum is greater than $x$.
Add unused negative terms, in their original order, until the partial sum is less than $x$.

Each stage takes finitely many terms. For the first stage, the positive terms have infinite total, so adding them to any fixed starting sum eventually takes the partial sum above $x$. For the second stage, the negative magnitudes have infinite total, so their addition eventually takes the partial sum below $x$. Each stage uses at least one term. Thus the process continues indefinitely, and it uses every positive and every negative term: each sign’s terms are consumed in their original order, and every stage of that sign uses at least one.

At the end of a positive stage, the partial sum has crossed $x$ from below. Its excess over $x$ is no greater than the last positive term added, because immediately before that term was added the sum was at most $x$. At the end of a negative stage, the partial sum has crossed $x$ from above, and its shortfall below $x$ is no greater than the magnitude of the last negative term added. The terms used at these crossings tend to zero: each sign’s terms are taken in their original order, and $a_n\to0$. Therefore the partial sums at the ends of the positive and negative stages both tend to $x$.

This also gives convergence of all the intervening partial sums, not just the sums at stage endpoints. During a positive stage, the partial sums increase from the preceding negative-stage endpoint to the positive-stage endpoint. During a negative stage, they decrease from the positive-stage endpoint to the next negative-stage endpoint. Once both endpoints are within $\varepsilon$ of $x$, every partial sum between them is also within $\varepsilon$ of $x$. Hence the rearranged partial sums converge to $x$.

If the original series has zero terms, the construction rearranges its nonzero terms; insert all zero terms into the resulting list, in any order and at any positions. They change no partial sum, and the list then contains every term of the original series exactly once. This completes the proof. $\square$

Worked Example: Rearranging the Alternating Harmonic Series

Worked Example: First Stages of a Rearrangement Toward One

The alternating harmonic series is conditionally convergent, so the Riemann Rearrangement Theorem guarantees a rearrangement converging to $1$. The construction begins by adding positive terms until the sum exceeds $1$: $$ 1+\frac13=\frac43>1. $$ Next add negative terms until the sum is below $1$: $$ \frac43-\frac12=\frac56<1. $$ Then add positive terms until the sum exceeds $1$: $$ \frac56+\frac15=\frac{31}{30}>1. $$ Adding the next negative term gives $$ \frac{31}{30}-\frac14=\frac{47}{60}<1. $$

For the next positive stage, $$ \frac{47}{60}+\frac17=\frac{389}{420}<1, $$ so another positive term is needed. Adding $1/9$ gives $$ \frac{389}{420}+\frac19 =\frac{1167}{1260}+\frac{140}{1260} =\frac{1307}{1260}>1. $$ The next negative term then gives $$ \frac{1307}{1260}-\frac16 =\frac{1097}{1260}<1. $$ The construction continues by alternating positive and negative stages. Each crossing error is bounded by the last term used in that stage, and those terms tend to zero. The theorem therefore guarantees that the entire rearranged series—not only these displayed partial sums—converges to $1$.

Why the Order Matters

The rearrangement theorem does not say that every rearrangement of a conditionally convergent series converges, or that every rearrangement has a different sum. It says that enough freedom exists to construct a rearrangement with any prescribed real sum. The proof depends on two features working together: each sign has infinite total magnitude, while individual terms become arbitrarily small. Infinite total magnitude lets each stage reach its target; small terms make the overshoots tend to zero.

A common pitfall is to assume that the signed sum is fixed independently of order simply because the original series converges. That conclusion is justified for absolutely convergent series, but not for conditionally convergent ones. In particular, rearranging terms requires a new convergence argument; the convergence of the original ordering alone does not establish convergence of the rearranged ordering.

Takeaway: Conditional convergence means convergence of the signed series but divergence of the magnitude series. Both positive and negative terms then have infinite total magnitude. By using each sign to cross a target in turn, with crossing errors tending to zero, one can rearrange the terms to converge to any real number.

Check Your Understanding

Use the definition, examples, and rearrangement theorem to answer the following questions.

What two conditions must hold for a series to be conditionally convergent?
Why must both the positive terms and the magnitudes of the negative terms have infinite sum?
Which two facts ensure that each stage of the rearrangement construction is finite and that its crossing error becomes small?
In the alternating harmonic example, why does the series converge while its series of magnitudes diverges?
Why must the proof control partial sums between stage endpoints, rather than only showing that the endpoints approach the target?

What You'll Learn