Two Different Questions About a Series
For a real series \(\sum_{n=1}^{\infty}a_n\), there are two related questions. Does the signed sequence of partial sums converge? And does the sequence of partial sums of the magnitudes, \(\sum_{n=1}^{N}|a_n|\), converge? The first question concerns the original series; the second asks whether the total size of its terms is finite, without allowing positive and negative terms to cancel.
These questions divide convergent series into two types. An absolutely convergent series remains summable when the signs are removed. A conditionally convergent series converges only in its signed form: its series of magnitudes diverges. The theorem Absolute Convergence Implies Convergence, established earlier, guarantees that absolute convergence is the stronger condition. The comparison between the two types is therefore not symmetric.
Every convergent series falls into exactly one of these two categories: its series of magnitudes either converges or diverges. If it diverges, the original series is conditionally convergent by definition. If it converges, the original series is absolutely convergent, and the earlier theorem also assures us that the signed series converges. Thus, for a convergent series, absolute convergence and conditional convergence are mutually exclusive alternatives.
Which Combinations of Behavior Are Possible?
It is useful to distinguish the behavior of the signed series from that of the series of magnitudes. Absolute convergence implies convergence of the signed series, but convergence of the signed series does not imply absolute convergence. The reverse implication rules out one apparent combination: a divergent signed series cannot have a convergent series of magnitudes.
| Signed series | Series of magnitudes | Classification |
|---|---|---|
| Converges | Converges | Absolutely convergent |
| Converges | Diverges | Conditionally convergent |
| Diverges | Diverges | Neither absolutely nor conditionally convergent |
| Diverges | Converges | Impossible |
The last row is impossible by Absolute Convergence Implies Convergence: if \(\sum |a_n|\) converges, then \(\sum a_n\) converges. This is a useful logical check when classifying a series. The fact that its terms tend to zero does not settle either question; the Necessary Condition for Series Convergence is necessary, not sufficient.
Worked Example: A Conditionally Convergent p-Series Variant
Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2/3}}. $$ The positive magnitudes \(b_n=1/n^{2/3}\) decrease to zero. Indeed, \(n^{2/3}\) increases with \(n\), so its reciprocal decreases, and \(1/n^{2/3}\to0\). The Alternating Series Test therefore shows that the signed series converges.
Its series of magnitudes is $$ \sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{n^{2/3}}\right| =\sum_{n=1}^{\infty}\frac{1}{n^{2/3}}. $$ This is a p-series with \(p=2/3\leq1\), so it diverges by the p-Series Convergence Criterion. The signed series is therefore conditionally convergent: it converges, but its magnitudes have an infinite sum.
Here cancellation is essential. The Alternating Series Test controls the signed partial sums by using the alternating signs and the decrease of the magnitudes. It does not show that the sum of the magnitudes is finite; that separate question has a different answer.
Worked Example: Absolute Convergence by Comparison
Consider the series $$ \sum_{n=1}^{\infty}(-1)^n\frac{1}{n^2+5n}. $$ To test absolute convergence, remove the signs: $$ \sum_{n=1}^{\infty}\left|(-1)^n\frac{1}{n^2+5n}\right| =\sum_{n=1}^{\infty}\frac{1}{n^2+5n}. $$ For every \(n\geq1\), \(n^2+5n\geq n^2>0\), and hence $$ 0\leq\frac{1}{n^2+5n}\leq\frac{1}{n^2}. $$ The comparison series \(\sum 1/n^2\) converges because it is a p-series with \(p=2>1\). The Comparison Principle for Positive-Term Series implies that the series of magnitudes converges. Thus the displayed signed series is absolutely convergent, and Absolute Convergence Implies Convergence gives convergence of the signed series as well.
This method does not require us to determine the sum of the series or to use its alternating signs. Once a summable upper bound for the magnitudes is found, the conclusion follows.
What Absolute Convergence Guarantees
The definition makes clear that absolute convergence concerns the total amount of the terms, not their particular signs. One consequence is that multiplying terms by bounded factors cannot destroy absolute convergence. These factors may change signs, reduce terms, or set some terms equal to zero, but they cannot increase their magnitudes by more than a fixed multiplier.
Proof. For every \(n\), the bound on \(c_n\) gives $$ |c_na_n|=|c_n||a_n|\leq M|a_n|. $$ Since \(\sum |a_n|\) converges, the series \(\sum M|a_n|\) also converges; for \(M=0\), it is the zero series, and for \(M>0\), it is a constant multiple of a convergent positive-term series. The Comparison Principle for Positive-Term Series now shows that \(\sum |c_na_n|\) converges. This is precisely absolute convergence of \(\sum c_na_n\). \(\square\)
Taking \(c_n=-1\) for selected indices and \(c_n=1\) for the others shows that changing signs term by term preserves absolute convergence. Taking \(c_n=1\) on a chosen set of indices and \(c_n=0\) elsewhere shows that retaining only terms from a subseries also preserves absolute convergence. In both cases the coefficient bound holds with \(M=1\).
Worked Example: A Sign Pattern That Does Not Affect the Test
Define $$ a_n=\frac{(-1)^{n(n+1)/2}}{n^2+1},\qquad n\geq1. $$ The exponent \(n(n+1)/2\) is an integer, so the numerator is either \(1\) or \(-1\). In particular, $$ |a_n|=\frac{1}{n^2+1}. $$ For every \(n\geq1\), \(n^2+1\geq n^2\), and therefore $$ 0\leq |a_n|\leq\frac{1}{n^2}. $$ The p-series \(\sum 1/n^2\) converges, so comparison proves that \(\sum |a_n|\) converges. Consequently, \(\sum a_n\) is absolutely convergent, regardless of the pattern of signs.
The sign sequence here is not the regular term-by-term alternation \((-1)^n\) in the preceding example. That difference does not affect the absolute convergence test: after taking magnitudes, the sign pattern disappears entirely.
Adding Absolutely Convergent Series
Absolute convergence is also stable under addition and scalar multiplication. Linearity of convergent series, established earlier, concerns signed sums. The result below gives the stronger conclusion that combining absolutely convergent series preserves absolute convergence.
Proof. The triangle inequality gives, for every \(n\), $$ |\alpha a_n+\beta b_n| \leq |\alpha||a_n|+|\beta||b_n|. $$ The series on the right, when summed over \(n\), is convergent: both \(\sum |a_n|\) and \(\sum |b_n|\) converge, and multiplying either by a fixed nonnegative constant preserves convergence. The Comparison Principle for Positive-Term Series therefore implies that $$ \sum_{n=1}^{\infty}|\alpha a_n+\beta b_n| $$ converges. This is the definition of absolute convergence for \(\sum(\alpha a_n+\beta b_n)\). The argument also covers \(\alpha=0\) or \(\beta=0\), since the corresponding terms on the right are then zero. \(\square\)
The bound is useful even when the terms \(\alpha a_n+\beta b_n\) have signs that vary irregularly. It is the summability of the magnitudes of the two original series, together with the triangle inequality, that controls the new series.
Worked Example: Combining Two Absolutely Convergent Series
Let $$ a_n=\frac{(-1)^n}{n^2}, \qquad b_n=\frac{1}{n^3}, \qquad u_n=3a_n-2b_n. $$ The magnitudes of the first series are \(1/n^2\), and those of the second are \(1/n^3\). Both p-series converge, so both original series are absolutely convergent. For each \(n\), $$ |u_n| =\left|\frac{3(-1)^n}{n^2}-\frac{2}{n^3}\right| \leq\frac{3}{n^2}+\frac{2}{n^3}. $$ The series of these upper bounds converges, since it is the sum of three times a convergent p-series with \(p=2\) and two times a convergent p-series with \(p=3\). Comparison proves that \(\sum |u_n|\) converges. Thus \(\sum u_n\) is absolutely convergent, even though its terms combine two different sign and size patterns.
Why the Distinction Matters
For an absolutely convergent series, the total magnitude of any tail can be made small, as expressed by the Absolute Tail Bound. For a conditionally convergent series, the signed partial sums still converge, but the series of magnitudes diverges. The alternating \(p\)-series example illustrates the contrast: the signed terms settle through cancellation, while removing their signs produces a divergent p-series.
A common pitfall is to infer absolute convergence from convergence of the original series, or from the fact that its terms tend to zero. Neither inference is valid. To establish absolute convergence, test \(\sum |a_n|\), often by comparison with a known positive series. To establish conditional convergence, first prove convergence of the signed series and then prove divergence of the series of magnitudes. The Alternating Series Test followed by the p-Series Convergence Criterion is one useful combination, but the two conclusions must be checked separately.
Check Your Understanding
Use the definitions, comparisons, and proved results to answer the following questions.
- What additional series must be tested to determine whether a convergent series is absolutely or conditionally convergent?
- Why is it impossible for a signed series to diverge while its series of magnitudes converges?
- How does the bounded-multiplier theorem apply when the coefficients are \(1\) or \(-1\)?
- What two separate facts must be established to prove that a series is conditionally convergent?
- Why does the triangle inequality help prove that a linear combination of absolutely convergent series is absolutely convergent?