Why the Absolute-Value Series Matters
A series can converge because its positive and negative terms cancel. That cancellation may be enough for the original series to have a finite sum, even when the sum of the term magnitudes is infinite. To classify such a series correctly, separate two questions: does the series of absolute values converge, and, if not, does the original series still converge?
This distinction is especially useful in BC free-response questions. A conclusion such as “the alternating series converges” does not by itself say whether it converges absolutely or conditionally. A complete classification explains what happens both before and after taking absolute values.
These categories are determined by the original series and its absolute-value series together. A series cannot be conditionally convergent if its original sum diverges, even if some separate test establishes that the absolute-value series diverges too.
The Two-Stage Classification
Begin with \(\sum |a_n|\). If it converges, the original series converges absolutely. If it diverges, do not stop: test \(\sum a_n\) itself. It may still converge, in which case convergence is conditional, or it may diverge.
Write \(\sum |a_n|\) and choose an appropriate convergence test for it.
Absolute convergence guarantees convergence of the original series.
Use a suitable test on \(\sum a_n\). Convergence now means conditional convergence; divergence means the series diverges.
The tests used in the first or second stage may include the \(n\)th-Term Test, geometric-series test, Ratio Test, Integral Test, Comparison Test, Limit Comparison Test, or Alternating Series Test, provided that each test’s conditions are satisfied. In particular, taking absolute values can remove alternating signs, so a test that works on the original terms may not answer the absolute-convergence question.
Why Absolute Convergence Guarantees Convergence
The implication from absolute convergence to convergence follows from the Cauchy criterion for series. A series converges precisely when its partial sums approach a finite limit; equivalently, for every \(\varepsilon>0\), the sum over any sufficiently far-out finite block has magnitude less than \(\varepsilon\).
Proof. Since the series of nonnegative terms \(\sum |a_n|\) converges, its tails can be made arbitrarily small. Let \(\varepsilon>0\). There is an integer \(N\) such that whenever \(q\ge p\ge N\),
The triangle inequality then gives, for the same \(p\) and \(q\),
Thus the partial sums of \(\sum a_n\) satisfy the Cauchy criterion. The real numbers are complete, so every Cauchy sequence of real numbers converges. Therefore the partial sums converge, and \(\sum a_n\) converges. This proves the implication.
The reverse implication is not guaranteed: convergence of \(\sum a_n\) does not necessarily imply convergence of \(\sum |a_n|\). Alternating series provide familiar examples where cancellation allows the original series to converge although the absolute-value series does not.
Worked Examples
Worked Example: Absolute Convergence of an Alternating \(p\)-Series
Classify \(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\).
First take absolute values:
This is a \(p\)-series with \(p=2>1\), so it converges. By the absolute-convergence theorem, the original series converges. Its classification is therefore absolutely convergent.
As a check on the sign removal, \(|(-1)^{n+1}|=1\) for every positive integer \(n\), so the absolute-value terms really are \(1/n^2\), not an alternating sequence.
Worked Example: Conditional Convergence of the Alternating Harmonic Series
Classify \(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\).
The absolute-value series is
This is the harmonic series, which diverges. Absolute convergence fails.
Now test the original series. It has the form \(\sum (-1)^{n+1}b_n\), where \(b_n=1/n\). The terms are positive, \(b_{n+1}=1/(n+1)\leq 1/n=b_n\), and \(\lim_{n\to\infty}b_n=0\). The Alternating Series Test therefore establishes convergence of the original series.
It converges but does not converge absolutely, so it is conditionally convergent. The check is that the two tests answer different questions: the harmonic series test concerns magnitudes, while the Alternating Series Test concerns the signed original terms.
Worked Example: Alternating Terms That Still Diverge
Classify \(\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}\).
Its absolute-value terms satisfy
In particular, the absolute-value series diverges by the \(n\)th-Term Test, because its terms do not approach zero.
Check the original series using the same necessary condition for convergence:
For example, the even-indexed terms approach \(1\), while the odd-indexed terms approach \(-1\). A convergent series must have terms approaching zero, so the original series diverges. Its classification is divergent, not conditionally convergent. Alternating signs alone do not guarantee convergence.
Using the Alternating Series Test Carefully
For an alternating series \(\sum (-1)^n b_n\) or \(\sum (-1)^{n+1}b_n\), the Alternating Series Test applies when \(b_n\geq 0\), the sequence \(b_n\) is eventually nonincreasing, and \(\lim_{n\to\infty}b_n=0\). Under these conditions the original alternating series converges. The test does not establish absolute convergence: that requires a separate test on \(\sum b_n\).
Sometimes the monotonicity condition holds only after a finite number of terms. That is sufficient: adding or removing finitely many terms does not change whether an infinite series converges. State the range on which the conditions hold, especially if a formula’s behavior near its first few terms differs from its eventual behavior.
A Common Pitfall: Stopping After the Absolute-Value Test
A common error is to see that \(\sum |a_n|\) diverges and immediately conclude that the original series diverges. That conclusion is not valid. The alternating harmonic series is a counterexample: its absolute-value series is the divergent harmonic series, but its original series converges by the Alternating Series Test. The correct conclusion is conditional convergence.
A second error is to use “alternating” as if it were itself a convergence test. For example, in the third worked example, the magnitudes \(n/(n+1)\) approach \(1\), not \(0\), so the necessary term condition fails. Before applying the Alternating Series Test, verify the size and monotonicity conditions; before classifying a convergent original series as conditional, verify that its absolute-value series diverges.
Writing a Complete Classification in an FRQ
A concise but justified response identifies the series being tested and states the test result. For absolute convergence, name the test applied to \(\sum |a_n|\). If that series diverges, state that fact and then give a separate argument for the original series. Finish with one of the three classifications, rather than leaving the reader to infer it.
Keep necessary conditions distinct from sufficient tests. The \(n\)th-Term Test can prove divergence when terms fail to approach zero, but a zero term limit alone does not prove convergence. Likewise, failure of absolute convergence does not prove divergence of the original series. Each conclusion requires the appropriate second step.
Check Your Understanding
For each question, explain which series you test and state the conclusion justified by that test.
- What must be true about \(\sum |a_n|\) for \(\sum a_n\) to converge absolutely?
- If \(\sum |a_n|\) diverges but the Alternating Series Test proves \(\sum a_n\) converges, how should the original series be classified?
- Why does divergence of \(\sum |a_n|\) not by itself show that \(\sum a_n\) diverges?
- For \(\sum_{n=1}^{\infty}(-1)^{n+1}/n^3\), identify the absolute-value series and classify the original series.
- What conclusion follows from the \(n\)th-Term Test if \(\lim_{n\to\infty}a_n\neq 0\)?
