Why the Absolute-Value Series Comes First
A series with positive and negative terms can behave differently from the series formed by taking the absolute value of every term. Cancellation may make the original series converge even when the sum of the absolute values diverges. To classify a series correctly, test for absolute convergence first; if that fails, test the original series separately.
This order gives a complete decision process. The absolute-value series can establish absolute convergence, but its divergence alone cannot determine whether the original series converges or diverges.
These categories are mutually exclusive. In particular, conditional convergence requires two separate conclusions: the original series converges, and the absolute-value series diverges. Showing only one of those facts is not enough.
The Two-Stage Classification
For a given series, first form \(\sum |a_n|\). Because its terms are nonnegative, use an appropriate positive-term test, such as a comparison, a limit comparison, a \(p\)-series test, or the ratio test when its conditions are met. If this series converges, the original series is absolutely convergent. If it diverges, do not stop: test \(\sum a_n\) itself.
Determine whether \(\sum |a_n|\) converges using a valid test for nonnegative-term series.
Absolute convergence implies convergence of the original series.
If \(\sum a_n\) converges, the convergence is conditional; if it does not, the original series diverges.
The ratio test can be useful at the first stage. For a series with nonzero terms, if the limit of \(\left|a_{n+1}/a_n\right|\) is less than \(1\), the series converges absolutely; if the limit is greater than \(1\), the series diverges. A limit of \(1\) gives no conclusion, so a different test is needed. In particular, a ratio-test result of \(1\) does not classify the original series.
Why Absolute Convergence Implies Convergence
This implication is the link that makes the first stage conclusive. The idea is that the absolute-value series bounds every tail of the original series. If the absolute-value tails can be made arbitrarily small, then the original partial sums satisfy the Cauchy criterion for convergence.
Proof. Let \(S_N=\sum_{n=1}^{N}a_n\) be the \(N\)th partial sum. Since \(\sum |a_n|\) converges, the Cauchy criterion for series says that for every \(\varepsilon>0\), there is an integer \(N_0\) such that whenever \(m>n\geq N_0\),
For those same integers \(m\) and \(n\), the triangle inequality gives
Therefore the partial sums \((S_N)\) satisfy the Cauchy criterion. The real numbers are complete, so every Cauchy sequence of real numbers converges. Hence \((S_N)\) converges, which means \(\sum a_n\) converges. This proves the theorem.
The proof also explains why the implication works: absolute values prevent cancellation, so their tail sums provide an upper bound for the size of every tail in the original series. The converse is not guaranteed. A convergent original series may rely on cancellation and have a divergent absolute-value series.
Worked Example: Absolute Convergence Despite an Inconclusive Ratio Test
Worked Example: Classifying a Reciprocal-Square Series
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\).
Step 1: Test the absolute-value series. Taking absolute values gives
This is a \(p\)-series with \(p=2\). A \(p\)-series \(\sum 1/n^p\) converges when \(p>1\), so the absolute-value series converges.
Step 2: Conclude for the original series. By the theorem on absolute convergence, the original series converges. Since its absolute-value series also converges, the original series is absolutely convergent.
Check: The classification uses the absolute-value series \(\sum 1/n^2\), not just the alternating signs. As an additional check on the choice of test, the ratio of consecutive absolute terms is
The ratio test is inconclusive here, but that does not conflict with the \(p\)-series result. The \(p\)-series test establishes convergence directly.
When Absolute Convergence Fails
If \(\sum |a_n|\) diverges, the original series may still converge. For alternating series, the alternating series test can often establish this second conclusion. In its standard form, if \(b_n\geq 0\), \(b_n\) is nonincreasing, and \(\lim_{n\to\infty} b_n=0\), then \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) converges. When the absolute-value series diverges as well, the convergence is conditional.
Each condition matters: the terms must alternate in the required pattern, the positive magnitudes must be nonincreasing, and those magnitudes must approach zero. A series cannot be declared convergent by this test just because its signs alternate.
Worked Example: Conditional Convergence of the Alternating Harmonic Series
Worked Example: Testing the Original Series After Absolute Divergence
Classify \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\).
Step 1: Test absolute convergence.
The series on the right is the harmonic series, the \(p\)-series with \(p=1\). Since a \(p\)-series diverges when \(p\leq 1\), the absolute-value series diverges. Absolute convergence has not been established, so the original series must still be tested.
Step 2: Test the original series. It has alternating terms with \(b_n=1/n\). For every \(n\geq1\), \(b_n>0\), and
The alternating series test therefore shows that the original series converges.
Conclusion and check: The original series converges, while the absolute-value series diverges. By definition, the series is conditionally convergent. Both required facts have been checked independently.
Worked Example: A Different Conditional Series
Worked Example: Verifying the Conditions Before Using the Alternating Series Test
Classify \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}\).
Step 1: Show that the absolute-value series diverges. Its terms are \(b_n=n/(n^2+1)>0\). Compare them with \(1/n\) using limit comparison:
The harmonic series diverges, and this limit is a finite positive number. The limit comparison test therefore shows that \(\sum b_n\), the absolute-value series, diverges.
Step 2: Verify the alternating series test for the original series. First, \(b_n>0\) for every \(n\geq1\). Next, compare consecutive terms:
The denominator is positive. For \(n\geq1\), the numerator satisfies \(n^2+n-1\geq1>0\). Thus \(b_n>b_{n+1}\), so the magnitudes decrease. Finally,
The alternating series test proves that the original series converges. Since its absolute-value series diverges, the classification is conditional convergence.
Check: The proof did not assume that alternating signs alone guarantee convergence. It verified positivity, decreasing magnitudes, and limit zero, then separately established divergence of the absolute-value series.
Worked Example: Divergence Even Though the Signs Alternate
Worked Example: Applying the Nth-Term Test to the Original Series
Classify \(\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}\).
Step 1: Check the absolute-value series.
The terms of the absolute-value series do not approach zero. By the nth-term test for divergence, the absolute-value series diverges.
Step 2: Check the original terms. The original term is \(a_n=(-1)^n n/(n+1)\). Its magnitude approaches \(1\), so \(a_n\) does not approach zero. More explicitly, its even-indexed terms approach \(1\), while its odd-indexed terms approach \(-1\); a sequence with these two different subsequential limits cannot have limit zero.
The nth-term test for divergence says that if the terms of a series do not approach zero, then the series diverges. Therefore the original series diverges. Conclusion: this series is neither absolutely nor conditionally convergent; it is divergent.
A Common Mistake: Treating Absolute Divergence as Proof of Divergence
A tempting but invalid argument is: “The absolute-value series diverges, so the original series diverges.” The alternating harmonic series shows exactly why that conclusion fails. Its absolute-value series is \(\sum 1/n\), which diverges, but its original series converges by the alternating series test.
The correction is to keep the two tests separate. Divergence of \(\sum |a_n|\) rules out absolute convergence; it does not settle convergence of \(\sum a_n\). Apply a valid test to the original series. If it converges, classify it as conditionally convergent. If it diverges, classify it as divergent.
Writing a Complete Justification
A strong written classification names the series tested, states the test and the conditions that make it applicable, and ends with the classification. A useful structure is:
- Write the absolute-value series \(\sum |a_n|\).
- Use a valid test to state whether it converges or diverges.
- If it converges, conclude that the original series is absolutely convergent.
- If it diverges, test \(\sum a_n\) itself and state the result.
- Conclude conditional convergence only when the original converges and the absolute-value series diverges.
For an alternating series, explicitly verify the conditions of the alternating series test rather than writing only “it alternates.” For a divergent series, the nth-term test is valid when the terms fail to approach zero; if the terms do approach zero, that test gives no conclusion and another argument is needed.
Check Your Understanding
For each question, identify which series must be tested and what conclusion the stated information supports.
- If \(\sum |a_n|\) converges, what can you conclude about \(\sum a_n\), and what is the classification?
- If \(\sum |a_n|\) diverges but \(\sum a_n\) converges, how is \(\sum a_n\) classified?
- Why does divergence of \(\sum |a_n|\) not prove that \(\sum a_n\) diverges?
- State the three conditions needed to apply the alternating series test to \(\sum (-1)^{n-1}b_n\).
- If the terms of the original series do not approach zero, what can you conclude? What if they do approach zero?
