Determining Absolute or Conditional Convergence: Study Guide (Calculus BC 10.9)

BC-only material. Absolute and conditional convergence are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

This topic pulls together the tests from earlier in Unit 10. If any is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
nth term test	10.3	Spotting divergence quickly when the terms do not approach 0
Integral test, p-series, harmonic series	10.4, 10.5	Deciding whether Σ |aₙ| converges
Comparison and limit comparison tests	10.6	Comparing Σ |aₙ| with a known series
Alternating series test	10.7	Showing Σ aₙ converges when Σ |aₙ| does not
Ratio test	10.8	Testing Σ |aₙ| when terms contain powers or factorials

Notation on this page: Σ aₙ means the infinite series a₁ + a₂ + a₃ + …, and Σ |aₙ| is the same series with every term replaced by its absolute value.

Two series hiding inside one

Most of the tests you have learned (integral, comparison, limit comparison) only work for series with positive terms. The alternating series test handles one special sign pattern. So what do you do with a series whose terms have mixed signs?

Every series Σ aₙ comes with a partner: Σ |aₙ|, where every negative term is flipped to positive. The partner has positive terms, so all your positive-term tests apply to it. Comparing the two series gives three possible outcomes:

Σ |aₙ|	Σ aₙ	Name
converges	converges (always)	absolutely convergent
diverges	converges	conditionally convergent
diverges	diverges	divergent

There is no fourth row. A series cannot have Σ |aₙ| converging while Σ aₙ diverges. The next section explains why.

“Conditionally” is a good word for the middle row. The series converges, but only because its positive and negative terms cancel each other. Make all the terms positive and the cancellation disappears, and so does the convergence.

Why absolute convergence implies convergence

Suppose Σ |aₙ| converges. For any number a, the value a + |a| is either 0 (if a is negative) or 2a (if a is positive or zero). So

0 ≤ aₙ + |aₙ| ≤ 2|aₙ| for every n.

• The series Σ 2|aₙ| converges (it is twice a convergent series).
• By the comparison test, Σ (aₙ + |aₙ|) converges, because its terms are non-negative and no bigger than 2|aₙ|.
• Now aₙ = (aₙ + |aₙ|) − |aₙ|. The difference of two convergent series converges, so Σ aₙ converges.

So you never need a separate test for Σ aₙ once you know Σ |aₙ| converges. The reverse is false: Σ |aₙ| can diverge while Σ aₙ still converges.

A decision path

Two practical points:

• The nth term test first saves time. If aₙ does not approach 0, the series diverges and there is nothing more to do.
• When the ratio test is used on |aₙ| and gives L > 1, the series Σ aₙ diverges as well, not just Σ |aₙ|. A ratio above 1 means the sizes |aₙ| eventually grow, so aₙ cannot approach 0.

A reference family: Σ (−1)ⁿ⁺¹/nᵖ

This one family shows all three outcomes, and you can use it for comparisons.

Value of p	Σ |aₙ| = Σ 1/nᵖ	Σ (−1)ⁿ⁺¹/nᵖ	Classification
p > 1	converges (p-series)	converges	absolutely convergent
0 < p ≤ 1	diverges (p-series)	converges (alternating series test)	conditionally convergent
p ≤ 0	diverges	diverges (terms do not → 0)	divergent

For example, Σ (−1)ⁿ⁺¹/n² is absolutely convergent, Σ (−1)ⁿ⁺¹/√n is conditionally convergent, and Σ (−1)ⁿ⁺¹ (where p = 0) diverges.

Worked example 1: a conditionally convergent series

Question. Determine whether Σ from n = 1 to ∞ of (−1)ⁿ (n + 2)/(n² + 3) converges absolutely, converges conditionally or diverges. Justify your answer.

1. Terms approach 0? (n + 2)/(n² + 3) → 0 as n → ∞ (degree 1 over degree 2). So the nth term test gives no conclusion; keep going.
2. Test Σ |aₙ| = Σ (n + 2)/(n² + 3). For large n this behaves like n/n² = 1/n. Use the limit comparison test with bₙ = 1/n: lim as n → ∞ of [(n + 2)/(n² + 3)] ÷ (1/n) = lim of (n² + 2n)/(n² + 3) = 1. The limit is finite and positive, and Σ 1/n is the harmonic series, which diverges. So Σ |aₙ| diverges. The series is not absolutely convergent.
3. Test Σ aₙ itself with the alternating series test. Let bₙ = (n + 2)/(n² + 3) > 0. Check all three conditions:
   • The signs alternate because of (−1)ⁿ. ✓
   • bₙ → 0 (step 1). ✓
   • bₙ is decreasing. Let f(x) = (x + 2)/(x² + 3). Then f′(x) = (−x² − 4x + 3)/(x² + 3)². At x = 1 the numerator is −1 − 4 + 3 = −2, and −x² − 4x + 3 only gets more negative as x grows. So f′(x) < 0 for x ≥ 1 and the terms decrease. ✓ (Check: b₁ = 3/4, b₂ = 4/7 ≈ 0.571, b₃ = 5/12 ≈ 0.417, b₄ = 6/19 ≈ 0.316.) So Σ aₙ converges.
4. Conclusion. The series converges, but the series of absolute values diverges. So Σ (−1)ⁿ (n + 2)/(n² + 3) is conditionally convergent.

What to write. Name each test, check every condition, and state both results: “Σ |aₙ| diverges by limit comparison with the harmonic series; Σ aₙ converges by the alternating series test; therefore conditionally convergent.”

Worked example 2: signs that do not alternate

Question. Does Σ from n = 1 to ∞ of cos(n)/n^(3/2) converge? Classify it.

1. Look at the signs. cos 1 ≈ 0.540, cos 2 ≈ −0.416, cos 3 ≈ −0.990, cos 4 ≈ −0.654. For n = 1 to 10 the sign pattern is + − − − + + + − − −. This is not an alternating series, so the alternating series test cannot be used.
2. Use absolute values instead. Since |cos n| ≤ 1 for every n, 0 ≤ |cos(n)/n^(3/2)| ≤ 1/n^(3/2).
3. Compare. Σ 1/n^(3/2) is a p-series with p = 3/2 > 1, so it converges. By the comparison test, Σ |cos(n)/n^(3/2)| converges.
4. Conclusion. The series is absolutely convergent, so it converges.

Why this example matters. Without absolute convergence there would be no test on this course that could handle these irregular signs. Testing Σ |aₙ| turns a mixed-sign problem into a positive-term problem.

Worked example 3: the ratio test on |aₙ|

The ratio test (Topic 10.8) uses |aₙ₊₁/aₙ|, so it is really a test on Σ |aₙ|. When it gives L < 1, it proves absolute convergence.

(a) Classify Σ from n = 1 to ∞ of (−1)ⁿ⁺¹ n²/3ⁿ.

• |aₙ₊₁/aₙ| = [(n + 1)²/3ⁿ⁺¹] × [3ⁿ/n²] = (n + 1)²/(3n²).
• As n → ∞, this approaches 1/3.
• L = 1/3 < 1, so Σ |aₙ| converges. The series is absolutely convergent.

(b) Classify Σ from n = 1 to ∞ of (−1)ⁿ 3ⁿ/n².

• |aₙ₊₁/aₙ| = [3ⁿ⁺¹/(n + 1)²] × [n²/3ⁿ] = 3n²/(n + 1)², which approaches 3.
• L = 3 > 1. The sizes |aₙ| grow without bound (3, 9/4, 3, 81/16, 243/25, …), so aₙ does not approach 0. The series diverges by the nth term test. It is not conditionally convergent: there is no cancellation that could rescue it.

Regrouping and reordering terms

For a finite sum, order never matters. For an infinite series, it can.

Absolutely convergent series behave like finite sums. If Σ aₙ converges absolutely, then you can regroup the terms (add brackets) or rearrange them (change their order), and the new series converges to the same value. This is why manipulating an absolutely convergent series is safe.

Conditionally convergent series do not. Their convergence depends on the positive and negative terms cancelling in a particular order. Change the order and the cancellation changes.

Background (beyond what is assessed). The alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges to ln 2 ≈ 0.6931. Use exactly the same terms, but take two positive terms, then one negative term: 1 + 1/3 − 1/2 + 1/5 + 1/7 − 1/4 + … . This rearranged series converges to (3/2) ln 2 ≈ 1.0397. Same terms, different order, different sum. In fact, a theorem of Riemann says that a conditionally convergent series can be reordered to give any sum you like, or to diverge.

What you need for the exam is the safe direction: absolute convergence guarantees that regrouping or reordering does not change the value.

Common misconceptions

• “If Σ |aₙ| diverges, then Σ aₙ diverges.” False. Σ (−1)ⁿ⁺¹/n converges although Σ 1/n diverges. You must test Σ aₙ separately.
• “If Σ aₙ converges, then Σ |aₙ| converges.” False for the same reason. Only the other direction is true.
• “Conditionally convergent” means “might converge”. No. A conditionally convergent series does converge. The word describes why it converges (cancellation), not whether.
• Using the alternating series test on irregular signs. The test needs strictly alternating signs. For a series like Σ cos(n)/n^(3/2), test the absolute values instead.
• Forgetting the absolute value bars. Write Σ |aₙ| and |aₙ₊₁/aₙ|. A ratio test with signs left in can give a negative “limit”.
• Stopping after Σ |aₙ| diverges. That only rules out absolute convergence. The classification is not finished until you have tested Σ aₙ.
• Skipping the conditions of the alternating series test. State that the terms decrease and approach 0, and show why.
• Reordering a conditionally convergent series. Its sum can change. Only reorder or regroup when the series converges absolutely.

Where this leads

Next, Topic 10.10 shows how to bound the error when you stop an alternating series after a finite number of terms. Continue with Topic 10.10, Alternating Series Error Bound. Absolute and conditional convergence return in Topic 10.13, where the endpoints of an interval of convergence are often exactly this kind of question. See the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.