AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/10/10-9)
Unit 10 · Topic 10.9
BC only10.9 Determining Absolute or Conditional Convergence
BC only. A series converges absolutely if the series of absolute values converges, and absolute convergence guarantees the original series converges too. If a series converges but its absolute-value series diverges, it converges conditionally, like the alternating harmonic series.
Key terms
- absolute convergence
- conditional convergence
- absolute value
- alternating harmonic series
Three possibilities
This whole unit is BC only. Every series falls into exactly one of these categories.
| Type | Meaning | Example |
|---|---|---|
| absolutely convergent | the series of absolute values converges | Σ (−1)ⁿ/n² |
| conditionally convergent | the series converges, but the absolute values don't | Σ (−1)ⁿ⁺¹/n |
| divergent | the series itself diverges | Σ (−1)ⁿ n/(n + 1) |
Absolute convergence implies convergence
If Σ |aₙ| converges, then Σ aₙ converges. Making some terms negative can only help them cancel, so the sum stays finite. This is powerful for series whose signs follow no pattern, like Σ sin(n)/n². They aren't alternating, so the alternating series test can't help, but their absolute values are easy to compare.
The converse is false: a convergent series need not converge absolutely. The alternating harmonic series converges, but Σ 1/n diverges.
How to classify a series
- Check the absolute values first: does Σ |aₙ| converge? Use p-series, comparison, integral or ratio tests (they need positive terms, which |aₙ| has). If yes, the series converges absolutely. Stop.
- If Σ |aₙ| diverges, check the original series. If it's alternating, try the alternating series test. If it passes, the series converges conditionally.
- If the terms don't go to 0, the series diverges by the nth term test.
Why absolute convergence is enough
Here's the short reason. For every term, 0 ≤ aₙ + |aₙ| ≤ 2|aₙ|, since aₙ + |aₙ| is either 0 (if aₙ is negative) or 2|aₙ| (if it's positive). If Σ |aₙ| converges, then Σ 2|aₙ| does too, so Σ (aₙ + |aₙ|) converges by direct comparison. Subtracting the convergent Σ |aₙ| leaves Σ aₙ, which therefore converges.
Rearranging terms
Absolutely convergent series behave like finite sums: you can rearrange or regroup their terms in any order, and the sum stays the same. Conditionally convergent series are fragile. Rearranging their terms can change the sum, or even make the series diverge. You won't need to construct a rearrangement, but you may be asked which kind of series allows it.
The ratio test (10.8) proves absolute convergence directly, because it uses |aₙ₊₁/aₙ|. So if the ratio test gives L < 1, you can say “converges absolutely” right away.
Power series often converge absolutely inside the interval of convergence and conditionally at an endpoint, as Σ (x − 3)ⁿ/(n·2ⁿ) does at x = 1 (10.13).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Absolutely convergent
Classify Σ (n = 1 to ∞) (−1)ⁿ/n².
Show the solutionHide the solution
- Step 1: Absolute values: Σ 1/n², a p-series with p = 2 > 1, converges.
- Step 2: So the series converges absolutely.
Answer: Absolutely convergent.
- Example 2
Conditionally convergent
Classify Σ (n = 1 to ∞) (−1)ⁿ⁺¹/√n.
Show the solutionHide the solution
- Step 1: Absolute values: Σ 1/√n is a p-series with p = 1/2 ≤ 1, so it diverges. Not absolutely convergent.
- Step 2: Original series: alternating, with bₙ = 1/√n positive, decreasing and approaching 0. It converges by the alternating series test.
- Step 3: It converges, but not absolutely: conditionally convergent.
Answer: Conditionally convergent.
- Example 3
Trap: signs that don't alternate
Does Σ (n = 1 to ∞) sin(n)/n² converge?
Show the solutionHide the solution
- Step 1: sin(n) is positive for some n and negative for others, but not in a strict alternating pattern. The alternating series test doesn't apply.
- Step 2: Look at absolute values: |sin(n)/n²| ≤ 1/n², because |sin n| ≤ 1.
- Step 3: Σ 1/n² converges (p = 2), so Σ |sin(n)/n²| converges by direct comparison.
- Step 4: Absolute convergence implies convergence.
Answer: Converges (absolutely).
Common mistakes
- Calling a series conditionally convergent without showing that it actually converges.
- Applying the alternating series test to a series whose signs don't strictly alternate.
- Thinking “converges” and “converges absolutely” mean the same thing.
- Checking only the absolute values and calling the series divergent when they diverge. The original series might still converge conditionally.
On the exam
- Free-response questions often ask whether a series (or a power series at an endpoint) converges absolutely, conditionally or diverges. Answer with both checks: the absolute-value series and the original series.
- Multiple-choice questions may list several series and ask which is conditionally convergent. Look for alternating series whose terms behave like 1/nᵖ with p ≤ 1.
Connected topics
Videos
Check yourself
4 questions on 10.9 Determining Absolute or Conditional Convergence. Pick an answer to see if you got it, and why.
Which of the following series converges conditionally?
The series Σ (n = 1 to ∞) aₙ converges conditionally. Which of the following must be true? I. Σ (n = 1 to ∞) |aₙ| diverges. II. Σ (n = 1 to ∞) (aₙ)² converges. III. lim (n→∞) aₙ = 0
Which of the following correctly describes the series Σ (n = 2 to ∞) (−1)ⁿ/(n ln n)?
Which of the following correctly describes the series Σ (n = 1 to ∞) (cos n)/n²?
0 of 4 answered