AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/10/10-6)
Unit 10 · Topic 10.6
BC only10.6 Comparison Tests for Convergence
BC only. Comparison tests judge a series with positive terms against a benchmark you already know, usually a geometric series or a p-series. Direct comparison uses inequalities between terms; limit comparison uses the limit of the ratio of terms.
Key terms
- direct comparison test
- limit comparison test
- positive terms
- benchmark series
Direct comparison test
This whole unit is BC only. Suppose 0 ≤ aₙ ≤ bₙ for all n (or for all n past some point).
If Σ bₙ converges, then Σ aₙ converges. Smaller than something finite is finite.
If Σ aₙ diverges, then Σ bₙ diverges. Bigger than something infinite is infinite.
The other two directions tell you nothing. Being smaller than a divergent series, or bigger than a convergent one, doesn't settle anything.
Limit comparison test
Suppose aₙ > 0 and bₙ > 0. If lim (n → ∞) aₙ/bₙ = L, where L is a finite positive number (0 < L < ∞), then Σ aₙ and Σ bₙ both converge or both diverge.
Pick bₙ by keeping only the dominant terms of aₙ; then the limit is usually a simple number like 1 or 2. This is easier to use than direct comparison when the inequality goes the wrong way. It just asks whether the two series' terms behave alike for large n.
Choosing the benchmark
- Keep only the highest powers of n in the numerator and denominator. (2n + 1)/(n³ + 4) behaves like 2n/n³ = 2/n², so compare with Σ 1/n².
- For exponentials, compare with a geometric series: 1/(3ⁿ + 1) behaves like (1/3)ⁿ.
- Logs grow slower than any power: ln n < n for n ≥ 1, and ln n > 1 for n ≥ 3.
- Bounded pieces like sin² n or 1/(2 + cos n) can be bounded by constants for a direct comparison.
- Example with a geometric benchmark: 0 < 1/(3ⁿ + 1) < (1/3)ⁿ, and Σ (1/3)ⁿ converges (r = 1/3), so Σ 1/(3ⁿ + 1) converges.
Writing the justification
For direct comparison, state the inequality and why it holds, name the benchmark and say whether it converges. For example: “0 < 1/(n² + 3) < 1/n² for all n ≥ 1, and Σ 1/n² is a convergent p-series (p = 2), so Σ 1/(n² + 3) converges by the direct comparison test.”
For limit comparison, show the limit computation, state that L is positive and finite, and name what the benchmark does. Both tests require positive terms. For series with negative terms, look at absolute values (10.9) or use the alternating series test.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Direct comparison: converges
Does Σ (n = 1 to ∞) 1/(n² + 3) converge?
Show the solutionHide the solution
- Step 1: For all n ≥ 1, n² + 3 > n², so 0 < 1/(n² + 3) < 1/n².
- Step 2: Σ 1/n² is a p-series with p = 2 > 1, so it converges.
- Step 3: By the direct comparison test, Σ 1/(n² + 3) converges.
Answer: Converges (direct comparison with Σ 1/n²).
- Example 2
Direct comparison: diverges
Does Σ (n = 3 to ∞) (ln n)/n converge?
Show the solutionHide the solution
- Step 1: For n ≥ 3, ln n > 1, so (ln n)/n > 1/n > 0.
- Step 2: Σ 1/n (from n = 3) diverges, since it's the harmonic series without its first two terms.
- Step 3: The given series is bigger term by term than a divergent series, so it diverges by the direct comparison test.
Answer: Diverges (direct comparison with Σ 1/n).
- Example 3
Trap: the inequality points the wrong way
Does Σ (n = 1 to ∞) 1/(n² − n + 1) converge?
Show the solutionHide the solution
- Step 1: Tempting: compare with 1/n². But for n ≥ 2, n² − n + 1 < n², so 1/(n² − n + 1) > 1/n². Being bigger than a convergent series proves nothing.
- Step 2: Use limit comparison with bₙ = 1/n² instead.
- Step 3: lim (n → ∞) [1/(n² − n + 1)] / [1/n²] = lim n²/(n² − n + 1) = 1.
- Step 4: L = 1 is positive and finite, and Σ 1/n² converges, so the given series converges by the limit comparison test.
Answer: Converges (limit comparison with Σ 1/n², L = 1).
Common mistakes
- Using a direct comparison in the useless direction: smaller than divergent, or bigger than convergent.
- Concluding from a limit comparison when L = 0 or L = ∞. The basic test needs 0 < L < ∞.
- Applying comparison tests to series with negative terms.
- Not stating the inequality or not naming the benchmark.
On the exam
- On free response, comparison arguments are graded on the inequality (or limit), the benchmark's behavior and the conclusion. Include all three.
- If you can't get an inequality to go the right way quickly, switch to limit comparison.
Connected topics
Videos
Check yourself
4 questions on 10.6 Comparison Tests for Convergence. Pick an answer to see if you got it, and why.
Which of the following correctly applies the limit comparison test to the series Σ (n = 1 to ∞) (2n + 1)/(n³ − n + 4)?
Which of the following is a correct use of the direct comparison test?
Which of the following correctly applies the limit comparison test to the series Σ (n = 1 to ∞) (√n + 1)/(n² + 3)?
Which of the following correctly describes the series Σ (n = 1 to ∞) (2 + sin n)/n?
0 of 4 answered