Comparison tests

About this lesson

Calculus BC (AP-aligned) › Infinite series and convergence tests

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Comparison tests

Warm-up

Three quick questions from earlier lessons.

Discover

0<1n2+5<1n20 < \dfrac{1}{n^2 + 5} < \dfrac{1}{n^2} and ∑1n2\sum \dfrac{1}{n^2} converges. What can you say about ∑1n2+5\sum \dfrac{1}{n^2 + 5}?

  1. It converges
  2. Nothing

Key idea

Compare with a series you know.

Direct comparison (0≤an≤bn0 \le a_n \le b_n): if ∑bn\sum b_n converges, so does ∑an\sum a_n; if ∑an\sum a_n diverges, so does ∑bn\sum b_n. Limit comparison: if an,bn>0a_n, b_n > 0 and an/bn→ca_n/b_n \to c with 0<c<∞0 < c < \infty, both series converge or both diverge.

Worked example

∑3n+2n3−n+4\displaystyle\sum \dfrac{3n + 2}{n^3 - n + 4}.

It behaves like 3n2\dfrac{3}{n^2}
Leading terms: 3n/n³.
Limit comparison with ∑1n2\sum \dfrac{1}{n^2}
Ratio with 1/n² → 3.
Converges
p = 2 > 1.

Your turn, guided

Your turn: ∑nn2+1\displaystyle\sum \dfrac{n}{n^2 + 1} behaves like ∑1np\sum \dfrac{1}{n^{p}} with…

p=1p = 1: diverges
n/n² = 1/n: diverges.

Practice ★

  1. Does ∑n=1∞1n2+7\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 7} converge?
    1. Yes: it is smaller than the convergent ∑ 1/n²
    2. No: it is smaller than the divergent ∑ 1/n
    3. No test applies
  1. Does ∑n=1∞2n+1n3+8\displaystyle\sum_{n=1}^{\infty} \dfrac{2n + 1}{n^3 + 8} converge?
    1. Yes: it behaves like ∑ 2/n², a convergent p-series
    2. No: the top grows
    3. No: it behaves like ∑ 1/n

Watch out!

Practice ★★

  1. Does ∑n=1∞1n+8\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n} + 8} converge?
    1. No: it behaves like ∑ 1/√n, which diverges (limit comparison)
    2. Yes: it is smaller than ∑ 1/√n
    3. Yes: its terms tend to 0
  2. Does ∑n=1∞1n+6\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n} + 6} converge?
    1. No: it behaves like ∑ 1/√n, which diverges (limit comparison)
    2. Yes: it is smaller than ∑ 1/√n
    3. Yes: its terms tend to 0

Exit ticket

Three questions to check you have got it.

  1. Does ∑n=1∞1n2+7\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 7} converge?
    1. Yes: it is smaller than the convergent ∑ 1/n²
    2. No: it is smaller than the divergent ∑ 1/n
    3. No test applies
  2. Does ∑n=1∞1n2+8\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 8} converge?
    1. Yes: it is smaller than the convergent ∑ 1/n²
    2. No: it is smaller than the divergent ∑ 1/n
    3. No test applies
  3. Does ∑n=1∞1n2+9\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 9} converge?
    1. Yes: it is smaller than the convergent ∑ 1/n²
    2. No: it is smaller than the divergent ∑ 1/n
    3. No test applies

Summary card

Comparison tests

  • Below convergent: converges. Above divergent: diverges.
  • Same leading behavior: limit comparison.

Watch out: below a divergent series tells you nothing.

End of the lesson

Next lesson: Alternating series and the error bound Back to the chapter

Answers: Comparison tests

A

  1. Yes: it is smaller than the convergent ∑ 1/n²

B

  1. Yes: it behaves like ∑ 2/n², a convergent p-series

C

  1. No: it behaves like ∑ 1/√n, which diverges (limit comparison)
  2. No: it behaves like ∑ 1/√n, which diverges (limit comparison)

D

  1. Yes: it is smaller than the convergent ∑ 1/n²
  2. Yes: it is smaller than the convergent ∑ 1/n²
  3. Yes: it is smaller than the convergent ∑ 1/n²

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