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Calculus II (Integration & Series) Unit 8: Sequences & convergence tests

An infinite series adds infinitely many terms. Convergence tests tell you whether the sum is a finite number.

Covers: Sequences, geometric, nth-term, integral, p-series, comparison, ratio, alternating series tests

Study guide

A sequence is an ordered list a₁, a₂, …; it converges if its terms approach a finite limit. A series adds the terms of a sequence: Σ aₙ. Its partial sums Sₙ form a sequence, and the series converges if those partial sums approach a finite value.

Several tests decide convergence. The nth-term test: if lim aₙ ≠ 0, the series diverges (but if the limit is 0, the test tells you nothing). Geometric series Σ arⁿ converge to a/(1 − r) when |r| 1; the harmonic series Σ 1/n diverges.

The integral test compares a positive, decreasing series to the integral of the matching function. The direct comparison and limit comparison tests compare a series to a known one, usually a p-series or geometric series. The ratio test computes L = lim |aₙ₊₁/aₙ|: the series converges absolutely if L 1, and the test is inconclusive if L = 1. It works best with factorials and exponentials.

The alternating series test says Σ (−1)ⁿbₙ converges if bₙ decreases to 0, and the error from stopping is at most the next term. A series converges absolutely if Σ |aₙ| converges, and conditionally if it converges but Σ |aₙ| doesn't.

When you use a test, state its conditions. The alternating series test needs terms that are positive, decreasing and approaching 0. The integral test needs f positive, continuous and decreasing. Limit comparison: if aₙ > 0, bₙ > 0 and lim aₙ/bₙ = c with 0 < c < ∞, then Σaₙ and Σbₙ both converge or both diverge. Compare with the dominant terms, such as 1/n² for (n + 1)/(n³ + 2).

To find the limit of a sequence, treat n as a continuous variable and use limit techniques, including L'Hospital's rule. Useful facts: n!/nⁿ → 0 and (1 + 1/n)ⁿ → e. A geometric series sums to (first term)/(1 − r), where the first term is whatever the first term actually is. For example, Σₙ₌₂^∞ (1/3)ⁿ = (1/9)/(1 − 1/3) = 1/6.

Key ideas

Formulas and rules

Geometric sum: a/(1 − r), |r| < 1
p-series: converges ⇔ p > 1
Ratio test: L = lim |aₙ₊₁/aₙ|
Alternating error ≤ first omitted term

Common mistakes

Words to know

nth-term test
If lim aₙ ≠ 0, the series diverges.
Harmonic series
Σ 1/n; it diverges.
P-series
Σ 1/nᵖ; converges when p > 1.
Ratio test (L = 1)
Inconclusive; try another test.
Ratio test
L = lim |aₙ₊₁/aₙ|: L 1 diverges. Good for factorials and exponentials.
Conditional convergence
The series converges but the series of absolute values diverges.
Geometric series
Σ arⁿ converges to a/(1 − r) when |r| < 1.
Sequence
An ordered list a₁, a₂, …; converges if lim aₙ is finite.
Series
Σ aₙ: the limit of the partial sums.
Partial sum
Sₙ = a₁ + a₂ + … + aₙ
Integral test
For positive decreasing aₙ = f(n), Σ aₙ and ∫ f(x) dx both converge or both diverge.
Direct comparison test
Compare to a known series: smaller than convergent converges; bigger than divergent diverges.
Limit comparison test
If lim aₙ/bₙ is positive and finite, both series do the same thing.
Alternating series test
Σ (−1)ⁿbₙ converges if bₙ decreases to 0.
Absolute convergence
Σ |aₙ| converges.

Sample practice questions

29 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

Σ (1/2)ⁿ for n = 0 to ∞

  1. Diverges
  2. 1/2
  3. 1
  4. 2
Show the answer and steps

Answer: D) 2

  1. Geometric: a = 1, r = 1/2.
  2. Sum = 1/(1 − 1/2) = 2.
Question 2 (level 1)

Find Σ from n = 0 to ∞ of 9(−3/4)ⁿ.

  1. Diverges
  2. 36/7
  3. 36
  4. −27/4
Show the answer and steps

Answer: B) 36/7

  1. Geometric series with first term 9 and ratio −3/4; |r| < 1 so it converges.
  2. Sum = 9/(1 − (−3/4)) = 36/7.
Question 3 (level 3)

Σ 1/n²

  1. Converges (p-series with p = 2)
  2. Test is inconclusive
  3. Diverges by the nth-term test
  4. Diverges like the harmonic series
Show the answer and steps

Answer: A) Converges (p-series with p = 2)

  1. Σ 1/n² is a p-series with p = 2.
  2. p > 1, so it converges.