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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
- nth-term test: if aₙ doesn't go to 0, the series diverges. (If it does, the test says nothing.)
- Geometric series Σ arⁿ converge when |r| < 1.
- p-series Σ 1/nᵖ converge when p > 1.
- Comparison and limit comparison: compare to a series you already know.
- Ratio test: lim |aₙ₊₁/aₙ| 1 diverges, = 1 inconclusive. Great for factorials and exponentials.
- Alternating series test: terms decrease to 0 ⇒ converges. Absolute vs conditional convergence.
Formulas and rules
Common mistakes
- Using the nth-term test to prove convergence.
- Forgetting the harmonic series Σ 1/n diverges.
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.
Σ (1/2)ⁿ for n = 0 to ∞
- Diverges
- 1/2
- 1
- 2
Show the answer and steps
Answer: D) 2
- Geometric: a = 1, r = 1/2.
- Sum = 1/(1 − 1/2) = 2.
Find Σ from n = 0 to ∞ of 9(−3/4)ⁿ.
- Diverges
- 36/7
- 36
- −27/4
Show the answer and steps
Answer: B) 36/7
- Geometric series with first term 9 and ratio −3/4; |r| < 1 so it converges.
- Sum = 9/(1 − (−3/4)) = 36/7.
Σ 1/n²
- Converges (p-series with p = 2)
- Test is inconclusive
- Diverges by the nth-term test
- Diverges like the harmonic series
Show the answer and steps
Answer: A) Converges (p-series with p = 2)
- Σ 1/n² is a p-series with p = 2.
- p > 1, so it converges.