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Infinite Series

9.5 Alternating Series

Focus on alternating series test, alternating error bound, absolute and conditional convergence.

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Ideas in this lesson

01Alternating Series TestWorked steps · five questions02Alternating error boundWorked steps · five questions03Absolute and conditional convergenceWorked steps · five questions
Idea 1 of 3

Alternating Series Test

Decreasing term magnitudes tending to zero guarantee convergence of an alternating series.

Worked example 1

Classify ∑(n=1\sum (n = 1 to ∞)2⋅−1n+1n\infty ) \dfrac{2 \cdot -1^{n + 1}}{n}.
  1. bn=1nb_{n} = \dfrac{1}{n} ↓ 00; bn+1<bnb_{n+1} < b_{n}
  2. ∑(−1)n+1bn\sum (-1)^{n+1}b_{n} converges by the Alternating Series Test
  3. ∑∣an∣=∑1n\sum |a_{n}| = \sum \dfrac{1}{n} diverges (p≤1)  ⟹  (p \le 1) \implies Conditionally convergent

Answer: Conditionally convergent

Worked example 2

Classify ∑(n=1\sum (n = 1 to ∞)3⋅−1n+1n2\infty ) \dfrac{3 \cdot -1^{n + 1}}{n^{2}}.
  1. bn=1n2b_{n} = \dfrac{1}{n^{2}} ↓ 00; bn+1<bnb_{n+1} < b_{n}
  2. ∑(−1)n+1bn\sum (-1)^{n+1}b_{n} converges by the Alternating Series Test
  3. ∑∣an∣=∑1n2\sum |a_{n}| = \sum \dfrac{1}{n^{2}} converges (p>1)  ⟹  (p > 1) \implies Absolutely convergent

Answer: Absolutely convergent

Idea 2 of 3

Alternating error bound

The error after nn terms is no larger than the magnitude of term n+1n + 1 under the test’ss hypotheses.

Worked example 1

After 22 terms of ∑−1n+1n\sum \dfrac{-1^{n + 1}}{n}, give the Alternating Series error bound.
  1. ∣R2∣≤b3|R_2| \le b_3
  2. ∣R2∣≤13=13≈0.33|R_2| \le \dfrac{1}{3} = \dfrac{1}{3} \approx 0.33

Answer: 13≈0.33\dfrac{1}{3} \approx 0.33

Worked example 2

After 33 terms of ∑−1n+1n2\sum \dfrac{-1^{n + 1}}{n^{2}}, give the Alternating Series error bound.
  1. ∣R3∣≤b4|R_3| \le b_4
  2. ∣R3∣≤142=116≈0.06|R_3| \le \dfrac{1}{4^{2}} = \dfrac{1}{16} \approx 0.06

Answer: 116≈0.06\dfrac{1}{16} \approx 0.06

Idea 3 of 3

Absolute and conditional convergence

Test the absolute-value series separately. Conditional convergence means the signed series converges but the absolute series does not.

Worked example 1

Classify ∑\sum from n=1n = 1 to ∞\infty of 2(−1)n+1n\dfrac{2(-1)^{n + 1}}{n}.
  1. bn=2nb_{n} = \dfrac{2}{n}; bn+1≤bnb_{n+1} \le b_{n}; lim⁡bn=0  ⟹  \displaystyle \lim b_{n} = 0 \implies alternating series converges
  2. ∑∣an∣=∑2n\sum |a_{n}| = \sum \dfrac{2}{n} diverges (p=1)(p = 1)
  3. Classification == Conditionally convergent

Answer: Conditionally convergent

Worked example 2

Classify ∑\sum from n=1n = 1 to ∞\infty of 3(−1)n+1n2\dfrac{3(-1)^{n + 1}}{n^{2}}.
  1. bn=3n2b_{n} = \dfrac{3}{n^{2}}; bn+1≤bnb_{n+1} \le b_{n}; lim⁡bn=0  ⟹  \displaystyle \lim b_{n} = 0 \implies alternating series converges
  2. ∑∣an∣=∑3n2\sum |a_{n}| = \sum \dfrac{3}{n^{2}} converges (p=2>1)(p = 2 > 1)
  3. Classification == Absolutely convergent

Answer: Absolutely convergent

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.5,p9.5, p. 619619. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.