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Infinite Series

9.8 Power Series

Focus on radius of convergence, endpoint tests, coefficients and geometric expansions.

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Ideas in this lesson

01Radius of convergenceWorked steps · five questions02Endpoint testsWorked steps · five questions03Coefficients and geometric expansionsWorked steps · five questions
Idea 1 of 3

Radius of convergence

A ratio or root test usually identifies the open interval ∣x−a∣<R|x - a| < R. The radius does not decide endpoints.

Worked example 1

Find the radius of convergence of ∑(n=0\sum (n = 0 to ∞)(x−1)n2n\infty ) \dfrac{(x - 1)^{n}}{2^{n}}.
  1. r=x−12r = \dfrac{x - 1}{2}
  2. ∣r∣<1|r| < 1 ⇔ ∣x−1∣<2|x - 1| < 2
  3. Radius R=2R = 2

Answer: 22

Worked example 2

Find the radius of convergence of ∑(n=0\sum (n = 0 to ∞)(x−2)n3n\infty ) \dfrac{(x - 2)^{n}}{3^{n}}.
  1. r=x−23r = \dfrac{x - 2}{3}
  2. ∣r∣<1|r| < 1 ⇔ ∣x−2∣<3|x - 2| < 3
  3. Radius R=3R = 3

Answer: 33

Idea 2 of 3

Endpoint tests

Substitute each endpoint into the original series and apply an appropriate convergence test.

Worked example 1

Find the interval of convergence of ∑(n=1\sum (n = 1 to ∞)(x−1)nn2n\infty ) \dfrac{(x - 1)^{n}}{n2^{n}}.
  1. ∣x−1∣<2  ⟹  −1<x<3|x - 1| < 2 \implies -1 < x < 3
  2. x=−1  ⟹  ∑(−1)nnx = -1 \implies \sum \dfrac{(-1)^{n}}{n} converges
  3. x=3  ⟹  ∑1nx = 3 \implies \sum \dfrac{1}{n} diverges
  4. Interval of convergence: x∈[−1,3)x \in [-1, 3)

Answer: [−1,3)[-1, 3)

Worked example 2

Find the interval of convergence of ∑(n=1\sum (n = 1 to ∞)(x−2)nn3n\infty ) \dfrac{(x - 2)^{n}}{n3^{n}}.
  1. ∣x−2∣<3  ⟹  −1<x<5|x - 2| < 3 \implies -1 < x < 5
  2. x=−1  ⟹  ∑(−1)nnx = -1 \implies \sum \dfrac{(-1)^{n}}{n} converges
  3. x=5  ⟹  ∑1nx = 5 \implies \sum \dfrac{1}{n} diverges
  4. Interval of convergence: x∈[−1,5)x \in [-1, 5)

Answer: [−1,5)[-1, 5)

Idea 3 of 3

Coefficients and geometric expansions

Use 11−u=∑un\frac{1}{1 - u} = \sum u^{n} inside ∣u∣<1|u| < 1. Replacing uu changes both coefficients and the convergence interval.

Worked example 1

In the Maclaurin expansion of 11−2x\dfrac{1}{1 - 2x}, find the coefficient of x2x^{2}.
  1. 11−2x=∑(2x)n,∣x∣<12\dfrac{1}{1 - 2x} = \sum (2x)^{n}, |x| < \dfrac{1}{2}
  2. (2x)n=2nxn(2x)^{n} = 2^{n}x^{n}
  3. Coefficient of x2=22=4x^{2} = 2^{2} = 4

Answer: 44

Worked example 2

In the Maclaurin expansion of 11−3x\dfrac{1}{1 - 3x}, find the coefficient of x3x^{3}.
  1. 11−3x=∑(3x)n,∣x∣<13\dfrac{1}{1 - 3x} = \sum (3x)^{n}, |x| < \dfrac{1}{3}
  2. (3x)n=3nxn(3x)^{n} = 3^{n}x^{n}
  3. Coefficient of x3=33=27x^{3} = 3^{3} = 27

Answer: 2727

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.8,p9.8, p. 647647. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.