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Infinite Series

9.2 Series and Convergence

Focus on zero term limit is inconclusive, geometric series, telescoping sums, nth-term divergence test.

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Ideas in this lesson

01Zero term limit is inconclusiveWorked steps · five questions02Geometric seriesWorked steps · five questions03Telescoping sumsWorked steps · five questions04Nth-term divergence testWorked steps · five questions
Idea 1 of 4

Zero term limit is inconclusive

A zero term limit is necessary for convergence but does not prove it. The harmonic series is a counterexample.

Worked example 1

For an=2n,lim⁡an=0\displaystyle a_{n} = \dfrac{2}{n}, \lim a_{n} = 0. What does the nth-term test conclude about ∑\sum an_{n}?
  1. lim⁡n→∞2n=0\displaystyle \lim\limits_{n\to \infty } \dfrac{2}{n} = 0
  2. lim⁡an=0\displaystyle \lim a_{n} = 0 does not imply ∑\sum an_{n} converges
  3. ∑2n\sum \dfrac{2}{n} diverges by the pp-Series Test (p=1)(p = 1)

Answer: Inconclusive

Worked example 2

For an=3n,lim⁡an=0\displaystyle a_{n} = \dfrac{3}{n}, \lim a_{n} = 0. What does the nth-term test conclude about ∑\sum an_{n}?
  1. lim⁡n→∞3n=0\displaystyle \lim\limits_{n\to \infty } \dfrac{3}{n} = 0
  2. lim⁡an=0\displaystyle \lim a_{n} = 0 does not imply ∑\sum an_{n} converges
  3. ∑3n\sum \dfrac{3}{n} diverges by the pp-Series Test (p=1)(p = 1)

Answer: Inconclusive

Idea 2 of 4

Geometric series

For ∣r∣<1|r| < 1, the sum is first term divided by 1−r1 - r. Check the starting index to identify the first term.

Worked example 1

Find the sum of ∑∞(n=0\sum ^{\infty } (n = 0 onward )2(12)n) 2(\dfrac{1}{2})^{n}.
  1. a0=2a_{0} = 2; r=12r = \dfrac{1}{2}; ∣r∣<1|r| < 1
  2. S=21−12=4S = \dfrac{2}{1 - \dfrac{1}{2}} = 4

Answer: 44

Worked example 2

Find the sum of ∑∞(n=0\sum ^{\infty } (n = 0 onward )3(13)n) 3(\dfrac{1}{3})^{n}.
  1. a0=3a_{0} = 3; r=13r = \dfrac{1}{3}; ∣r∣<1|r| < 1
  2. S=31−13=92≈4.5S = \dfrac{3}{1 - \dfrac{1}{3}} = \dfrac{9}{2} \approx 4.5

Answer: 92≈4.5\dfrac{9}{2} \approx 4.5

Idea 3 of 4

Telescoping sums

Write partial sums before taking a limit. Cancellation leaves a small number of boundary terms.

Worked example 1

Find ∑(n=1\sum (n = 1 to 2)2(1n−1n+1)2) 2(\dfrac{1}{n} - \dfrac{1}{n + 1}).
  1. S2=2[1−12+12−13+S_2 = 2[1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + … +12−13]+ \dfrac{1}{2} - \dfrac{1}{3}]
  2. S2=2(1−13)=43≈1.33S_2 = 2(1 - \dfrac{1}{3}) = \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Worked example 2

Find ∑(n=1\sum (n = 1 to 3)3(1n−1n+1)3) 3(\dfrac{1}{n} - \dfrac{1}{n + 1}).
  1. S3=3[1−12+12−13+S_3 = 3[1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + … +13−14]+ \dfrac{1}{3} - \dfrac{1}{4}]
  2. S3=3(1−14)=94≈2.25S_3 = 3(1 - \dfrac{1}{4}) = \dfrac{9}{4} \approx 2.25

Answer: 94≈2.25\dfrac{9}{4} \approx 2.25

Idea 4 of 4

Nth-term divergence test

A nonzero or nonexistent term limit proves divergence. A zero term limit does not prove convergence.

Worked example 1

For ∑(n=1\sum (n = 1 to ∞)2nn+1\infty ) \dfrac{2n}{n + 1}, what does the nth-term test show?
  1. lim⁡n→∞2nn+1=2≠0\displaystyle \lim\limits_{n\to \infty } \dfrac{2n}{n + 1} = 2 \ne 0
  2. ∑\sum an_{n} convergent   ⟹  lim⁡an=0\displaystyle \implies \lim a_{n} = 0; 2≠0  ⟹  2 \ne 0 \implies divergent

Answer: The series diverges

Worked example 2

For ∑(n=1\sum (n = 1 to ∞)3nn+1\infty ) \dfrac{3n}{n + 1}, what does the nth-term test show?
  1. lim⁡n→∞3nn+1=3≠0\displaystyle \lim\limits_{n\to \infty } \dfrac{3n}{n + 1} = 3 \ne 0
  2. ∑\sum an_{n} convergent   ⟹  lim⁡an=0\displaystyle \implies \lim a_{n} = 0; 3≠0  ⟹  3 \ne 0 \implies divergent

Answer: The series diverges

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.2,p9.2, p. 595595. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.