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Infinite Series

9.3 The Integral Test and p-Series

Focus on p-series, integral test remainder, integral and series convergence.

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Ideas in this lesson

01p-SeriesWorked steps · five questions02Integral Test remainderWorked steps · five questions03Integral and series convergenceWorked steps · five questions
Idea 1 of 3

p-Series

The series ∑1n\sum \frac{1}{n}ᵖ converges exactly for p>1p > 1. The harmonic series is the divergent boundary case p=1p = 1.

Worked example 1

Classify ∑(n=1\sum (n = 1 to ∞)2n2\infty ) \dfrac{2}{n^{2}}.
  1. p=2p = 2
  2. ∑1np\sum \dfrac{1}{nᵖ} converges ⇔ p>1p > 1
  3. p>1  ⟹  p > 1 \implies Converges

Answer: Converges

Worked example 2

Classify ∑(n=1\sum (n = 1 to ∞)3n\infty ) \dfrac{3}{n}.
  1. p=1p = 1
  2. ∑1np\sum \dfrac{1}{nᵖ} converges ⇔ p>1p > 1
  3. p≤1  ⟹  p \le 1 \implies Diverges

Answer: Diverges

Idea 2 of 3

Integral Test remainder

For positive decreasing continuous ff, bound a convergent tail between integrals starting at n+1n + 1 and nn.

Worked example 1

Use the Integral Test upper bound for the tail after n=2n = 2 of ∑1n2\sum \dfrac{1}{n^{2}}.
  1. ∫3∞x−2 dx≤R2≤∫2∞x−2 dx\int_{3}^{\infty } x^{-2}\,dx \le R_2 \le \int_{2}^{\infty } x^{-2}\,dx
  2. f(x)=x−2>0f(x) = x^{-2} > 0; f′(x)=−2x−3<0f'(x) = -2x^{-3} < 0
  3. Upper bound =2−11=12≈0.5= \dfrac{2^{-1}}{1} = \dfrac{1}{2} \approx 0.5

Answer: 12≈0.5\dfrac{1}{2} \approx 0.5

Worked example 2

Use the Integral Test upper bound for the tail after n=4n = 4 of ∑1n3\sum \dfrac{1}{n^{3}}.
  1. ∫5∞x−3 dx≤R4≤∫4∞x−3 dx\int_{5}^{\infty } x^{-3}\,dx \le R_4 \le \int_{4}^{\infty } x^{-3}\,dx
  2. f(x)=x−3>0f(x) = x^{-3} > 0; f′(x)=−3x−4<0f'(x) = -3x^{-4} < 0
  3. Upper bound =4−22=132≈0.03= \dfrac{4^{-2}}{2} = \dfrac{1}{32} \approx 0.03

Answer: 132≈0.03\dfrac{1}{32} \approx 0.03

Idea 3 of 3

Integral and series convergence

Check positivity, continuity and decreasing terms before comparing the integral and series. Their convergence behavior agrees; their values need not.

Worked example 1

Use the Integral Test for ∑\sum from n=1n = 1 to ∞\infty of 2n\dfrac{2}{n}.
  1. f(x)=2x>0f(x) = \dfrac{2}{x} > 0; f′(x)=−2x2<0f'(x) = -\dfrac{2}{x^{2}} < 0
  2. ∫1t2xdx=2ln⁡t→∞\int _{1}^{t} \dfrac{2}{x} dx = 2 \ln t \to \infty
  3. p=1  ⟹  p = 1 \implies Integral Test: Diverges

Answer: Diverges

Worked example 2

Use the Integral Test for ∑\sum from n=1n = 1 to ∞\infty of 3n2\dfrac{3}{n^{2}}.
  1. f(x)=3x2>0f(x) = \dfrac{3}{x^{2}} > 0; f′(x)=−6x3<0f'(x) = -\dfrac{6}{x^{3}} < 0
  2. ∫1∞3x2dx=31<∞\int _{1}^\infty \dfrac{3}{x^{2}} dx = \dfrac{3}{1} < \infty
  3. p=2  ⟹  p = 2 \implies Integral Test: Converges

Answer: Converges

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.3,p9.3, p. 605605. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.