Aim for FiveCalculus BC

Unit 10 · Topic 10.4

BC only

10.4 Integral Test for Convergence

BC only. If a series' terms come from a function f that is positive, continuous and decreasing, then the series and the improper integral of f either both converge or both diverge. The integral's value is not the series' sum; it only tells you whether the series converges.

Key terms

  • integral test
  • improper integral
  • positive, continuous, decreasing

The test

This whole unit is BC only. Suppose aₙ = f(n), where f is positive, continuous and decreasing on [k, ∞) for some k. Then Σ (n = k to ∞) aₙ and ∫ from k to ∞ of f(x) dx either both converge or both diverge.

All three conditions matter, and you should state them when you use the test. Decreasing only needs to hold eventually (for x beyond some point); you can start the integral there.

Why it works

Draw the graph of f and rectangles of width 1 at x = 1, 2, 3, … . With heights f(1), f(2), f(3), … (left endpoints), the rectangles sit above the decreasing curve, so the sum is bigger than the area under the curve. With heights f(2), f(3), … placed one step to the left (right endpoints), they sit below the curve. The sum is trapped near the integral. If the area is finite, the sum is finite, and if the area is infinite, so is the sum.

Checking the conditions

  • Positive: check the sign of f(x) for x ≥ k.
  • Continuous: no division by zero or other breaks for x ≥ k.
  • Decreasing: show f′(x) < 0 for x ≥ k, or argue that the denominator grows while the numerator stays fixed.
  • Then evaluate the improper integral with a limit (6.13).

Setting up the integral

Replace n with x to get f(x), and start the integral at the series' starting index (or at the point where f begins to decrease). Then write the improper integral as a limit. For example, Σ (n = 2 to ∞) 1/(n ln n) becomes lim (b → ∞) ∫ from 2 to b of 1/(x ln x) dx. Whatever you conclude about the integral, you conclude about the series.

The sum isn't the integral

∫ from 1 to ∞ of 1/(x² + 1) dx = π/4 ≈ 0.785, but Σ (n = 1 to ∞) 1/(n² + 1) ≈ 1.077. The test tells you only that the series converges, not what it converges to.

The integral test is the reason p-series behave the way they do (10.5). It works best when f has an antiderivative you can find, like 1/(x ln x) or x e^(−x²). For terms with factorials or (−1)ⁿ, use a different test.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Converges

    Use the integral test to decide whether Σ (n = 1 to ∞) 1/(n² + 1) converges.

    Show the solution
    1. Step 1: Let f(x) = 1/(x² + 1). For x ≥ 1, f is positive, continuous and decreasing (the denominator increases).
    2. Step 2: ∫ from 1 to ∞ of 1/(x² + 1) dx = lim (b → ∞) [arctan b − arctan 1] = π/2 − π/4 = π/4.
    3. Step 3: The integral converges, so the series converges by the integral test.
    4. Step 4: Note: the sum is about 1.077, not π/4.

    Answer: Converges (integral test; the integral equals π/4).

  2. Example 2

    Diverges

    Does Σ (n = 2 to ∞) 1/(n ln n) converge or diverge?

    Show the solution
    1. Step 1: f(x) = 1/(x ln x) is positive and continuous for x ≥ 2, and decreasing because x ln x increases.
    2. Step 2: ∫ from 2 to ∞ of 1/(x ln x) dx: let u = ln x, du = dx/x. The antiderivative is ln(ln x).
    3. Step 3: lim (b → ∞) [ln(ln b) − ln(ln 2)] = ∞.
    4. Step 4: The integral diverges, so the series diverges.

    Answer: Diverges (integral test).

  3. Example 3

    Trap: decreasing only eventually

    Does Σ (n = 1 to ∞) n e^(−n²) converge?

    Show the solution
    1. Step 1: f(x) = x e^(−x²). f′(x) = e^(−x²)(1 − 2x²), which is negative for x > 1/√2 ≈ 0.707. So f is decreasing for x ≥ 1, and also positive and continuous there.
    2. Step 2: ∫ from 1 to ∞ of x e^(−x²) dx = lim (b → ∞) [−½e^(−b²) + ½e^(−1)] = 1/(2e).
    3. Step 3: The integral converges, so the series converges.
    4. Step 4: Students sometimes skip the decreasing check because the function isn't decreasing near 0. You only need it on the interval you use.

    Answer: Converges (integral test; f is positive, continuous and decreasing for x ≥ 1).

Common mistakes

  • Saying the series converges to the integral's value.
  • Not checking or not stating that f is positive, continuous and decreasing.
  • Using the integral test on an alternating series. The terms must be positive.
  • Treating ∞ as a number instead of writing a limit.

On the exam

  • When you use the integral test on free response, state the three conditions, evaluate the improper integral with limit notation, and then make the conclusion about the series.
  • Multiple-choice questions may ask which integral could be used to test a given series. Match f(n) = aₙ and the starting value.

Connected topics

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Check yourself

4 questions on 10.4 Integral Test for Convergence. Pick an answer to see if you got it, and why.

Question 1 of 4

The function f(x) = 1/(x(ln x)²) is positive, continuous, and decreasing for x ≥ 2. Which of the following correctly describes the series Σ (n = 2 to ∞) 1/(n(ln n)²)?

Question 2 of 4

For which of the following series can the integral test NOT be used to determine convergence?

Question 3 of 4

Which of the following correctly describes the series Σ (n = 2 to ∞) 1/(n ln n)?

Question 4 of 4

For which values of p does the series Σ (n = 2 to ∞) 1/(n(ln n)ᵖ) converge?

0 of 4 answered