Integral Test for Convergence: Study Guide (Calculus BC 10.4)

BC-only material. The integral test is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need four earlier ideas. If any is shaky, revisit it first. The Calculus BC roadmap shows where each one sits.

Prerequisite	Topic	What you use it for here
Improper integrals	6.13	The test compares a series with ∫ from N to ∞ of f(x) dx; see the Topic 6.13 study guide
Substitution and parts	6.9, 6.11	Finding the antiderivative
Decreasing functions	5.3	Showing f′(x) < 0 to check a condition
The nth term test	10.3	The quick first check; see the Topic 10.3 study guide

Notation on this page: ∑ from n = 1 to ∞ of aₙ is a series and ∫ from 1 to ∞ of f(x) dx means the limit as b → ∞ of ∫ from 1 to b of f(x) dx. The integral converges if that limit is a finite number and diverges otherwise.

The idea: a series is a sum of rectangle areas

The nth term test (Topic 10.3) can only prove divergence. When the terms approach 0, you need something sharper. The integral test uses a picture.

Suppose the terms come from a function: aₙ = f(n). Draw a rectangle of width 1 and height aₙ for each term. The total area of the rectangles is the sum of the series. If f is positive and decreasing, the rectangles hug the curve y = f(x), either just above it or just below it.

Why the two pictures settle the question. Let I = ∫ from 1 to ∞ of f(x) dx.

• If I is finite. The right-hand picture says a₂ + a₃ + … + aₙ ≤ I for every n. The partial sums are increasing (all terms are positive) and they never pass a₁ + I. An increasing sequence with a ceiling has a finite limit, so the series converges.
• If I is infinite. The left-hand picture says a₁ + a₂ + … + aₙ is at least ∫ from 1 to n + 1 of f(x) dx. That integral grows without bound, so the partial sums do too. The series diverges.

Both cases need f to be positive (so that areas and partial sums only grow) and decreasing (so that the rectangles really stay on one side of the curve).

The integral test

Integral test. Suppose aₙ = f(n), where f is positive, continuous and decreasing for x ≥ N. Then ∑ from n = N to ∞ of aₙ and ∫ from N to ∞ of f(x) dx either both converge or both diverge.

Three points to notice.

1. “Decreasing for x ≥ N” is enough. The first few terms can do anything. Adding or removing finitely many terms changes the sum but never changes whether the series converges.
2. The test gives the behaviour, not the sum. In Figure 1 the integral is about 0.785 and the sum is about 1.077. They are different numbers.
3. Always check the conditions in writing. In a free-response answer, a correct integral with no mention of the conditions is an incomplete justification.

How to show “decreasing”. Usually find f′(x) and show it is negative for x ≥ N. Sometimes a quick argument works: if the denominator of a positive fraction increases and the numerator is constant, the fraction decreases.

Worked example 1: a convergent series, using substitution

Question. Use the integral test to decide whether ∑ from n = 1 to ∞ of n e^(−n²) converges.

1. Choose f. f(x) = x e^(−x²), so that f(n) = aₙ.
2. Check the conditions for x ≥ 1.
   • Positive: x > 0 and e^(−x²) > 0. ✓
   • Continuous: a product of continuous functions. ✓
   • Decreasing: f′(x) = e^(−x²) − 2x² e^(−x²) = e^(−x²)(1 − 2x²). For x ≥ 1, 1 − 2x² ≤ −1 < 0, so f′(x) < 0. ✓
3. Evaluate the improper integral. Substitute u = x², du = 2x dx: ∫ x e^(−x²) dx = −½ e^(−x²) + C. ∫ from 1 to b of x e^(−x²) dx = [−½ e^(−x²)] from 1 to b = ½e⁻¹ − ½e^(−b²).
4. Take the limit. As b → ∞, e^(−b²) → 0, so the integral converges to ½e⁻¹ = 1/(2e) ≈ 0.184.
5. Conclude. Since f is positive, continuous and decreasing on [1, ∞) and ∫ from 1 to ∞ of f(x) dx converges, ∑ n e^(−n²) converges by the integral test.

Check. The first terms are a₁ ≈ 0.368, a₂ ≈ 0.0366, a₃ ≈ 0.00037. They shrink very fast, so convergence is believable. The sum is about 0.405, not 0.184: the integral only told us that it converges.

Worked example 2: a divergent series, decreasing only from n = 2

Question. Does ∑ from n = 1 to ∞ of 2n/(n² + 4) converge or diverge? Justify your answer.

1. Try the nth term test first. 2n/(n² + 4) → 0 (higher power on the bottom). Inconclusive, so a sharper test is needed.
2. Choose f. f(x) = 2x/(x² + 4). It is positive and continuous for x ≥ 1.
3. Is it decreasing? By the quotient rule, f′(x) = [2(x² + 4) − 2x · 2x]/(x² + 4)² = (8 − 2x²)/(x² + 4)². f′(x) < 0 only when x² > 4, that is, for x > 2. Check the terms: a₁ = 2/5 = 0.4, a₂ = 1/2 = 0.5, a₃ = 6/13 ≈ 0.462. The terms rise and then fall. So f is decreasing for x ≥ 2, and we apply the test with N = 2.
4. Evaluate the improper integral from 2. Substitute u = x² + 4: ∫ from 2 to b of 2x/(x² + 4) dx = [ln(x² + 4)] from 2 to b = ln(b² + 4) − ln 8.
5. Take the limit. As b → ∞, ln(b² + 4) → ∞. The integral diverges.
6. Conclude. f is positive, continuous and decreasing for x ≥ 2, and ∫ from 2 to ∞ of f(x) dx diverges, so ∑ from n = 2 to ∞ of aₙ diverges by the integral test. Adding the single term a₁ = 0.4 cannot make it converge, so ∑ from n = 1 to ∞ of 2n/(n² + 4) diverges.

Interpretation. The integral from 2 to b grows like a logarithm: about 2.57 at b = 10, 7.13 at b = 100 and 11.74 at b = 1000. Slow, but without limit. The terms do approach 0, which is exactly the case where the nth term test is silent.

Worked example 3: using integration by parts

Question. Use the integral test to show that ∑ from n = 1 to ∞ of (ln n)/n³ converges.

1. First term. a₁ = (ln 1)/1 = 0, so the series has the same behaviour (and the same sum) as ∑ from n = 2 to ∞ of (ln n)/n³.
2. Choose f and check conditions for x ≥ 2. f(x) = (ln x)/x³.
   • Positive: ln x > 0 for x > 1. ✓
   • Continuous: for x > 0. ✓
   • Decreasing: f′(x) = [(1/x) · x³ − ln x · 3x²]/x⁶ = (1 − 3 ln x)/x⁴. This is negative when ln x > 1/3, that is, x > e^(1/3) ≈ 1.40. So f is decreasing for x ≥ 2. ✓
3. Antiderivative by parts. u = ln x (du = dx/x), dv = x⁻³ dx (v = −1/(2x²)): ∫ (ln x)/x³ dx = −(ln x)/(2x²) + ∫ 1/(2x³) dx = −(ln x)/(2x²) − 1/(4x²) + C.
4. Evaluate from 2 to b. [−(ln x)/(2x²) − 1/(4x²)] from 2 to b = −(ln b)/(2b²) − 1/(4b²) + (ln 2)/8 + 1/16.
5. Take the limit. (ln b)/b² → 0 by L’Hospital’s Rule: (1/b)/(2b) = 1/(2b²) → 0. And 1/(4b²) → 0. So the integral converges to (ln 2)/8 + 1/16 ≈ 0.149.
6. Conclude. The conditions hold for x ≥ 2 and ∫ from 2 to ∞ of f(x) dx converges, so ∑ (ln n)/n³ converges by the integral test.

Check. d/dx [−(ln x)/(2x²) − 1/(4x²)] = −1/(2x³) + (ln x)/x³ + 1/(2x³) = (ln x)/x³. ✓ The sum is about 0.198, which is between the integral (0.149) and the integral plus a₂ (0.149 + 0.087 = 0.236), as the rectangle pictures predict.

Background: trapping the sum between two integrals

This is not required by the course framework, but it follows straight from Figure 1 and is a good check on your work. If f is positive, continuous and decreasing for x ≥ 1 and the integral converges, then

∫ from 1 to ∞ of f(x) dx ≤ ∑ from n = 1 to ∞ of aₙ ≤ a₁ + ∫ from 1 to ∞ of f(x) dx.

For Worked example 1: 0.184 ≤ sum ≤ 0.368 + 0.184 = 0.552. The true sum, about 0.405, sits inside.

When the integral test does not fit

• Terms that change sign, such as (−1)ⁿ/n or (sin n)/n². f is not positive. Use the tests of Topics 10.7 and 10.9.
• Terms that wobble, such as (2 + cos n)/n². These are positive but not decreasing. A comparison test (Topic 10.6) is the right tool.
• No usable f or antiderivative, for example 1/n! (n! has no simple formula for non-whole x) or e^(−n²) (e^(−x²) has no elementary antiderivative). Even where the conditions hold, you cannot evaluate the integral. Choose another test.

Common misconceptions

• “The series converges to the value of the integral.” No. In Worked example 1 the integral is 1/(2e) ≈ 0.184 but the sum is about 0.405.
• Skipping the conditions. Write “f is positive, continuous and decreasing for x ≥ N” and show why, especially “decreasing”.
• Thinking “decreasing” must hold from n = 1. It only has to hold from some N onwards (Worked example 2).
• Using the test on terms that are not positive, such as an alternating series.
• Stopping at the antiderivative. The integral is improper. You must take the limit as b → ∞.
• Saying “the integral converges because f(x) → 0”. That is the same mistake as in Topic 10.3. In Worked example 2, f(x) → 0 but the integral diverges.
• Mixing up the conclusions. A divergent integral means a divergent series, and a convergent integral means a convergent series. Neither direction gives the sum.

Where this leads

The integral test proves the most useful family of results in the unit: the p-series rule, which says when ∑ 1/nᵖ converges. Continue with Topic 10.5, Harmonic Series and p-Series, or see the order on the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.