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10.4 - Integral Test for Convergence

Introduction

The integral test decides convergence or divergence of certain series by tying them to improper integrals.
Key idea: if you can integrate the function easily, that integral tells you what the series does.

Conditions for Using It

You can only apply the integral test when the function f(x)f(x) matching the terms ana_n satisfies three conditions on [k,∞)[k, \infty):
  1. Positive: f(x)>0f(x) > 0 for all x≥kx \geq k
  1. Continuous: f(x)f(x) is continuous for all x≥kx \geq k
  1. Decreasing: f(x)f(x) is decreasing for all x≥kx \geq k
10.4 - Integral Test for Convergence

Statement of the Test

Let an=f(n)a_n = f(n) where f(x)f(x) is positive, continuous, and decreasing for x≥kx \geq k. Then
∑n=k∞anand∫k∞f(x) dx\sum_{n=k}^{\infty} a_n \quad \text{and} \quad \int_{k}^{\infty} f(x) \, dx
both converge or both diverge together.
  • If ∫k∞f(x) dx\int_{k}^{\infty} f(x)\,dx converges, the series converges
  • If ∫k∞f(x) dx\int_{k}^{\infty} f(x)\,dx diverges, the series diverges

Proof Sketch

Take f(x)f(x) positive, continuous, and decreasing on [1,∞)[1, \infty).
10.4 - Integral Test for Convergence
Left rectangles (overestimate). Each rectangle has width 1 and height f(n)=anf(n) = a_n. Since ff is decreasing, the left endpoint sits above the curve, so the rectangles overshoot:
∑n=1Nan≥∫1N+1f(x) dx\sum_{n=1}^{N} a_n \geq \int_{1}^{N+1} f(x) \, dx
Right rectangles (underestimate).
10.4 - Integral Test for Convergence
∑n=2N+1an≤∫1N+1f(x) dx\sum_{n=2}^{N+1} a_n \leq \int_{1}^{N+1} f(x) \, dx
Combine. Add a1a_1 to both sides of the underestimate:
∑n=1N+1an≤a1+∫1N+1f(x) dx\sum_{n=1}^{N+1} a_n \leq a_1 + \int_{1}^{N+1} f(x) \, dx
Put it together with the overestimate:
∫1N+1f(x) dx≤∑n=1Nan≤a1+∫1Nf(x) dx\int_{1}^{N+1} f(x) \, dx \leq \sum_{n=1}^{N} a_n \leq a_1 + \int_{1}^{N} f(x) \, dx
Take the limit as N→∞N \to \infty:
∫1∞f(x) dx≤∑n=1∞an≤a1+∫1∞f(x) dx\int_{1}^{\infty} f(x) \, dx \leq \sum_{n=1}^{\infty} a_n \leq a_1 + \int_{1}^{\infty} f(x) \, dx
The series is sandwiched between the integral and the integral plus a constant. So if the integral is finite, the series is bounded and converges. If the integral blows up, so does the series.
Important: they converge or diverge together, but not necessarily to the same value.

Step-by-Step

  1. Verify the three conditions. Is f(x)f(x) positive, continuous, and decreasing for x≥kx \geq k? Graph it if you’re unsure.
  1. Set up the improper integral: lim⁡b→∞∫kbf(x) dx\lim_{b \to \infty} \int_{k}^{b} f(x)\,dx
  1. Evaluate it. Find the antiderivative, apply the limits, see whether the result is finite.
  1. Conclude. Integral converges means series converges; integral diverges means series diverges.

Example 1: A Divergent Series

Test ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n}.
Step 1: Check conditions for f(x)=1xln⁡xf(x) = \frac{1}{x\ln x} on x≥2x \geq 2.
10.4 - Integral Test for Convergence
Positive, continuous (since ln⁡x\ln x is defined and nonzero), and decreasing. All good.
Step 2: Set up.
lim⁡b→∞∫2b1xln⁡x dx\lim_{b \to \infty} \int_{2}^{b} \frac{1}{x\ln x} \, dx
Step 3: Evaluate with uu-substitution. Let u=ln⁡xu = \ln x, so du=1xdxdu = \frac{1}{x}dx.
=lim⁡b→∞∫ln⁡2ln⁡b1u du=lim⁡b→∞[ln⁡(ln⁡b)−ln⁡(ln⁡2)]= \lim_{b \to \infty} \int_{\ln 2}^{\ln b} \frac{1}{u} \, du = \lim_{b \to \infty} \left[\ln(\ln b) - \ln(\ln 2)\right]
As b→∞b \to \infty, ln⁡b→∞\ln b \to \infty, so ln⁡(ln⁡b)→∞\ln(\ln b) \to \infty. The integral diverges.
Step 4: Conclude. The series ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n\ln n} diverges by the integral test.

Example 2: A Convergent Series

Test ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}.
Step 1: Check conditions for f(x)=1x2f(x) = \frac{1}{x^2} on x≥1x \geq 1.
10.4 - Integral Test for Convergence
Positive, continuous, decreasing.
Step 2 and 3: Set up and evaluate.
lim⁡b→∞∫1bx−2 dx=lim⁡b→∞[−1x]1b=lim⁡b→∞(−1b+1)=1\lim_{b \to \infty} \int_{1}^{b} x^{-2} \, dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_{1}^{b} = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1
Step 4: Conclude. The integral converges to 1, so the series converges by the integral test.
Note: the series converges, but not to 1. The integral is 1; the series is actually π26≈1.645\frac{\pi^2}{6} \approx 1.645. The test tells you whether it converges, not what it converges to.

Example 3: Series with an Exponential

Test ∑n=1∞ne−n2\sum_{n=1}^{\infty} ne^{-n^2}.
Step 1: Check conditions for f(x)=xe−x2f(x) = xe^{-x^2} on x≥1x \geq 1. Positive, continuous, and decreasing (you can confirm with the derivative).
Step 2 and 3: Set up and evaluate with u=−x2u = -x^2, so du=−2x dxdu = -2x\,dx and x dx=−12dux\,dx = -\frac{1}{2}du.
lim⁡b→∞∫1bxe−x2 dx=−12lim⁡b→∞[eu]−1−b2=−12lim⁡b→∞(e−b2−e−1)\lim_{b \to \infty} \int_{1}^{b} xe^{-x^2}\,dx = -\frac{1}{2}\lim_{b \to \infty}\left[e^{u}\right]_{-1}^{-b^2} = -\frac{1}{2}\lim_{b\to\infty}\left(e^{-b^2} - e^{-1}\right)
As b→∞b \to \infty, e−b2→0e^{-b^2} \to 0:
=−12(0−e−1)=12e= -\frac{1}{2}\left(0 - e^{-1}\right) = \frac{1}{2e}
Step 4: Conclude. The integral converges to 12e\frac{1}{2e}, so the series converges by the integral test.

When the Function Isn’t Decreasing at First

10.4 - Integral Test for Convergence
The integral test still works as long as the function is eventually positive, continuous, and decreasing.
If f(x)f(x) satisfies the three conditions for all x≥kx \geq k (for some kk), then
∑n=1∞anand∫k∞f(x) dx\sum_{n=1}^{\infty} a_n \quad \text{and} \quad \int_{k}^{\infty} f(x)\,dx
behave the same way, even though the series starts at n=1n = 1.
Why? The first few terms are just a finite sum tacked onto both sides. A finite number can’t change whether something converges.

Common Candidates

pp-Series

∑n=1∞1np converges if p>1, diverges if p≤1\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges if } p > 1, \text{ diverges if } p \leq 1
For instance, ∑1n3\sum \frac{1}{n^3} converges, and the harmonic series ∑1n\sum \frac{1}{n} diverges.

Good Candidates for the Integral Test

  • Rational functions: 1n2\frac{1}{n^2}, 1n3+1\frac{1}{n^3 + 1}
  • Functions with logarithms: 1nln⁡n\frac{1}{n\ln n}
  • Exponentials: ne−nne^{-n}

Poor Candidates (use something else)

  • Factorials: 1n!\frac{1}{n!}
  • Alternating series: (−1)nn\frac{(-1)^n}{n}
  • Anything that’s a nightmare to integrate

Strategy Hierarchy

When testing a series:
  1. Try the nth term test first, it’s the fastest way to show divergence
  1. Check if it’s geometric, since then you get an easy formula
  1. Use the integral test if the function is easy to integrate and the conditions hold
  1. Use other tests for the messier cases (later lessons)

Key Takeaways

  • Integral test: if ff is positive, continuous, and decreasing, then ∑an\sum a_n and ∫f(x) dx\int f(x)\,dx behave the same
  • Three conditions required: positive, continuous, decreasing. Verify before using.
  • Both converge or both diverge, but not necessarily to the same value
  • Set it up as a limit: lim⁡b→∞∫kbf(x) dx\lim_{b \to \infty} \int_{k}^{b} f(x)\,dx
  • Reach for uu-substitution when needed
  • “Eventually decreasing” is enough, the conditions don’t have to hold from n=1n = 1
  • Only use it when the integration is manageable
  • State it as “the series converges/diverges by the integral test”
  • The proof uses rectangles: left rectangles overestimate, right rectangles underestimate
  • pp-series rule: ∑1np\sum \frac{1}{n^p} converges if p>1p > 1, diverges if p≤1p \leq 1

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