10.4 Integral Test for Convergence

What the Integral Test says

If f(x) is positive, continuous, and decreasing for x ≥ 1, and a_n = f(n), then the series Σaₙ and the improper integral ∫₁^∞ f(x) dx either both converge or both diverge. This lets you use integration techniques to determine convergence of a series when a direct comparison or ratio test is hard to apply.

Checking the conditions

Before applying the test, verify all three conditions: positive (f(x) > 0), continuous, and decreasing on the interval. Decreasing can be shown by checking f′(x) < 0 or by reasoning about the function's behavior. Skipping this check is a common error — the test doesn't apply otherwise!

Classic application: p-series

The integral test proves the p-series rule: Σ(1/n^p) converges if p > 1 and diverges if p ≤ 1. For example, ∫₁^∞ (1/x²) dx converges (p=2>1), matching the convergence of Σ1/n². This test is especially useful for series resembling functions with known antiderivatives, like 1/(x ln x).

Key terms

Integral Test conditions

f(x) must be positive, continuous, and decreasing for x ≥ 1, with aₙ = f(n).

Integral Test conclusion

Σaₙ and ∫₁^∞ f(x) dx either both converge or both diverge.

p-series convergence (via Integral Test)

Σ1/n^p converges if p > 1, diverges if p ≤ 1.

Harmonic series

Σ1/n diverges, shown by ∫₁^∞ (1/x) dx diverging (equals ln x → ∞).

Good candidate for Integral Test

A series whose terms match a function that's positive, continuous, decreasing, and has a computable antiderivative, e.g., 1/(x ln x).

Try an AP-style question

Using the integral test, what does Σₙ₌₂^∞ 1/(n ln n) do?

1. Diverges, since ∫ 1/(x ln x) dx = ln(ln x) → ∞
2. The integral test cannot be used
3. Converges, since the terms approach 0
4. Converges, since it's smaller than 1/n