Mathwords: Integral Test

Key Formula

\text{If } f(x) \text{ is continuous, positive, and decreasing for } x \geq 1, \text{ and } a_n = f(n), \text{ then:}

\sum_{n=1}^{\infty} a_n \text{ and } \int_1^{\infty} f(x)\,dx \text{ either both converge or both diverge.}

Where:

$f(x)$ = A continuous, positive, decreasing function defined for x ≥ 1
$a_n$ = The nth term of the series, where a_n = f(n)
$\sum_{n=1}^{\infty} a_n$ = The infinite series being tested for convergence
$\int_1^{\infty} f(x)\,dx$ = The corresponding improper integral

Worked Example

Problem: Use the Integral Test to determine whether the series converges or diverges: ∑(n=1 to ∞) 1/n².

Step 1: Define f(x) = 1/x² and verify the conditions. This function is continuous, positive, and decreasing for x ≥ 1. All three conditions are satisfied.

f(x) = \frac{1}{x^2}, \quad x \geq 1

Step 2: Set up the corresponding improper integral.

\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty} \int_1^{b} x^{-2}\,dx

Step 3: Evaluate the antiderivative.

\lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^{b} = \lim_{b \to \infty} \left(-\frac{1}{b} + \frac{1}{1}\right)

Step 4: Compute the limit. As b → ∞, the term −1/b approaches 0.

= 0 + 1 = 1

Step 5: Since the improper integral converges (to 1), the Integral Test tells us the series also converges.

\int_1^{\infty} \frac{1}{x^2}\,dx = 1 \quad \Rightarrow \quad \sum_{n=1}^{\infty} \frac{1}{n^2} \text{ converges}

Answer: The series ∑(1/n²) converges by the Integral Test. Note: the integral equals 1, but the actual sum of the series is π²/6 ≈ 1.6449. The integral and the series do not have the same value.

Another Example

This example differs from the first in two ways: the series starts at n = 2 instead of n = 1 (which is fine — the starting index does not affect convergence), and the integral requires a u-substitution rather than a straightforward power rule. It also demonstrates the test proving divergence rather than convergence.

Problem: Use the Integral Test to determine whether the series converges or diverges: ∑(n=1 to ∞) 1/(n ln(n)), starting at n = 2.

Step 1: Define f(x) = 1/(x ln x) for x ≥ 2. This function is continuous, positive, and decreasing on [2, ∞) since both x and ln x are increasing, making their product increase and the reciprocal decrease.

f(x) = \frac{1}{x \ln x}, \quad x \geq 2

Step 2: Set up the improper integral and use the substitution u = ln x, so du = (1/x) dx.

\int_2^{\infty} \frac{1}{x \ln x}\,dx = \lim_{b \to \infty} \int_2^{b} \frac{1}{x \ln x}\,dx = \lim_{b \to \infty} \int_{\ln 2}^{\ln b} \frac{1}{u}\,du

Step 3: Evaluate the resulting integral.

\lim_{b \to \infty} \left[\ln|u|\right]_{\ln 2}^{\ln b} = \lim_{b \to \infty} \left(\ln(\ln b) - \ln(\ln 2)\right)

Step 4: Compute the limit. As b → ∞, ln(ln b) → ∞, so the integral diverges.

\lim_{b \to \infty} \ln(\ln b) = \infty

Answer: The series ∑(n=2 to ∞) 1/(n ln n) diverges by the Integral Test, because the corresponding improper integral diverges.

Frequently Asked Questions

Does the Integral Test tell you the exact sum of the series?

No. The Integral Test only tells you whether the series converges or diverges. The value of the improper integral is generally not equal to the sum of the series. For example, ∫₁^∞ 1/x² dx = 1, but ∑(1/n²) = π²/6 ≈ 1.645. However, the integral can be used to estimate the sum through integral test remainder bounds.

What are the conditions for the Integral Test?

The function f(x) must satisfy three conditions on the interval [N, ∞) for some integer N: it must be (1) continuous, (2) positive, and (3) decreasing. If any of these conditions fails, the Integral Test cannot be applied. The series terms must satisfy aₙ = f(n).

When should you use the Integral Test instead of another convergence test?

The Integral Test is most useful when the series terms involve functions you can easily integrate, such as powers, logarithms, or exponentials. If you can quickly compute the improper integral, this test works well. For series that resemble geometric series or factorials, the Ratio Test or Comparison Test may be simpler.

Integral Test vs. Comparison Test

	Integral Test	Comparison Test
How it works	Compares the series to an improper integral of a related function	Compares the series term-by-term to a known convergent or divergent series
Requirements	f(x) must be continuous, positive, and decreasing; you must evaluate an integral	You need a known benchmark series with terms that bound yours from above or below
Best used when	The terms involve functions that are easy to integrate (powers, logs, exponentials)	The terms closely resemble a p-series or geometric series
Gives the sum?	No, but can provide remainder estimates	No

Why It Matters

The Integral Test is one of the first convergence tests taught in Calculus II, and it connects two major topics: series and integration. It is the standard way to prove that the p-series ∑1/nᵖ converges for p > 1 and diverges for p ≤ 1, a result used throughout the study of series. The remainder estimate from the Integral Test also provides a practical way to approximate the sum of a convergent series to a desired accuracy.

Common Mistakes

Mistake: Assuming the value of the improper integral equals the sum of the series.

Correction: The Integral Test only determines convergence or divergence. The numerical value of the integral is not the sum of the series. For example, ∫₁^∞ 1/x² dx = 1, but ∑(1/n²) = π²/6.

Mistake: Applying the Integral Test when the function is not decreasing on the entire interval.

Correction: You must verify that f(x) is eventually decreasing (for all x ≥ N for some integer N). If the function increases on part of the interval, the test does not apply. A common check is to show f'(x) < 0 for x ≥ N.

Related Terms

Convergence Tests — The Integral Test is one of several convergence tests
Positive Series — The test applies to series with positive terms
Integral Test Remainder — Estimates how far a partial sum is from the true sum
Improper Integral — The integral ∫₁^∞ f(x) dx used in the test
Convergent Series — One of the two possible outcomes of the test
Divergent Series — The other possible outcome of the test
p-Series — Convergence of p-series is proven via the Integral Test
Term — Each aₙ = f(n) is a term of the series