Integral Test

An infinite series is the sum of infinitely many terms, and one of the central questions in mathematics is whether such a sum converges to a finite value or diverges to infinity. The Integral Test is a powerful tool that connects the behavior of a series to the behavior of a related continuous function — specifically, by comparing the discrete sum to a definite integral. If we can evaluate (or estimate) an improper integral, we gain direct information about the corresponding infinite series. The key insight is geometric: the terms of a series can be visualized as the areas of rectangles on a graph, and a smooth curve through those rectangles is an integrable function. If the area under the curve (an improper integral) is finite, the rectangles — and therefore the series — must also sum to something finite. This geometric bridge between calculus and discrete sums is what makes the Integral Test so elegant and useful. The Integral Test applies to series whose terms are defined by a function f(n) that is continuous, positive, and decreasing on some interval [N, ∞). Classic applications include the p-series ∑ 1/n^p and logarithm-related series that resist simpler tests. Once you confirm the three conditions are met, evaluating a single improper integral settles the convergence question for the entire series.

Key takeaways

State the conditions
Set up the comparison inequality
Apply the convergence conclusion
Check the conditions carefully
Evaluate the improper integral

State the conditions

Let f(x) be a function defined on [1, ∞). For the Integral Test to apply, f must satisfy three conditions on [N, ∞) for some positive integer N: (1) f is continuous, (2) f(x) > 0 for all x ≥ N (positive), and (3) f is decreasing — that is, f'(x) ≤ 0 for all x ≥ N. When these hold, the series ∑ f(n) and the integral ∫ f(x) dx share the same convergence behavior.

Set up the comparison inequality

Because f is decreasing, on each interval [n, n+1] the function satisfies f(n+1) ≤ f(x) ≤ f(n). Integrating across [n, n+1] gives f(n+1) ≤ ∫ from n to n+1 of f(x) dx ≤ f(n). Summing from n = 1 to N yields the key double inequality: ∑_{n=2}^{N+1} f(n) ≤ ∫_1^{N+1} f(x) dx ≤ ∑_{n=1}^{N} f(n). This sandwich shows the partial sums and integral are bounded by each other.

Apply the convergence conclusion

If the improper integral ∫_1^∞ f(x) dx converges to a finite value L, then the partial sums of ∑ f(n) are bounded above by f(1) + L, so by the Monotone Convergence Theorem (since all terms are positive), the series converges. Conversely, if the integral diverges to infinity, the partial sums are bounded below by the divergent integral, so the series also diverges. The test does NOT give the exact sum of the series, only whether it converges.

Check the conditions carefully

Always verify all three conditions before applying the test. For example, f(x) = sin²(x)/x² is positive but NOT monotonically decreasing everywhere, so the test cannot be applied naively. A common trap: the function f(x) must be eventually decreasing — it is fine if f increases initially, as long as from some index N onward it is decreasing. You may start your integral at N rather than 1; convergence behavior is unaffected by finitely many terms.

Evaluate the improper integral

Compute ∫_N^∞ f(x) dx = lim_{t→∞} ∫_N^t f(x) dx. If this limit equals a finite real number, the series converges. If the limit is +∞ or does not exist (as a finite value), the series diverges. For many standard functions — powers, logarithms, exponentials — this integral can be evaluated explicitly using antiderivatives and limit techniques from calculus.

The Key Equation

\sum_{n=1}^{\infty} f(n) \text{ converges} \iff \int_{1}^{\infty} f(x)\, dx < \infty

f(n) — Series term function (dimensionless): The function evaluated at positive integers n; must be continuous, positive, and decreasing as a function of x
\int_{1}^{\infty} f(x)\, dx — Improper integral (dimensionless): The limit lim_{t→∞} ∫_1^t f(x) dx; finite value indicates convergence, infinite value indicates divergence
\iff — If and only if (logic): Indicates that the series and integral have exactly the same convergence or divergence behavior

Worked Example

Problem. Determine whether the series ∑_{n=1}^{∞} 1/n² converges or diverges using the Integral Test.

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

Answer. The series converges (the integral equals 1, confirming convergence).

Figure

Diagram

Origins · 1 minColin Maclaurin and how this was discovered

Colin Maclaurin

1698–1746 · Scottish

Although the Integral Test is often attributed to Cauchy, Colin Maclaurin independently discovered essentially the same comparison between sums and integrals in his 1742 Treatise of Fluxions. He showed that partial sums of series can be bounded above and below by corresponding integrals — the geometric argument at the heart of the modern Integral Test. This work appeared as he developed techniques to handle series convergence with rigour unusual for his era. Maclaurin was one of the most gifted mathematical prodigies in history — he became a professor of mathematics in Aberdeen at age 19 and was personally recommended for the post by Isaac Newton himself.

p-Series Convergence — The p-series ∑ 1/nᵖ is the most direct application of the Integral Test, and understanding the test clarifies exactly why p > 1 is required for convergence.
Improper Integrals — The Integral Test relies on evaluating ∫_1^∞ f(x) dx as a limit; mastering improper integrals is prerequisite knowledge for applying the test.
Comparison Test for Series — Like the Integral Test, the Comparison Test uses inequalities between sequences to determine convergence; the two tests complement each other for different types of series.
The Harmonic Series — The divergence of ∑ 1/n is the p-series case p = 1, and applying the Integral Test to ∫ 1/x dx = ln x → ∞ gives the cleanest proof of this famous divergence.
Monotone Convergence Theorem — The final step of the Integral Test proof invokes the fact that an increasing sequence bounded above must converge — this theorem is the logical backbone of the argument.