Integral Test for Convergence¶

Proposition¶

Let $f\colon[1,\infty)\to\mathbb{R}$ be continuous, nonnegative, and nonincreasing. Then

\sum_{n=1}^{\infty}f(n) \quad\text{and}\quad \int_1^{\infty}f(x)\,\mathrm{d}x

converge or diverge together. Moreover, the sequence

\left(\sum_{i=1}^n f(i)-\int_1^n f(x)\,\mathrm{d}x\right)_{n=1}^{\infty}

converges.

Interpretation¶

The integral test compares a series of nonnegative terms with an improper integral of the same shape. The terms $f(n)$ are the heights of rectangles of width $1$ along the graph of $f$ , so the partial sums and the integrals $\int_1^N f(x)\,\mathrm{d}x$ grow together, and one stays bounded if and only if the other does.

Rectangles of heights 1, 1/2, 1/3 and so on on unit intervals, lying above the curve 1/x, so the sum of 1/n is at least the integral of 1/x, which diverges.

For $f(x)=\frac{1}{x}$ the rectangles have areas $1,\frac{1}{2},\frac{1}{3},\dots$ and lie above the graph, so $\displaystyle\sum_{n=1}^{N}\frac{1}{n}\ge\int_1^{N+1}\frac{\mathrm{d}x}{x}$ $\int_{1}^{N + 1} \frac{d x}{x}$ , which tends to $\infty$ .

The rectangles of height $f(i)$ and the area under $f$ grow together. Their difference settles for every $p$ , which is why the series and the integral converge or diverge together.

The rectangles give a precise comparison. Since $f$ is nonincreasing, $f(n+1)\le f(x)\le f(n)$ for $x\in[n,n+1]$ $1]$ ; integrating over $[n,n+1]$ and adding gives

\sum_{n=2}^{N}f(n)\le\int_1^N f(x)\,\mathrm{d}x \le\sum_{n=1}^{N-1}f(n).

Because $f\ge0$ $0$ , both the partial sums and the integrals increase with $N$ , so each converges if and only if it is bounded, and the inequalities transfer boundedness in both directions. Without monotonicity the test fails: $f(x)=\sin^2(\pi x)$ is continuous and nonnegative with $f(n)=0$ for every $n$ , so the series converges, while $\int_1^\infty\sin^2(\pi x)\,\mathrm{d}x$ diverges.

The test is how the $p$ -series are settled from the $p$ -test. The second statement says more: for $f(x)=\frac{1}{x}$ the differences $\displaystyle \sum_{i=1}^n\frac{1}{i}-\ln n$ converge, with limit about $0.577$ , so the partial sums of the harmonic series grow like $\ln n$ .

A common slip is to take the integral as the value of the series. The series $\displaystyle \sum\frac{1}{n^2}$ and $\int_1^\infty\frac{\mathrm{d}x}{x^2}$ converge together, but the sum is $\frac{\pi^2}{6}\approx1.645$ while the integral is $1$ .

Key Results¶

The test applies only to continuous, nonnegative, nonincreasing functions.
Convergence of the improper integral implies convergence of the series.
Divergence of the improper integral implies divergence of the series.
The test provides a comparison, not the value of the sum.

Examples¶

The series $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ converges because $\displaystyle \int_1^\infty \frac{1}{x^2}\,\mathrm{d}x$ converges.
The series $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ diverges because $\displaystyle \int_1^\infty \frac{1}{x}\,\mathrm{d}x$ diverges.

Prerequisites¶

Sources¶

Math I: Calculus · Lecture 12

Aliases: integral test, Cauchy integral test, Maclaurin-Cauchy test Symbols: element of, infinity, integral, pi, real numbers, sum

Last updated: 22 October 2026 19:04