The Integral Test — AP® Calculus BC Notes

The nth term test can only prove divergence. The integral test can prove either convergence or divergence, for series whose terms are positive and shrinking. It connects a series to the improper integrals of Topic 6.13.

The picture

Suppose $a_n = f(n)$ for a function $f$ that is positive and decreasing. Draw each term as a box of width 1 and height $a_n$ . Its area is $a_n$ .

Left: the curve f(x) = 1/x² with boxes of width 1 drawn under it; box n spans x = n − 1 to n and has height 1/n², for n = 2 to 7. The boxes fit under the curve from x = 1 on, so 1/2² + 1/3² + ... is less than the integral from 1 to infinity of 1/x², which is 1: the series converges. Right: the curve f(x) = 1/x with boxes of height 1/n from x = n to n + 1, for n = 1 to 6, sticking out above the curve, so 1 + 1/2 + ... + 1/N is more than the integral from 1 to N + 1 of 1/x, which is ln(N + 1) and grows without bound: the series diverges. — **Figure 1.** Boxes under $y = \frac{1}{x^2}$ and over $y = \frac{1}{x}$ . LevlPrep original diagram.

Boxes under the curve. Put box $n$ on $[n - 1, n]$ . Since $f$ is decreasing, its height $f(n)$ is the lowest value of $f$ there, so the box fits under the curve. Then $a_2$ $+ a_3$ $+ \cdots$ $+ a_n \le \int_1^{n} f(x)\,dx$ . If $\int_1^{\infty} f(x)\,dx$ is finite, the partial sums are bounded and keep rising, so they approach a limit: the series converges.
Boxes over the curve. Put box $n$ on $[n, n + 1]$ . Now $f(n)$ is the highest value there, so the box covers the curve. Then $a_1$ $+ \cdots$ $+ a_n \ge \int_1^{n+1} f(x)\,dx$ . If that integral grows without bound, so do the partial sums: the series diverges.

The figure shows both: $\sum \frac{1}{n^2}$ is held down by a finite area, and $\sum \frac{1}{n}$ is pushed up by $\int_1^{N+1} \frac{dx}{x}$ $= \ln(N + 1)$ , which grows without bound.

The test

Integral test. If $f$ is positive, continuous and decreasing on $[k, \infty)$ and $a_n = f(n)$ for $n \ge k$ , then

$\sum_{n=k}^{\infty} a_n \text{ and } \int_k^{\infty} f(x)\,dx \text{ both converge or both diverge.}$

All three conditions matter, and on the exam you must say they hold. Positive and continuous are usually clear from the formula. Decreasing usually needs a derivative: show $f'(x) < 0$ for $x \ge k$ .

Worked example 1. Does $\sum_{n=1}^{\infty} \frac{n}{e^{n^2}}$ converge?

Conditions: $f(x) = xe^{-x^2}$ is positive and continuous for $x \ge 1$ . $f'(x)$ $= e^{-x^2}(1 - 2x^2) < 0$ for $x \ge 1$ , so $f$ is decreasing.

Integral (substitute $u = x^2$ ): $\int_1^{\infty} xe^{-x^2}\,dx$ $= \lim_{b \to \infty} \Big[-\frac{1}{2}e^{-x^2}\Big]_1^{b}$ $= \lim_{b \to \infty} \left(-\frac{1}{2}e^{-b^2} + \frac{1}{2e}\right)$ $= \frac{1}{2e}$ .

Conclusion: the integral converges, so the series converges by the integral test.

Worked example 2. Does $\sum_{n=2}^{\infty} \frac{1}{n\ln n}$ converge?

Conditions: $f(x) = \frac{1}{x\ln x}$ is positive and continuous for $x \ge 2$ , and decreasing because its denominator $x\ln x$ increases.

Integral (substitute $u = \ln x$ ): $\int_2^{b} \frac{dx}{x\ln x}$ $= \ln(\ln b) - \ln(\ln 2)$ , which grows without bound as $b \to \infty$ .

Conclusion: the integral diverges, so the series diverges by the integral test, even though its terms approach 0.

Three cautions

Caution	Why
The integral is not the sum	$\int_1^{\infty} \frac{dx}{x^2}$ $= 1$ , but $\sum_{n=1}^{\infty} \frac{1}{n^2}$ $\approx 1.645$ ; the boxes and the curve have different areas
Decreasing may start late	$\frac{\ln x}{x}$ rises until $x = e$ ; run the test from $n = 3$ , since the first terms never change convergence
Keep limit notation	write $\lim_{b \to \infty} \int_1^{b}$ ; never substitute $\infty$ as a number

When the conditions fail, for example $\frac{\sin^2 n}{n^2}$ , whose function keeps dropping to 0 and rising again, the integral test cannot be used at all; Topic 10.6 gives tests that handle such series. Next, in 10.5, the integral test settles a whole family at once: $\sum \frac{1}{n^p}$ .