The Integral Test

Most series don’t telescope and aren’t geometric, so you can’t find their sums. But you can often still decide whether they converge. The integral test does this by comparing a series with an improper integral. Since you already know how to evaluate improper integrals, this turns a question about infinitely many terms into a calculus problem you can solve.

Key ideas

The test

Suppose $a_n = f(n)$, where for $x \ge k$ the function $f$ is

• positive,
• continuous, and
• decreasing.

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

The conditions only need to hold eventually, from some $x = k$ onward. A few odd terms at the start never affect convergence.

Why it works

Draw boxes of width $1$ whose heights are the terms. Each box has area $a_n$, so the total area of the boxes is the sum of the series.

Left: the boxes for $\sum_{n=2}^{\infty} \frac{1}{n^2}$ fit under $y = \frac{1}{x^2}$, so their total is finite. Right: the boxes for $\sum_{n=1}^{\infty} \frac{1}{n}$ cover the area under $y = \frac{1}{x}$, which is infinite.

• If the boxes fit under the curve and the area under the curve is finite, the boxes’ total is finite too, so the series converges.
• If the boxes cover the area under the curve and that area is infinite, the series diverges.

“Decreasing” is what makes the boxes line up with the curve this neatly. That’s why it’s a hypothesis.

The sum is not the integral

The test tells you only whether the series converges, not its sum. For example,

$$\int_1^{\infty} \frac{1}{x^2}\,dx = 1, \qquad\text{but}\qquad \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645 .$$

The boxes and the region under the curve have different areas.

Checking that f is decreasing

Often you can see it: $\dfrac{1}{x^2}$ clearly gets smaller as $x$ grows. When it isn’t obvious, show that $f'(x) \lt 0$ for $x \ge k$.

On the AP exam, name the test and state the hypotheses, for example: ”$f(x) = \dfrac{1}{x^2}$ is positive, continuous, and decreasing for $x \ge 1$, and $\int_1^{\infty} \frac{1}{x^2}\,dx$ converges, so $\sum \frac{1}{n^2}$ converges by the integral test.”

Worked examples

Example 1: A convergent series

Use the integral test to show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.

Solution. Let $f(x) = \dfrac{1}{x^2}$. For $x \ge 1$, $f$ is positive, continuous, and decreasing. Evaluate the improper integral as a limit:

$$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(1 - \frac{1}{b}\right) = 1$$

The integral converges, so the series converges by the integral test. (Its sum is not $1$. It’s $\frac{\pi^2}{6}$.)

Example 2: A divergent series

Does $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ converge or diverge?

Solution. Let $f(x) = \dfrac{1}{x \ln x}$. For $x \ge 2$, both $x$ and $\ln x$ are positive and increasing, so their product is positive and increasing, and $f$ is positive, continuous, and decreasing.

Use the substitution $u = \ln x$, $du = \dfrac{1}{x}\,dx$:

$$\int_2^{\infty} \frac{1}{x \ln x}\,dx = \lim_{b \to \infty} \Big[\ln(\ln x)\Big]_2^b = \lim_{b \to \infty} \big(\ln(\ln b) - \ln(\ln 2)\big) = \infty$$

The integral diverges, so the series diverges by the integral test. (Notice the terms do go to $0$, so the $n$th term test couldn’t have settled this.)

Example 3: Showing f is decreasing with a derivative

Does $\displaystyle\sum_{n=1}^{\infty} n e^{-n^2}$ converge or diverge?

Solution. Let $f(x) = x e^{-x^2}$, which is positive and continuous for $x \ge 1$. By the product rule,

$$f'(x) = e^{-x^2} + x \cdot \left(-2x e^{-x^2}\right) = e^{-x^2}\left(1 - 2x^2\right)$$

For $x \ge 1$, $1 - 2x^2 \lt 0$, so $f'(x) \lt 0$ and $f$ is decreasing. Now integrate with $u = -x^2$:

$$\int_1^{\infty} x e^{-x^2}\,dx = \lim_{b \to \infty} \left[-\frac{1}{2}e^{-x^2}\right]_1^b = \lim_{b \to \infty} \left(\frac{1}{2e} - \frac{1}{2}e^{-b^2}\right) = \frac{1}{2e}$$

The integral converges, so the series converges by the integral test.

Example 4: Decreasing only eventually

Does $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ converge or diverge?

Solution. Let $f(x) = \dfrac{\ln x}{x}$. By the quotient rule,

$$f'(x) = \frac{x \cdot \frac{1}{x} - \ln x}{x^2} = \frac{1 - \ln x}{x^2}$$

This is negative when $\ln x \gt 1$, that is, when $x \gt e \approx 2.718$. So $f$ is positive, continuous, and decreasing for $x \ge 3$. That’s enough: apply the test from $n = 3$.

With $u = \ln x$:

$$\int_3^{\infty} \frac{\ln x}{x}\,dx = \lim_{b \to \infty} \left[\frac{(\ln x)^2}{2}\right]_3^b = \infty$$

So $\sum_{n=3}^{\infty} \frac{\ln n}{n}$ diverges, and adding back the first two terms doesn’t change that. The series diverges.

Common mistakes

Skipping the hypotheses. AP graders look for “positive, continuous, and decreasing.” If $f$ isn’t obviously decreasing, show $f'(x) \lt 0$. If a hypothesis fails (for example, the terms aren’t all positive), you can’t use this test.

Saying the sum equals the integral. In Example 1 the integral is $1$ but the sum is about $1.645$. The integral test answers “converge or diverge?” and nothing more.

Using it on series with negative or alternating terms. For $\sum \frac{(-1)^n}{n}$, $f$ would have to be positive, and it isn’t. Use the alternating series test instead.

Treating the improper integral as a regular one. Write the limit: $\displaystyle\lim_{b \to \infty} \int_k^b f(x)\,dx$. Writing "$\infty$" into an antiderivative as if it were a number loses marks and leads to errors.

Choosing the integral test when something easier works. If the series is geometric or a p-series, just say so. Save the integral test for terms you can integrate, like $\frac{1}{n \ln n}$ or $n e^{-n^2}$.

Practice

1. (Warm-up) Use the integral test to decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2n + 3}$ converges.

Solution

$f(x) = \dfrac{1}{2x + 3}$ is positive, continuous, and decreasing for $x \ge 1$.

$$\int_1^{\infty} \frac{1}{2x + 3}\,dx = \lim_{b \to \infty} \left[\frac{1}{2}\ln(2x + 3)\right]_1^b = \infty$$

The integral diverges, so the series diverges by the integral test.

2. (Warm-up) Use the integral test to show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ converges.

Solution

$f(x) = \dfrac{1}{x^3}$ is positive, continuous, and decreasing for $x \ge 1$.

$$\int_1^{\infty} x^{-3}\,dx = \lim_{b \to \infty} \left[-\frac{1}{2x^2}\right]_1^b = \lim_{b \to \infty} \left(\frac{1}{2} - \frac{1}{2b^2}\right) = \frac{1}{2}$$

The integral converges, so the series converges.

3. (Warm-up) A student says that, by Question 2, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3} = \frac{1}{2}$. Explain why that can’t be right.

Solution

The integral test only tells you the series converges, not its sum. In fact the first term alone is $1$, and every term is positive, so the sum is more than $1$. (It’s about $1.202$.)

4. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converge or diverge?

Solution

$f(x) = \dfrac{1}{x^2 + 1}$ is positive, continuous, and decreasing for $x \ge 1$ (the denominator increases).

$$\int_1^{\infty} \frac{1}{x^2 + 1}\,dx = \lim_{b \to \infty} \Big[\arctan x\Big]_1^b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$

The integral converges, so the series converges by the integral test.

5. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ converge or diverge? Show that the hypotheses are met.

Solution

$f(x) = \dfrac{x}{x^2 + 1}$ is positive and continuous for $x \ge 1$. By the quotient rule,

$$f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}$$

which is $\le 0$ for $x \ge 1$ (and negative for $x \gt 1$), so $f$ is decreasing. With $u = x^2 + 1$:

$$\int_1^{\infty} \frac{x}{x^2 + 1}\,dx = \lim_{b \to \infty} \left[\frac{1}{2}\ln(x^2 + 1)\right]_1^b = \infty$$

The integral diverges, so the series diverges by the integral test.

6. (Core) Does $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ converge or diverge?

Solution

$f(x) = \dfrac{1}{x(\ln x)^2}$ is positive, continuous, and decreasing for $x \ge 2$ (the denominator is a product of positive increasing factors). With $u = \ln x$, $du = \dfrac{1}{x}\,dx$:

$$\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx = \lim_{b \to \infty} \left[-\frac{1}{\ln x}\right]_2^b = \lim_{b \to \infty} \left(\frac{1}{\ln 2} - \frac{1}{\ln b}\right) = \frac{1}{\ln 2}$$

The integral converges, so the series converges. Compare this with Example 2: squaring the $\ln n$ made the difference.

7. (Core) Does $\displaystyle\sum_{n=1}^{\infty} n e^{-n}$ converge or diverge?

Solution

$f(x) = x e^{-x}$ is positive and continuous for $x \ge 1$, and

$$f'(x) = e^{-x} - x e^{-x} = (1 - x)e^{-x} \lt 0 \quad\text{for } x \gt 1,$$

so $f$ is decreasing for $x \ge 1$. Use integration by parts with $u = x$, $dv = e^{-x}\,dx$:

$$\int x e^{-x}\,dx = -x e^{-x} - e^{-x} + C$$$$\int_1^{\infty} x e^{-x}\,dx = \lim_{b \to \infty} \Big[-x e^{-x} - e^{-x}\Big]_1^b = 0 - \left(-\frac{1}{e} - \frac{1}{e}\right) = \frac{2}{e}$$

(Here $b e^{-b} \to 0$ by L’Hôpital’s rule.) The integral converges, so the series converges.

8. (Challenge) For which values of $p \gt 0$ does $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ converge?

Solution

$f(x) = \dfrac{1}{x(\ln x)^p}$ is positive, continuous, and decreasing for $x \ge 2$ when $p \gt 0$. With $u = \ln x$, the integral becomes

$$\int_2^{\infty} \frac{1}{x(\ln x)^p}\,dx = \int_{\ln 2}^{\infty} \frac{1}{u^p}\,du$$

This converges when $p \gt 1$ and diverges when $p \le 1$ (for $p = 1$ it’s $\ln u \to \infty$; for $p \lt 1$ it’s $\frac{u^{1-p}}{1-p} \to \infty$). So the series converges exactly when $p \gt 1$.

9. (Challenge) Use the box pictures to show that

$$\frac{1}{2} \le \sum_{n=1}^{\infty} \frac{1}{n^3} \le \frac{3}{2} .$$

Solution

Boxes of height $\frac{1}{n^3}$ drawn to the right of $x = n$ sit above the curve $y = \frac{1}{x^3}$ (as in the right-hand picture), so their total is at least the area under the curve:

$$\sum_{n=1}^{\infty} \frac{1}{n^3} \ge \int_1^{\infty} \frac{1}{x^3}\,dx = \frac{1}{2}$$

Boxes for $n \ge 2$ drawn to the left of $x = n$ fit under the curve (as in the left-hand picture), so

$$\sum_{n=2}^{\infty} \frac{1}{n^3} \le \int_1^{\infty} \frac{1}{x^3}\,dx = \frac{1}{2}$$

Adding the first term $1$ gives $\sum_{n=1}^{\infty} \frac{1}{n^3} \le \frac{3}{2}$. (The actual sum is about $1.202$, which fits.)