The integral test

Most series don't telescope and aren't geometric, so you can't find a formula for their partial sums. The integral test gets around this by comparing a series to an improper integral, which you can evaluate. It is also the tool that finally settles the harmonic series.

Series as areas

Picture the terms $a_n = f(n)$ as rectangles of width $1$: the rectangle over $[n, n+1]$ has height $f(n)$, so its area is $a_n$. The sum of the series is the total area of all the rectangles. If $f$ is decreasing, those rectangles either sit just above or just below the curve $y = f(x)$, depending on which endpoint you use for the height.

• Using left endpoints, the rectangles lie above the curve, so $\displaystyle\sum_{n=1}^{\infty} a_n \ge \int_1^{\infty} f(x)\,dx$.
• Using right endpoints (starting the rectangles one step later), they lie below the curve, so $\displaystyle\sum_{n=2}^{\infty} a_n \le \int_1^{\infty} f(x)\,dx$.

So if the integral is infinite, the bigger series is infinite too. If the integral is finite, the smaller series is finite, and adding back the single term $a_1$ keeps it finite. The series and the integral are tied together.

Checking the conditions

On a free-response question you must say why the test applies. Here is what to check.

1. Positive: usually clear from the formula.
2. Continuous: no division by zero or undefined logs on $[k, \infty)$.
3. Decreasing: either it is obvious (the denominator grows while the numerator stays fixed), or show $f'(x) < 0$ for $x \ge k$.

The conditions only need to hold eventually. If $f$ increases for a while and then decreases from $x = 3$ on, start the test at $k = 3$; the first few terms don't affect convergence.

The harmonic series diverges

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

$$\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b\to\infty} \ln b = \infty.$$

The integral diverges, so the harmonic series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, even though its terms go to $0$. The partial sums grow like $\ln n$: very slowly, but without bound. It takes more than 12,000 terms for the partial sum to pass $10$.

The same picture gives a useful estimate. The rectangles for $\frac12 + \frac13 + \cdots + \frac1n$ fit under the curve $y = \frac1x$ from $1$ to $n$, and the rectangles for $1 + \frac12 + \cdots + \frac{1}{n-1}$ sit on top of it, so

$$\ln n \;\le\; 1 + \frac12 + \frac13 + \cdots + \frac1n \;\le\; 1 + \ln n.$$

For example, the sum of the first million terms of the harmonic series is between $\ln(10^6) \approx 13.8$ and $14.8$. That is a striking amount of growth for a million terms, which shows how slowly this series diverges, but it does diverge: $\ln n$ has no upper bound.

When not to use the integral test

The integral test needs a function you can actually integrate, and it needs the terms to be positive and decreasing. Series with alternating signs like $\sum \frac{(-1)^n}{n}$, series with factorials like $\sum \frac{1}{n!}$ (there is no continuous "$x!$" that is easy to integrate), and series with oscillating terms like $\sum \frac{2 + \sin n}{n^2}$ (not decreasing) all call for other tests.

Worked examples

Practice

Practice 1

Which condition must $f$ satisfy on $[1, \infty)$ in order to apply the integral test to $\displaystyle\sum_{n=1}^{\infty} f(n)$?

A$f$ is positive, continuous and increasing.B$f$ is positive, continuous and decreasing.C$f$ is differentiable and $\displaystyle\lim_{x\to\infty} f(x) = 0$.D$f$ is continuous and $\displaystyle\int_1^{\infty} f(x)\,dx$ exists.

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Practice 2

Evaluate $\displaystyle\int_1^{\infty} \frac{1}{1 + x^2}\,dx$, which is used to test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{1 + n^2}$. Give an exact value.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 3

Evaluate $\displaystyle\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx$. Give an exact value.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 4

Based on the previous problem, what can you conclude about $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$?

AIt diverges, because the integral is positive.BIt converges to $\dfrac{1}{\ln 2}$.CIt converges, but its sum is not necessarily $\dfrac{1}{\ln 2}$.DNothing, because $\ln x$ is not positive on $[1, 2)$.

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Practice 5

Use the integral test to determine the behavior of $\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$.

AIt diverges, because $\displaystyle\int_1^{\infty} \frac{x}{x^2+1}\,dx$ diverges.BIt converges, because $\displaystyle\int_1^{\infty} \frac{x}{x^2+1}\,dx$ converges.CIt converges, because $\dfrac{n}{n^2+1} \to 0$.DThe integral test cannot be applied because the terms are not decreasing.

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Practice 6

Evaluate $\displaystyle\int_1^{\infty} x e^{-x^2}\,dx$. Give an exact value.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 7

Which series can be shown to converge using the integral test with $f(x) = \dfrac{\ln x}{x^2}$?

A$\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$B$\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{\ln (n+1)}$C$\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\ln n}$D$\displaystyle\sum_{n=2}^{\infty} \frac{\ln n}{n^2}$

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