Integral Test

The integral test says that if a_n = f(n), where f is positive, continuous and decreasing for x ≥ 1, then the series \sum_{n=1}^{\infty} a_n and the improper integral \int_1^{\infty} f(x)\,dx either both converge or both diverge. The integral decides convergence, but its value is not the sum.

Before this lesson: Improper Integrals, Infinite Series and Partial Sums and Divergence Test

Why a series behaves like an integral

Draw each term a_n as a rectangle of width 1 and height f(n), standing on the interval from n to n + 1. The total area of the rectangles is the sum of the series.

When f is decreasing, each rectangle starts at the curve and sits above it. So the rectangles cover the area under the curve: the series is at least as big as the integral. Shift the rectangles one unit left and they sit under the curve instead, so the series (minus its first term) is at most the integral.

The series and the integral are trapped close to each other. If the area is finite, the sum is finite too. If the area is infinite, so is the sum.

The integral test

Suppose f is positive, continuous and decreasing on [1, \infty), and a_n = f(n) for every n. Then:

\int_1^{\infty} f(x)\,dx \text{ converges} \quad \Longleftrightarrow \quad \sum_{n=1}^{\infty} a_n \text{ converges}

The conditions only need to hold from some point on. If f is decreasing for x ≥ N, start the integral at N; the first few terms do not change whether a series converges.

Remainder estimate. If the series converges and you stop after N terms, the error R_N = \sum_{n=N+1}^{\infty} a_n is trapped between two integrals:

\int_{N+1}^{\infty} f(x)\,dx \le R_N \le \int_N^{\infty} f(x)\,dx

How to use the integral test

1. Write a_n = f(n) and check that f(x) is positive, continuous and decreasing for x ≥ 1 (or from some point on). A negative derivative shows f is decreasing.
2. Evaluate \int_1^{\infty} f(x)\,dx as a limit, \lim_{b \to \infty} \int_1^{b} f(x)\,dx.
3. If the integral is finite, the series converges. If it is infinite, the series diverges.

Integral test examples

Example 1: The harmonic series

Use the integral test on \sum_{n=1}^{\infty} \dfrac{1}{n}.

f(x) = \dfrac{1}{x}

f(x) \gt 0,\ f'(x) = -\dfrac{1}{x^2} \lt 0

\int_1^{\infty} \dfrac{1}{x}\,dx = \lim_{b \to \infty} \int_1^{b} \dfrac{1}{x}\,dx

\int_1^{b} \dfrac{1}{x}\,dx = \ln b - \ln 1

\ln b - \ln 1 = \ln b

\lim_{b \to \infty} \ln b = \infty

Answer: the integral diverges, so the harmonic series diverges, even though its terms go to 0.

Example 2: A series that converges

Use the integral test on \sum_{n=1}^{\infty} \dfrac{1}{n^2 + 1}.

f(x) = \dfrac{1}{x^2 + 1}

f'(x) = -\dfrac{2x}{(x^2 + 1)^2} \lt 0

\int_1^{b} \dfrac{1}{x^2 + 1}\,dx = \arctan b - \arctan 1

\arctan b - \arctan 1 = \arctan b - \dfrac{\pi}{4}

\lim_{b \to \infty} \left(\arctan b - \dfrac{\pi}{4}\right) = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}

Answer: the integral equals π/4 ≈ 0.785, a finite number, so the series converges. The sum of the series is about 1.077, not π/4.

Example 3: A u-substitution integral

Use the integral test on \sum_{n=1}^{\infty} ne^{-n^2}.

f(x) = xe^{-x^2}

f'(x) = e^{-x^2} - 2x^2e^{-x^2}

f'(x) = e^{-x^2}\left(1 - 2x^2\right) \lt 0

u = -x^2,\ du = -2x\,dx

\int xe^{-x^2}\,dx = -\dfrac{1}{2}\int e^{u}\,du = -\dfrac{1}{2}e^{-x^2} + C

\int_1^{b} xe^{-x^2}\,dx = -\dfrac{1}{2}e^{-b^2} + \dfrac{1}{2}e^{-1}

\lim_{b \to \infty} \left(-\dfrac{1}{2}e^{-b^2} + \dfrac{1}{2e}\right) = \dfrac{1}{2e}

Answer: the integral equals 1/(2e) ≈ 0.184, so the series converges.

Example 4: Starting at n = 2

Use the integral test on \sum_{n=2}^{\infty} \dfrac{1}{n\ln n}.

f(x) = \dfrac{1}{x\ln x}

u = \ln x,\ du = \dfrac{1}{x}\,dx

\int \dfrac{1}{x\ln x}\,dx = \int \dfrac{1}{u}\,du = \ln|\ln x| + C

\int_2^{b} \dfrac{1}{x\ln x}\,dx = \ln(\ln b) - \ln(\ln 2)

\lim_{b \to \infty} \ln(\ln b) = \infty

Answer: the integral diverges, so the series diverges. It grows extremely slowly: the first million terms add up to only about 3.4.

Example 5: How many terms are enough? (word problem)

A student writes a program to estimate \sum_{n=1}^{\infty} \dfrac{1}{n^3}, a constant that shows up in physics. (a) Use the first 5 terms and the remainder estimate to trap the sum. (b) How many terms make the error less than 0.0005?

S_5 = 1 + \dfrac{1}{8} + \dfrac{1}{27} + \dfrac{1}{64} + \dfrac{1}{125} \approx 1.18566

\int_N^{\infty} \dfrac{1}{x^3}\,dx = \lim_{b \to \infty} \left(-\dfrac{1}{2b^2} + \dfrac{1}{2N^2}\right) = \dfrac{1}{2N^2}

\dfrac{1}{2(6)^2} \le R_5 \le \dfrac{1}{2(5)^2}

0.01389 \le R_5 \le 0.02

1.18566 + 0.01389 \le S \le 1.18566 + 0.02

1.19955 \le S \le 1.20566

\dfrac{1}{2N^2} \lt 0.0005

N^2 \gt 1000

N \gt 31.6

Answer: (a) the sum is between about 1.1996 and 1.2057 (its true value is about 1.2021). (b) 32 terms guarantee an error less than 0.0005.

Common mistakes

Related methods

p-series come next

Applying the integral test to \sum 1/n^p shows that it converges exactly when p is greater than 1. That fact, the p-series test, is the next lesson and gives you a library of series to compare with.

Comparing instead of integrating

When f(x) is hard to integrate, or is not decreasing, compare the series with one you already know. That is the idea behind the direct comparison test and the limit comparison test later in this unit.

Practice problems

1. Use the integral test on \sum_{n=1}^{\infty} \dfrac{1}{n^2}.
2. Use the integral test on \sum_{n=1}^{\infty} \dfrac{1}{2n + 3}.
3. Use the integral test on \sum_{n=1}^{\infty} e^{-n}.
4. Use the integral test on \sum_{n=1}^{\infty} \dfrac{n}{n^2 + 4}.
5. Use the integral test on \sum_{n=1}^{\infty} \dfrac{1}{(n + 1)^{3/2}}.
6. Use the integral test on \sum_{n=3}^{\infty} \dfrac{\ln n}{n}.
7. Use the integral test on \sum_{n=2}^{\infty} \dfrac{1}{n(\ln n)^2}.
8. Use the integral test on \sum_{n=1}^{\infty} \dfrac{\arctan n}{1 + n^2}.
9. Explain why the integral test cannot be used on \sum_{n=1}^{\infty} \dfrac{2 + \sin n}{n^2}.
10. A calculator adds terms of \sum_{n=1}^{\infty} \dfrac{1}{n^2} one at a time. Use the remainder estimate to find how many terms make the error less than 0.01.

f(x) = \dfrac{1}{x^2},\ f'(x) = -\dfrac{2}{x^3} \lt 0

\int_1^{b} x^{-2}\,dx = -\dfrac{1}{b} + 1

\lim_{b \to \infty} \left(1 - \dfrac{1}{b}\right) = 1

f(x) = \dfrac{1}{2x + 3}

\int_1^{b} \dfrac{1}{2x + 3}\,dx = \dfrac{1}{2}\ln(2b + 3) - \dfrac{1}{2}\ln 5

\lim_{b \to \infty} \dfrac{1}{2}\ln(2b + 3) = \infty

f(x) = e^{-x},\ f'(x) = -e^{-x} \lt 0

\int_1^{b} e^{-x}\,dx = -e^{-b} + e^{-1}

\lim_{b \to \infty} \left(-e^{-b} + e^{-1}\right) = \dfrac{1}{e}

f'(x) = \dfrac{(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = \dfrac{4 - x^2}{(x^2 + 4)^2}

\dfrac{4 - x^2}{(x^2 + 4)^2} \lt 0 \text{ for } x \gt 2

\int_2^{b} \dfrac{x}{x^2 + 4}\,dx = \dfrac{1}{2}\ln(b^2 + 4) - \dfrac{1}{2}\ln 8

\lim_{b \to \infty} \dfrac{1}{2}\ln(b^2 + 4) = \infty

f(x) = (x + 1)^{-3/2}

\int_1^{b} (x + 1)^{-3/2}\,dx = -2(b + 1)^{-1/2} + 2(2)^{-1/2}

\lim_{b \to \infty} \left(-\dfrac{2}{\sqrt{b + 1}} + \dfrac{2}{\sqrt{2}}\right) = \sqrt{2}

f'(x) = \dfrac{1 - \ln x}{x^2} \lt 0 \text{ for } x \gt e

\int_3^{b} \dfrac{\ln x}{x}\,dx = \dfrac{(\ln b)^2}{2} - \dfrac{(\ln 3)^2}{2}

\lim_{b \to \infty} \dfrac{(\ln b)^2}{2} = \infty

f(x) = \dfrac{1}{x(\ln x)^2}

\int_2^{b} \dfrac{1}{x(\ln x)^2}\,dx = -\dfrac{1}{\ln b} + \dfrac{1}{\ln 2}

\lim_{b \to \infty} \left(-\dfrac{1}{\ln b} + \dfrac{1}{\ln 2}\right) = \dfrac{1}{\ln 2}

f'(x) = \dfrac{1 - 2x\arctan x}{(1 + x^2)^2} \lt 0 \text{ for } x \ge 1

u = \arctan x,\ du = \dfrac{dx}{1 + x^2}

\int_1^{b} \dfrac{\arctan x}{1 + x^2}\,dx = \dfrac{(\arctan b)^2}{2} - \dfrac{(\pi/4)^2}{2}

\lim_{b \to \infty} \left(\dfrac{(\arctan b)^2}{2} - \dfrac{\pi^2}{32}\right) = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{32}

\dfrac{\pi^2}{8} - \dfrac{\pi^2}{32} = \dfrac{3\pi^2}{32}

f(x) = \dfrac{2 + \sin x}{x^2}

f(5) \approx 0.0416,\ f(7.8) \approx 0.0493

R_N \le \int_N^{\infty} \dfrac{1}{x^2}\,dx = \dfrac{1}{N}

\dfrac{1}{N} \lt 0.01

N \gt 100

Integral test FAQ

What are the conditions for the integral test?

The function f with f(n) = a_n must be positive, continuous and decreasing for x ≥ 1, or at least from some point on. If any condition fails, the test does not apply.

Does the integral give the sum of the series?

No. The integral and the series converge or diverge together, but their values are usually different. For example, \int_1^{\infty} \frac{dx}{x^2} = 1, while \sum \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645.

Why does the harmonic series diverge?

The integral of 1/x from 1 to b is ln b, which grows without bound. By the integral test, the harmonic series diverges too.

When should I use the integral test?

Use it when the terms come from a function you can integrate easily, like \frac{1}{n\ln n} or ne^{-n^2}, and that function is positive and decreasing. For powers of n, the p-series test is quicker.

Summary

• If f is positive, continuous and decreasing with f(n) = a_n, then \sum a_n and \int_1^{\infty} f(x)\,dx both converge or both diverge.
• Check the three conditions, then evaluate the improper integral as a limit.
• The integral’s value is not the sum of the series.
• The remainder after N terms is between \int_{N+1}^{\infty} f(x)\,dx and \int_N^{\infty} f(x)\,dx.
• The harmonic series diverges because \int_1^{\infty} \frac{dx}{x} diverges.

Previous lesson: Telescoping Series

Next lesson: p-Series

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