The Integral Test and p-Series

Quick answer

When f is positive, continuous and decreasing, the series of values f(n) and the integral of f from 1 to ∞ either both converge or both diverge: the terms of the series are the areas of rectangles hugging the curve. This settles the p-series: the sum of 1/nᵖ converges exactly when p > 1. The harmonic series, the sum of 1/n, diverges, though its terms shrink to 0. The integral also bounds the error of stopping after n terms.

What you'll learn

Apply the integral test and check its conditions
Show that the harmonic series diverges
Classify p-series as convergent or divergent
Bound the error of a partial sum with an integral

A series next to an integral

A series with positive terms converges exactly when its partial sums stay bounded: they only go up, so they either level off or climb forever. To decide which, compare the series with an area you can compute. If $a_n = f(n)$ for a positive, decreasing function $f$ , each term is the area of a rectangle of width $1$ standing at the curve.

The integral test

Suppose $f$ is positive, continuous and decreasing for $x \ge 1$ , and $a_n = f(n)$ . Then $\sum_{n=1}^{\infty} a_n$ and $\int_1^{\infty} f(x)\,dx$ either both converge or both diverge.

Why the rectangles decide

Stand each rectangle of height $a_n$ on the interval $[n, n + 1]$ . Since $f$ is decreasing, its left edge is the tallest point of the curve over that interval, so the rectangles cover the region under the curve:

a_1 + a_2 + \cdots + a_n \ge \int_1^{n+1} f(x)\,dx

y = 1/x

The harmonic series as rectangles over y = 1/x

Shift the rectangles one unit left, onto $[n - 1, n]$ , and they fit under the curve instead:

a_2 + a_3 + \cdots + a_n \le \int_1^{n} f(x)\,dx

y = 1/x²

The series of 1/n² as rectangles under y = 1/x²

If the integral is infinite, the first inequality drags the partial sums up without bound. If it is finite, the second keeps them below $a_1 + \int_1^{\infty} f(x)\,dx$ . The series and the integral measure nearly the same area, so they are finite together or infinite together.

p-series

The integral test settles a whole family at once, because the $p$ -integrals are already known.

p-series

$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$ .

For $p \le 0$ the terms do not even approach $0$ . For $p > 0$ , $\tfrac{1}{x^p}$ is positive and decreasing, and $\int_1^{\infty} \tfrac{dx}{x^p}$ converges exactly when $p > 1$ .

Estimating the remainder

The same rectangles bound the error of stopping early. If the series passes the integral test, the part left out after $n$ terms satisfies

S - S_n = a_{n+1} + a_{n+2} + \cdots \le \int_n^{\infty} f(x)\,dx

Worked examples

Common mistakes

Practice problems

Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ converge?

Answer

Yes

Full solution

It is a $p$ -series with $p = 3 > 1$ .
Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{0.9}}$ converge?

Answer

No

Full solution

It is a $p$ -series with $p = 0.9 \le 1$ .
Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3}}$ converge?

Answer

Yes

Full solution

$\sqrt{n^3} = n^{3/2}$ , so $p = \tfrac{3}{2} > 1$ .
Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n + 1}$ converge?

Answer

No

Full solution

It is the harmonic series without its first term: $\tfrac{1}{2} + \tfrac{1}{3} + \cdots$ . Dropping one term does not change divergence. The integral test gives $\int_1^{\infty} \tfrac{dx}{x + 1} = \infty$ as well.
Use the integral test on $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ .

Answer

It converges.

Full solution

$\tfrac{1}{x^2 + 1}$ is positive and decreasing for $x \ge 1$ , and $\int_1^{\infty} \tfrac{dx}{x^2 + 1} = \tfrac{\pi}{2} - \tfrac{\pi}{4} = \tfrac{\pi}{4}$ is finite.
Use the integral test on $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ .

Answer

It diverges.

Full solution

$\tfrac{\ln x}{x}$ is positive and decreasing for $x \ge 3$ , and $\int_3^{b} \tfrac{\ln x}{x}\,dx = \tfrac{(\ln b)^2}{2} - \tfrac{(\ln 3)^2}{2}$ grows without bound. The first two terms do not affect convergence.
Use the integral test on $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ .

Answer

It converges.

Full solution

With $u = \ln x$ , $\int_2^{\infty} \tfrac{dx}{x(\ln x)^2} = \left[-\tfrac{1}{\ln x}\right]_2^{\infty} = \tfrac{1}{\ln 2}$ , which is finite.
Bound the error in $S_5$ for $\sum \tfrac{1}{n^3}$ .

Answer

At most $0.02$

Full solution

$\int_5^{\infty} \tfrac{dx}{x^3} = \tfrac{1}{2 \cdot 5^2} = 0.02$ .
For which $p$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ converge?

Answer

For $p > \tfrac{1}{2}$

Full solution

It is a $p$ -series with exponent $2p$ , which must exceed $1$ .
A student concludes that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \int_1^{\infty} \frac{dx}{x^2} = 1$ . What went wrong?

Hint

The first term of the series alone is $1$ .

Answer

The integral test compares convergence, not values. The sum is $\tfrac{\pi^2}{6} \approx 1.645$ .

Full solution

All the terms are positive and the first is $1$ , so the sum is more than $1$ . The integral is the area under the curve; the series is the area of rectangles that stick out above it, as in the first graph.

Frequently asked questions

What does the integral test say?

If f is positive, continuous and decreasing and aₙ = f(n), then the series of aₙ and the integral of f from 1 to ∞ both converge or both diverge.

When does a p-series converge?

The sum of 1/nᵖ converges when p > 1 and diverges when p ≤ 1.

Does the harmonic series converge?

No. Its partial sums grow like ln n: slowly, but without bound.

Is the sum of the series equal to the integral?

No. The test compares whether they are finite, not their values. The sum of 1/n² is π²/6 ≈ 1.645, while the integral of 1/x² from 1 to ∞ is 1.

How accurate is a partial sum?

For a series that passes the integral test, the error after n terms is at most the integral of f from n to ∞.

What to learn next

The Comparison Tests for SeriesJudge a series by comparing it with one you know: the direct comparison test, the limit comparison test, and how to choose a benchmark series.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

A series next to an integral

Why the rectangles decide

p-series

Estimating the remainder

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What does the integral test say?

When does a p-series converge?

Does the harmonic series converge?

Is the sum of the series equal to the integral?

How accurate is a partial sum?

What to learn next

More Calculus lessons