AP Calculus BC

Integral Test vs p-Series Test

Q: Why is p = 1 the boundary?

Because \int_1^{\infty} \frac{1}{x}\,dx is \ln x evaluated to infinity, which diverges, while any p > 1 leaves a negative power that decays fast enough to give a finite integral. The harmonic series sits on that boundary and diverges.

Q: Do I have to start the integral at 1?

No. Start anywhere past which f is positive, continuous, and decreasing. Changing the starting index changes the sum but never changes whether the series converges.

The p-series result is the integral test already carried out for terms of the form one over n to the power p: it converges when p is greater than 1 and diverges otherwise. Quote the p-series rule when the terms match that form exactly, and run the integral test yourself when they do not.

Integral test

Use when: The terms come from a positive, decreasing, continuous function you can actually antidifferentiate, such as one carrying a logarithm.

p-series test

Use when: The terms are exactly a reciprocal power of the index, so the answer is a one-line citation with no integration at all.

Side by side

	Integral test	p-series test
What you do	Evaluate $\int_1^{\infty} f(x)\,dx$	Read off $p$ and compare it to $1$
Applies to	Any positive, decreasing, continuous $f$ with $f(n) = a_n$	Only $\sum \frac{1}{n^p}$
Converges when	The improper integral converges	$p > 1$
Typical case	$\sum \frac{1}{n \ln n}$	$\sum \frac{1}{n^2}$
Common trap	Reporting the integral's value as the sum	Reading $p$ off terms that are not a pure power

The p-series rule is not a separate idea. Apply the integral test to $f(x) = x^{-p}$ , which is positive and decreasing exactly when $p > 0$ , and the improper integral $\int_1^{\infty} x^{-p}\,dx$ converges exactly when $p > 1$ ; at $p = 1$ the antiderivative is $\ln x$ , which grows without bound, so the harmonic series diverges. Once that work is done you never repeat it, which is what makes the p-series a shortcut rather than a test. For $p \le 0$ the terms do not even approach zero, so the nth term test disposes of those cases before the integral test is needed.

The integral test earns its keep when no shortcut fits. Its hypotheses are that $f$ is positive, continuous, and decreasing past some starting point, and that $f(n) = a_n$ . It reports convergence or divergence only. The integral and the sum are different numbers, though the integral does bound the tail left over after any partial sum.

The case the p-series cannot touch

For $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ there is no $p$ to read off. Substituting $u = \ln x$ gives $\int \frac{1}{x \ln x}\,dx = \ln(\ln x)$ , which grows without bound, so the series diverges. Change the denominator to $n(\ln n)^2$ and the same substitution converges, so that series converges.

Frequently asked questions

Does the integral test give the sum of the series?

No. It decides convergence and nothing more. The integral and the series are different numbers, although the integral of the tail does bound the remainder after a partial sum, which is how integral-test error estimates work.

Why is $p = 1$ the boundary?

Because $\int_1^{\infty} \frac{1}{x}\,dx$ is $\ln x$ evaluated to infinity, which diverges, while any $p > 1$ leaves a negative power that decays fast enough to give a finite integral. The harmonic series sits on that boundary and diverges.

Do I have to start the integral at 1?

No. Start anywhere past which $f$ is positive, continuous, and decreasing. Changing the starting index changes the sum but never changes whether the series converges.

In the CED: Unit 10: Infinite Sequences and Series (BC)