AP Calculus BC

Why p Must Exceed 1 for a p-Series

The p-series theorem says the sum of one over n to the p converges exactly when p is strictly greater than one. The boundary is not inclusive: at p equal to one you get the harmonic series, whose terms tend to zero and whose sum still runs away without limit.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   The exponent satisfies p > 1, strictly

   $$p > 1$$

   This is a genuine threshold rather than a rule of thumb, and it is sharp on both sides. The integral test is what locates it: the improper integral of $x^{-p}$ from 1 to infinity converges precisely when $p > 1$, and the series inherits that boundary exactly.

   Drop it and the theorem fails

   $$\sum_{n=1}^{\infty} \frac{1}{n^{1}} = \sum_{n=1}^{\infty} \frac{1}{n}$$

   The harmonic series, the boundary case p = 1

   At $p = 1$ the terms are $1/n$, which tend to zero as slowly as anything you are likely to meet. The check recomputes the exponent from the terms and confirms it is 1, and the theorem then gives divergence. This case cannot be settled by adding terms up: the partial sums grow like $\ln n$, so after a million terms the total is only about 14, which looks for all the world like it is settling.

2. 2

   The terms really are a power of n, not merely similar to one

   $$a_n = n^{-p} \text{ exactly}$$

   The theorem is about a specific family. Terms like $1/(n \ln n)$ sit between $1/n$ and $1/n^{1+\varepsilon}$ for every positive $\varepsilon$ and belong to no p-series at all, so this theorem cannot classify them; the integral test handles that one and returns divergence. For terms that merely behave like a p-series far out, the limit comparison test is the right tool. No counterexample is displayed here, because the failure is that the theorem does not apply rather than that it gives a wrong answer.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. The function $f(x) = x^{-p}$ is positive, continuous and decreasing on $[1, \infty)$ for $p > 0$, so the integral test applies.
2. For $p \ne 1$ the improper integral evaluates to $\lim_{b \to \infty} \frac{b^{1-p} - 1}{1 - p}$.
3. That limit is finite exactly when the exponent $1 - p$ is negative, which says $p > 1$.
4. For $p = 1$ the antiderivative is $\ln x$ instead, and $\ln b$ runs to infinity, so the integral diverges and the series with it. The two cases have different antiderivatives, which is why $p = 1$ has to be checked separately rather than read off the general formula.

What it does not say