AP Calculus BC

Why the nth Term Test Cannot Prove Convergence

The nth term test proves divergence and nothing else. If the terms tend to zero it is silent, because terms tending to zero is necessary for convergence but nowhere near sufficient. The harmonic series has terms tending to zero and diverges anyway.

nth Term Test for Divergence

lim⁡n→∞an≠0  ⟹  ∑an diverges\lim_{n \to \infty} a_n \ne 0 \implies \sum a_n \text{ diverges}

If the terms of a series do not tend to zero, the series diverges. If they do tend to zero, the test gives no information.

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 limit of the terms is nonzero, or fails to exist

    lim⁡n→∞an≠0\lim_{n \to \infty} a_n \ne 0

    This is the whole hypothesis, and the test is a one-way implication. Reading it backwards, to conclude convergence from terms that vanish, is the most common error in Unit 10, and the harmonic series is the standing refutation.

    Drop it and the theorem fails

    ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}

    The harmonic series, whose terms tend to zero

    The terms 1/n1/n tend to zero, so the nth term test is silent here. If vanishing terms implied convergence, this series would converge. It does not: it is a p-series with p=1p = 1, and the check recomputes that exponent from the terms themselves rather than trying to settle the question by adding them up, which no amount of computation could do.

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. Suppose the series converges, with partial sums SnS_n tending to some finite SS.
  2. Then Sn−1S_{n-1} also tends to SS, since it is the same sequence shifted by one place.
  3. The individual term is the difference: an=Sn−Sn−1a_n = S_n - S_{n-1}, which therefore tends to S−S=0S - S = 0.
  4. So convergence forces the terms to vanish. Contrapositive: terms that do not vanish rule out convergence. Notice the argument gives nothing in the other direction, which is exactly why the test cannot prove convergence.

What it does not say

Terms tending to zero means the series converges.

The single most common error in series. The harmonic series is the counterexample, and so is every p-series with p≤1p \le 1. Vanishing terms are necessary and not sufficient.

The test is inconclusive, so try it again more carefully.

When the terms vanish the test is finished, permanently. Move to a test that can prove convergence: comparison, integral, ratio, root, or alternating series.

It is called the nth term test, so it tests the nth term for convergence.

It tests the limit of the terms in order to say something about the SERIES. A sequence tending to zero and a series converging are different questions, and this test is the bridge that only carries traffic one way.

Frequently asked questions

When should I use the nth term test?

First, always. It is one limit and it either finishes the problem or costs you nothing. Only when the terms vanish do you need to reach for a real convergence test.

What if the limit of the terms does not exist?

That also proves divergence. The hypothesis is that the terms fail to tend to zero, and oscillating forever without settling is one way to fail. ∑(−1)n\sum (-1)^n diverges for this reason.