Skip to main content
United StatesCalculusSyllabus dot point

How does the nth term test show a series diverges, and why can it never prove convergence?

Topic 10.3 The nth Term Test for Divergence: use the limit of the terms to conclude divergence when the terms do not approach zero (BC).

A focused answer to AP Calculus BC Topic 10.3, using the nth term test to conclude divergence when the terms fail to approach zero, and understanding why it can never prove convergence, with worked examples.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. The test and its one-way logic
  3. Why it can only prove divergence
  4. A worked divergence conclusion
  5. When the limit does not exist
  6. Where it fits in the testing strategy

What this topic is asking

The College Board (Topic 10.3, BC only) gives the nn-th term test for divergence, the quickest divergence check and the one to try first on any series. It uses a simple necessary condition: if a series converges, its terms must shrink to zero, so terms that do not shrink to zero force divergence.

The test and its one-way logic

Why it can only prove divergence

The test rests on a theorem: if βˆ‘an\sum a_n converges, then lim⁑nβ†’βˆžan=0\lim_{n\to\infty} a_n = 0. Logically, the test is the contrapositive: if the terms do not go to zero, the series cannot converge, so it diverges. But the original theorem is a one-way implication; its converse ("terms go to zero implies convergence") is false. The decisive counterexample is the harmonic series βˆ‘1n\sum\frac{1}{n}, whose terms 1nβ†’0\frac{1}{n}\to 0 yet whose partial sums grow without bound. This is why lim⁑an=0\lim a_n = 0 leaves you no conclusion: the condition is necessary for convergence but not sufficient. Explaining this distinction is a recurring free-response request, as in the worked exam question above.

A worked divergence conclusion

When the limit does not exist

The test also fires when the terms have no limit at all, not just a nonzero one. For βˆ‘(βˆ’1)n\sum (-1)^n, the terms alternate βˆ’1,1,βˆ’1,1,…-1, 1, -1, 1, \ldots and never settle, so lim⁑an\lim a_n does not exist; the series diverges. Likewise βˆ‘cos⁑n\sum\cos n has terms that oscillate without approaching anything, giving divergence. The unifying statement is that convergence requires lim⁑an=0\lim a_n = 0; any failure of that, whether a nonzero limit or no limit, means divergence. So when you compute the limit of the terms, watch for oscillation as well as nonzero values, and treat both as immediate divergence.

Where it fits in the testing strategy

The nn-th term test is the first move in any convergence problem because it is cheap and decisive when it applies. If the terms obviously do not tend to zero (a ratio of equal-degree polynomials, a constant, an oscillation), you are done: divergence. If the terms do tend to zero, you have learned only that you need a real test, the integral test, comparison, ratio, alternating-series, or pp-series, depending on the form. Building the habit of checking lim⁑an\lim a_n first saves time and catches divergent series that a more elaborate test would also catch but more slowly. Just never write "the terms go to zero, so the series converges," which inverts the test illegitimately.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (BC, style)1 marksSection I (multiple choice, no calculator). The nth term test shows which series diverges? (A) βˆ‘n2n+1\sum \frac{n}{2n + 1} (B) βˆ‘1n2\sum \frac{1}{n^2} (C) βˆ‘1n\sum \frac{1}{n} (D) βˆ‘12n\sum \frac{1}{2^n}
Show worked answer β†’

The correct answer is (A), βˆ‘n2n+1\sum \frac{n}{2n+1}.

lim⁑nβ†’βˆžn2n+1=12β‰ 0\lim_{n\to\infty}\frac{n}{2n+1} = \frac{1}{2}\neq 0, so the nth term test gives divergence. The others have terms tending to 00, so the test is inconclusive for them (and B, D actually converge).

AP 2023 (BC, style)4 marksSection II (free response, no calculator). (a) State the nth term test for divergence. (b) Apply it to βˆ‘n=1∞3n2+1n2+5\sum_{n=1}^{\infty} \frac{3n^2 + 1}{n^2 + 5}. (c) Explain why the test cannot be used to prove a series converges.
Show worked answer β†’

A 4-point conceptual problem.

(a) (1 point) If lim⁑nβ†’βˆžanβ‰ 0\lim_{n\to\infty} a_n \neq 0 (or does not exist), then βˆ‘an\sum a_n diverges.
(b) (2 points) lim⁑nβ†’βˆž3n2+1n2+5=3β‰ 0\lim_{n\to\infty}\frac{3n^2 + 1}{n^2 + 5} = 3 \neq 0, so the series diverges.
(c) (1 point) lim⁑an=0\lim a_n = 0 is necessary but not sufficient; when the limit is 00 the test is inconclusive (e.g. βˆ‘1n\sum\frac{1}{n} has terms tending to 00 yet diverges).

Related dot points

Sources & how we know this