AP®MathsCollege BoardCalculus BCStudy GuidesUnit 10: Infinite Sequences and SeriesTests for Divergence & Convergencenth Term Test for Divergence

nth Term Test for Divergence (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 19 May 2026

nth term test

What is the nth term test?

The nth term test is a way to show that an infinite series diverges
- You cannot show that a series converges using this test
If a series converges, then the terms must become small and tend to zero
- This can be used to identify series that do not converge
The nth term test states that:
- Given a series $\sum_{n = 0}^{\infty} a_{n}$
- if $\lim_{n \to \infty} a_{n} \neq 0$ then the series diverges
Note that if $\lim_{n \to \infty} a_{n} = 0$ , this does not necessarily mean that the series converges
- Other tests must be used to determine convergence or divergence of such a series

Examiner Tips and Tricks

The nth term test is a good 'first test' to use when examining a series, as it is usually quick and simple to apply.

Worked Example

Determine whether the series $\sum_{n = 1}^{\infty} \frac{3 n^{2}}{1 - 5 n^{2}}$ converges or diverges.

Answer:

Consider the limit of the nth term, $\lim_{n \to \infty} \frac{3 n^{2}}{1 - 5 n^{2}}$

$\frac{3 n^{2}}{1 - 5 n^{2}} = \frac{3 n^{2}}{1 - 5 n^{2}} \cdot \frac{\frac{1}{n^{2}}}{\frac{1}{n^{2}}} = \frac{3}{\frac{1}{n^{2}} - 5}$

$\lim_{n \to \infty} (\frac{3 n^{2}}{1 - 5 n^{2}}) = \lim_{n \to \infty} (\frac{3}{\frac{1}{n^{2}} - 5}) = \frac{3}{0 - 5} = - \frac{3}{5}$

That limit is not equal to zero, so the series diverges

$\lim_{n \to \infty} \frac{3 n^{2}}{1 - 5 n^{2}} \neq 0$ so $\sum_{n = 1}^{\infty} \frac{3 n^{2}}{1 - 5 n^{2}}$ diverges by the nth term test

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