The nth term test for divergence is a quick check: if the terms of a series do not approach zero, the series diverges. It only proves divergence, so if the terms do go to zero, the test tells you nothing and you need a different test. This is a BC-only topic in AP Calculus.

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How Does the nth Term Test Work?

The nth term test asks whether the individual terms $a_n$ approach 0. If $\lim_{n\to\infty}a_n\neq 0$ or the limit does not exist, then $\sum a_n$ diverges; if $\lim_{n\to\infty}a_n=0$ , the test is inconclusive and you need another convergence test.

Why This Matters for the AP Calculus Exam

The nth term test is usually your first move when you see an infinite series on the AP Calculus BC exam. It is fast, and in multiple choice questions it can rule out divergent series in seconds. On free response questions that involve series, you may need to justify why a series diverges, and computing the limit of the general term is a clean way to do that. Knowing exactly what this test can and cannot conclude saves you from wasting time and from writing incorrect justifications.

Key Takeaways

The test in symbols: if $\displaystyle\lim_{n \to \infty} a_n \neq 0$ (or the limit does not exist), then $\sum a_n$ diverges.
If $\displaystyle\lim_{n \to \infty} a_n = 0$ , the test is inconclusive. The series could converge or diverge.
Terms going to zero is a necessary condition for convergence, not a sufficient one.
Three steps: write the limit of $a_n$ , evaluate it, then conclude based on whether it equals zero.
Use this as a preliminary check before reaching for the integral, comparison, ratio, or alternating series tests.
The harmonic series $\sum \frac{1}{n}$ shows why "terms to zero" is not enough: its terms go to zero, but the series still diverges.

What is the nth Term Test for Divergence?

The nth term test (also called the divergence test or term test) tells you when a series diverges. It states:

\text{if } \displaystyle\lim_{n \to \infty} a_n \neq 0,\text{ then } \sum a_n \text{ diverges}

If the nth term does not approach 0, the series diverges. If the nth term does approach 0, the test cannot decide anything. The series might converge, or it might still diverge. Everything hinges on whether the general term tends to zero.

This works because of a basic fact about convergent series: if a series converges, its terms must approach zero. The nth term test is the contrapositive of that statement. So passing the divergence test (meaning the terms go to zero) never proves convergence.

If $a_n \to 0$ , you cannot say the series converges. You only know the divergence test failed to show divergence.

How to Use This on the AP Calculus Exam

Problem Solving

There are three steps:

Write the limit of the general term, $\displaystyle\lim_{n \to \infty} a_n$ .
Evaluate the limit.
Conclude based on the nth term test.

Determine if the following series diverges.

\sum_{n=1}^{\infty} \dfrac{5n -1}{n-2}

Step 1: Write the limit.

\lim_{n \to \infty} \dfrac{5n -1}{n-2}

Step 2: Evaluate. Factor out the highest power of $n$ from the top and bottom.

\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})}

\lim_{n\to\infty}\dfrac{5-\frac{1}{n}}{1-\frac{2}{n}} = \frac{5}{1}

Recall that any finite number divided by $\infty$ is 0.

\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})} = \frac{5}{1} \neq 0

Step 3: Conclude.

\therefore \ \sum_{n=1}^{\infty} \dfrac{5n -1}{n-2} \ \ \text{diverges}

Because the limit of the terms is 5, not 0, the series diverges. You are applying a new test to limit work you already know.

Practice Problems

Try these two yourself.

1. \ \sum_{n=1}^{\infty} \dfrac{n^3+3n}{2n^3-5}

2. \ \sum_{n=1}^{\infty} \arctan(n)

Solution 1

\lim_{n \to \infty} \dfrac{n^3+3n}{2n^3-5}

\lim_{n \to \infty} \dfrac{n^3(1+\dfrac{3}{n^2})}{n^3(2-\dfrac{5}{n^3})}

\lim_{n\to\infty}\dfrac{1+\dfrac{3}{n^2}}{2-\dfrac{5}{n^3}} = \dfrac{1}{2}

The limit is $\frac{1}{2} \neq 0$ , so:

\therefore \ \sum_{n=1}^{\infty} \dfrac{n^3+3n}{2n^3-5} \ \ \text{diverges}

Solution 2

\lim_{n \to \infty} \arctan(n)

As $n$ goes to $\infty$ , $\arctan(n)$ approaches $\dfrac{\pi}{2}$ .

\ \lim_{n \to \infty} \arctan(n) = \dfrac{\pi}{2}

Since the limit is $\frac{\pi}{2} \neq 0$ :

\therefore \ \sum_{n=1}^{\infty} \arctan(n) \ \text{diverges}

Common Trap

When the limit of the terms is 0, stop and switch tests. Do not write a conclusion about convergence based on the nth term test alone. Reach for the integral test, comparison or limit comparison test, ratio test, or alternating series test instead.

Common Misconceptions

"If the terms go to zero, the series converges." This is false. The harmonic series $\sum \frac{1}{n}$ has terms that go to zero, yet it diverges. Terms going to zero only means the test is inconclusive.
"The nth term test can prove convergence." It cannot. It only ever proves divergence. When the limit is 0, the test gives no information.
"If the limit does not exist, the test fails." Actually, if the limit of $a_n$ does not exist (for example, $a_n = (-1)^n$ or $a_n = \sin(n)$ ), the terms do not approach 0, so the series diverges.
"I should always run the nth term test last." Run it first. It is the quickest way to spot an obviously divergent series before doing more work.
"Limit equals zero means I made a mistake." A limit of zero is normal and common. It just means you need another test to finish the problem.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
converges	A series converges when the sequence of partial sums approaches a finite limit as n approaches infinity.
diverges	A series diverges when the sequence of partial sums does not approach a finite limit as the number of terms increases indefinitely.
nth term test	A test for divergence that examines whether the limit of the nth term of a series equals zero; if the limit is not zero, the series diverges.
series	A sum of the terms of a sequence, often written as the sum of infinitely many terms.

Frequently Asked Questions

What is the nth term test for divergence?

The nth term test says that if the terms a_n of a series do not approach 0, or if the limit of a_n does not exist, then the series diverges.

Can the nth term test prove that a series converges?

No. The nth term test only proves divergence. If lim a_n = 0, the test is inconclusive and you need another convergence test.

What do you do if the nth term limit is zero?

If lim a_n = 0, stop using the nth term test and choose another test, such as p-series, geometric, integral, comparison, alternating series, or ratio test depending on the form.

Why does a series diverge when its terms do not go to zero?

For a series to converge, the added terms must eventually become small enough to approach 0. If the individual terms stay away from 0, the partial sums cannot settle to a finite value.

Should you use the nth term test first?

Usually yes. It is a quick preliminary check for divergence. If it does not prove divergence, move on to a test that can actually prove convergence or divergence for that series form.

How is the nth term test used on AP Calculus BC?

On AP Calculus BC, the nth term test is used to justify divergence quickly. A complete answer states the limit of a_n and explains that because the limit is not 0, the series diverges.