10.3 The nth Term Test for Divergence

TLDR

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.

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.

On multiple choice, the nth term test can rule out a divergent series in seconds. On free response, computing the limit of the general term is a clean way to justify that a series diverges.

What the nth Term Test Can and Cannot Prove

• 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.

Proving Divergence by Taking the Limit of Terms

Problem Solving: There are three steps:

1. Write the limit of the general term, $\displaystyle\lim_{n \to \infty} a_n$.
2. Evaluate the limit.
3. 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

• Misconception: If the terms go to zero, the series converges.
  Actually: 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.
• Misconception: The nth term test can prove convergence.
  Actually: It cannot. It only ever proves divergence. When the limit is 0, the test gives no information.
• Misconception: If the limit does not exist, the test fails.
  Actually: 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.
• Misconception: I should always run the nth term test last.
  Actually: Run it first. It is the quickest way to spot an obviously divergent series before doing more work.
• Misconception: Limit equals zero means I made a mistake.
  Actually: A limit of zero is normal and common. It just means you need another test to finish the problem.

Related AP Calculus Guides

• Unit 10 Review: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence
• 10.6 Comparison Tests for Convergence