The nth term test for divergence

AP Calculus BC· AP Calculus BC CED — Infinite Sequences and Series· 14 min read

1. Definition and Core Purpose★☆☆☆☆⏱ 3 min

The nth term test for divergence (often shortened to the nth term test) is the most fundamental first-step convergence test for infinite series, regularly tested in both multiple-choice and free-response sections of the AP Calculus BC exam, accounting for 1-3% of total exam weight. It leverages a core necessary condition for convergence to quickly eliminate obviously divergent series before you apply more complex, computationally intensive tests.

📘 Definition

nth term test for divergence

For an infinite series $\sum_{n = 1}^{\infty} a_{n}$ with general term $a_{n}$ , if $lim_{n \to \infty} a_{n}$ either does not exist, or exists and is not equal to zero, then the series $\sum_{n = 1}^{\infty} a_{n}$ must diverge. The test can only confirm divergence; it can never confirm convergence.

Exam tip:

Always apply the nth term test first when checking any series for convergence. A 10-second limit calculation can save you 5 minutes of unnecessary computation on a clearly divergent series.

2. Core Theorem and Underlying Logic★★☆☆☆⏱ 4 min

To understand why the nth term test works, we start with the definition of infinite series convergence: a series $\sum_{n = 1}^{\infty} a_{n}$ converges if and only if its sequence of partial sums $S_{k} = \sum_{n = 1}^{k} a_{n}$ converges to a finite limit $L$ as $k \to \infty$ . By definition of partial sums, the $k$ th general term can be written as:

a_k = S_k - S_{k-1}

If $lim_{k \to \infty} S_{k} = L$ and $lim_{k \to \infty} S_{k - 1} = L$ , we can take the limit of both sides:

\lim_{k \to \infty} a_k = \lim_{k \to \infty} (S_k - S_{k-1}) = L - L = 0

This proves that for any convergent series, the limit of the $n$ th term must be zero. This is a necessary condition for convergence—you cannot have a convergent series without it. The nth term test is just the logically equivalent contrapositive of this statement: if the limit of the $n$ th term is not zero, the series cannot be convergent, so it must diverge. This logical equivalence means the test is always valid when applied correctly.

📐 Worked Example

Use the nth term test to determine if $\sum_{n = 1}^{\infty} \frac{2 n ^{2} + 5 n}{3 n ^{2} - 1}$ diverges.

1
1. Identify the general term of the series: $a_{n} = \frac{2 n ^{2} + 5 n}{3 n ^{2} - 1}$ .
2
1. Evaluate $lim_{n \to \infty} a_{n}$ by dividing numerator and denominator by the highest power of $n$ , which is $n^{2}$ :
3
$\lim_{n \to \infty} \frac{2n^2 + 5n}{3n^2 - 1} = \lim_{n \to \infty} \frac{2 + \frac{5}{n}}{3 - \frac{1}{n^2}}$
4
1. As $n \to \infty$ , terms with $\frac{1}{n}$ and $\frac{1}{n ^{2}}$ approach 0, so the limit simplifies to $\frac{2}{3} \neq = 0$ .
5
1. Since $lim_{n \to \infty} a_{n} \neq = 0$ , the series diverges by the nth term test.

Exam tip:

Always evaluate the limit of the nth term before moving to any other convergence test.

3. Interpreting Inconclusive Test Results★★☆☆☆⏱ 4 min

The most commonly tested aspect of the nth term test is understanding what it means when $lim_{n \to \infty} a_{n} = 0$ . In this case, the test is inconclusive—it cannot tell you whether the series converges or diverges. This is because the condition that $lim_{n \to \infty} a_{n} = 0$ is necessary, but not sufficient, for convergence. In other words, all convergent series have $lim_{n \to \infty} a_{n} = 0$ , but not all series with $lim_{n \to \infty} a_{n} = 0$ converge. A famous example is the harmonic series $\sum_{n = 1}^{\infty} \frac{1}{n}$ : its general term approaches 0, but the series diverges. By contrast, the p-series $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ also has a general term that approaches 0, but it converges. When the limit is 0, you must always use a different test (like the integral test or comparison test) to resolve convergence.

📐 Worked Example

A student claims that since $lim_{n \to \infty} \frac{1}{n} = 0$ , the series $\sum_{n = 1}^{\infty} \frac{1}{n}$ must converge. Is the student correct? Justify your answer using the nth term test.

1
1. First, confirm the student’s limit calculation: $lim_{n \to \infty} \frac{1}{n} = 0$ , which is algebraically correct.
2
1. Recall the scope of the nth term test: the test only confirms divergence when the limit of the nth term is non-zero or does not exist. When the limit equals 0, the test is inconclusive, and cannot be used to confirm convergence.
3
1. While we need a p-series test to confirm divergence for this series (it is a p-series with $p = 1/2 < 1$ ), even with a limit of 0, the nth term test cannot prove convergence.
4
1. Conclusion: The student is incorrect, because a limit of zero does not guarantee convergence via the nth term test.

Exam tip:

On FRQ justifications, never write 'the series converges because $lim_{n \to \infty} a_{n} = 0$ '—this will always lose points.

4. Non-Existent Limits and Divergence★★★☆☆⏱ 3 min

Many students forget that the nth term test applies not just when the limit exists and is non-zero, but also when the limit of $a_{n}$ as $n \to \infty$ does not exist (DNE). Common cases are oscillating sequences that do not approach a single fixed value, and unbounded sequences that grow without bound. In both cases, the condition $lim_{n \to \infty} a_{n} = 0$ fails, so the necessary condition for convergence is not met, and the series diverges. This is a straightforward application that is often missed by students who only look for finite non-zero limits.

📐 Worked Example

Test the series $\sum_{n = 1}^{\infty} cos (\frac{nπ}{2})$ for divergence using the nth term test.

1
1. Write out the first few terms of $a_{n} = cos (\frac{nπ}{2})$ to identify its behavior: $a_{1} = 0$ , $a_{2} = - 1$ , $a_{3} = 0$ , $a_{4} = 1$ , $a_{5} = 0$ , $a_{6} = - 1$ , and so on.
2
1. Evaluate the limit as $n \to \infty$ : $a_{n}$ oscillates indefinitely between $- 1$ , $0$ , and $1$ , and never approaches a single fixed limit. Therefore, $lim_{n \to \infty} a_{n}$ does not exist.
3
1. The nth term test states that if the limit of the nth term does not exist, the necessary condition for convergence fails, so the series must diverge.
4
1. Conclusion: $\sum_{n = 1}^{\infty} cos (\frac{nπ}{2})$ diverges by the nth term test.

Exam tip:

Always check if the limit exists before concluding anything. Oscillating or unbounded general terms always result in divergence, even if terms equal zero infinitely often.

5. AP-Style Worked Practice Problems★★★☆☆⏱ 4 min

📐 Worked Example

Multiple Choice: Which of the following series diverge by the nth term test for divergence?

I. $\sum_{n = 1}^{\infty} \frac{5 n}{n ^{2} + 1}$ II. $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n} 3 n}{2 n + 7}$ III. $\sum_{n = 1}^{\infty} \frac{1}{n ^{3/2}}$

(A) II only (B) I and II only (C) II and III only (D) I, II, and III

1
Evaluate the limit of the nth term for each series individually: For I, $lim_{n \to \infty} \frac{5 n}{n ^{2} + 1} = 0$ , so the nth term test is inconclusive, and I does not diverge by this test.
2
For II, $lim_{n \to \infty} \frac{( - 1 ) ^{n} 3 n}{2 n + 7}$ does not exist (it oscillates between $- 3/2$ and $3/2$ ), so II diverges by the nth term test.
3
For III, $lim_{n \to \infty} \frac{1}{n ^{3/2}} = 0$ , so the test is inconclusive, and III does not diverge by this test.
4
Only II diverges by the nth term test. The correct answer is (A).

📐 Worked Example

Free Response: Consider the infinite series $\sum_{n = 1}^{\infty} a_{n}$ , where $a_{n} = \frac{n ^{2} - 4 n + 10}{2 n ^{2} - n + 3}$ .

(a) Evaluate $lim_{n \to \infty} a_{n}$ . (b) What does the nth term test for divergence tell you about this series? (c) Can the nth term test tell you anything about the convergence of $\sum_{n = 1}^{\infty} \frac{1}{a _{n}}$ ? Justify your answer.

1
(a) Divide numerator and denominator by $n^{2}$ to evaluate the limit:
2
$\lim_{n \to \infty} \frac{n^2 - 4n + 10}{2n^2 - n + 3} = \lim_{n \to \infty} \frac{1 - \frac{4}{n} + \frac{10}{n^2}}{2 - \frac{1}{n} + \frac{3}{n^2}} = \frac{1}{2}$
3
So $lim_{n \to \infty} a_{n} = \frac{1}{2}$ .
4
(b) By the nth term test, since $lim_{n \to \infty} a_{n} = \frac{1}{2} \neq = 0$ , the series $\sum_{n = 1}^{\infty} a_{n}$ diverges.
5
(c) Evaluate the limit of the new general term: $lim_{n \to \infty} \frac{1}{a _{n}} = \frac{1}{1/2} = 2 \neq = 0$ . By the nth term test, a non-zero limit confirms divergence, so yes, the nth term test can confirm that $\sum_{n = 1}^{\infty} \frac{1}{a _{n}}$ diverges.

📐 Worked Example

Real-World Application: An urban planner models the annual net change in the number of households in a growing city with an infinite series $\sum_{n = 0}^{\infty} Δ H_{n}$ , where $Δ H_{n} = 1200 \cdot \frac{0. 8 ^{n} + 1. 2 ^{n}}{1. 2 ^{n}}$ gives the net change (in households) in year $n$ after the start of the model. Will the total long-run number of new households approach a finite limit, according to this model? Justify your answer using the nth term test.

1
First simplify the general term:
2
$\Delta H_n = 1200 \left( \frac{0.8^n}{1.2^n} + \frac{1.2^n}{1.2^n} \right) = 1200 \left( \left(\frac{2}{3}\right)^n + 1 \right)$
3
Evaluate the limit as $n \to \infty$ : $(\frac{2}{3})^{n} \to 0$ , so $lim_{n \to \infty} Δ H_{n} = 1200 (0 + 1) = 1200$ , which is non-zero.
4
By the nth term test, the series diverges, meaning the total number of households grows without bound. According to this model, the total long-run number of new households does not approach a finite limit—it grows by a net 1200 households per year indefinitely.

6. Common Pitfalls

Wrong move:

Concluding a series converges because $lim_{n \to \infty} a_{n} = 0$

Why:

Students confuse the necessary condition for convergence with a sufficient condition, since the nth term test is the first convergence test they learn.

Correct move:

Whenever you get $lim_{n \to \infty} a_{n} = 0$ , explicitly note that the nth term test is inconclusive, then apply another convergence test to determine convergence.

Wrong move:

Applying the nth term test to the sequence of partial sums instead of the general term of the series

Why:

Students mix up the definition of series vs partial sums, especially on exam questions about partial sum behavior.

Correct move:

Always confirm you are taking the limit of the individual general term $a_{n}$ , not the partial sum $S_{n}$ , when applying the nth term test.

Wrong move:

Approximating the limit of $a_{n}$ for large $n$ as 'approximately zero' and concluding the test is inconclusive when the actual limit is non-zero

Why:

Students confuse small numerical values of $a_{n}$ for large finite $n$ with a limit of zero as $n \to \infty$ .

Correct move:

Always compute the exact limit algebraically, don't rely on approximate values of individual terms. Any non-zero limit, no matter how small, means the series diverges.

Wrong move:

Claiming the series diverges because $lim_{n \to \infty} \frac{a _{n + 1}}{a _{n}} \neq = 1$ , and misattributing this to the nth term test

Why:

Students mix up the nth term test with the ratio test early in the unit.

Correct move:

Keep test names straight: the nth term tests the limit of $a_{n}$ , not the ratio of consecutive terms. If you use the ratio test, name it correctly.

Wrong move:

Ignoring that the limit does not exist, and forcing a conclusion of 'inconclusive' for oscillating sequences because some terms are zero

Why:

Students see occasional zero terms so they incorrectly assume the limit is zero.

Correct move:

If the sequence $a_{n}$ oscillates between non-zero values or grows without bound, the limit does not exist, which immediately gives divergence by the nth term test.

7. Quick Reference Cheatsheet

Category	Statement/Formula	Key Notes
Core nth term test statement	For $\sum_{n = 1}^{\infty} a_{n}$ , if $lim_{n \to \infty} a_{n} \neq = 0$ or $lim_{n \to \infty} a_{n}$ DNE, then $\sum a_{n}$ diverges	Only proves divergence; never use to prove convergence
Necessary condition for convergence	If $\sum a_{n}$ converges, then $lim_{n \to \infty} a_{n} = 0$	Always true for all convergent series, no exceptions
Inconclusive test result	If $lim_{n \to \infty} a_{n} = 0$ , test is inconclusive	Series may converge or diverge; apply another test
Non-existent limit rule	Oscillating/unbounded $a_{n}$ means $lim_{n \to \infty} a_{n}$ DNE	DNE limit always implies divergence
Inconclusive example 1	$\sum_{n = 1}^{\infty} \frac{1}{n}$ , $lim_{n \to \infty} \frac{1}{n} = 0$	Series diverges, classic inconclusive result
Inconclusive example 2	$\sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ , $lim_{n \to \infty} \frac{1}{n ^{2}} = 0$	Series converges, second example of inconclusive result
First step rule	Always apply nth term test first	Saves time on obviously divergent series before complex tests

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · BC MCQ
Identify divergent series via nth term test
2019 · BC FRQ
Justify conclusion using nth term test

What's Next

The nth term test is the foundational first test for infinite series, and a required prerequisite for every other convergence test you will study next. Every time you test a series for convergence, you will start by applying this test to rule out obvious divergence before moving to more computationally intensive methods. Without mastering the logic of the nth term test and its limitations, you will almost certainly lose points on FRQ justifications, as AP readers regularly penalize incorrect claims that a limit of zero guarantees convergence. This test sets the stage for all subsequent work with infinite series, so mastering its scope and limitations is critical for success on the AP Calculus BC exam. Explore the follow-up resources below to continue your study of series convergence: