The nth Term Test for Divergence - AP Calculus BC

SECTION 1

Historical Context & Motivation

The study of infinite series stretches back to antiquity, but the rigorous tools for determining whether an infinite sum actually converges to a finite value only crystallized over the course of several centuries. Greek mathematicians such as Archimedes computed specific geometric sums, yet they lacked a general framework for deciding when adding infinitely many terms could yield a meaningful result. The development of the nth Term Test for Divergence arose naturally once mathematicians formalized the notion of limits, recognizing that a simple necessary condition could immediately rule out convergence for many series.

~250 BCE • Archimedes and Geometric Series

Archimedes computed the sum of the geometric series 1 + 1/4 + 1/16 + ⋯ by exhaustion methods. Though he lacked limit notation, he demonstrated that certain infinite processes converge to a finite area.

1668 • Mercator and the Harmonic Series

Nicolaus Mercator studied the logarithmic series and related sums. Pietro Mengoli and later Johann Bernoulli proved the harmonic series diverges despite its terms approaching zero — a pivotal counterexample showing that the converse of the nth Term Test is false.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy published his rigorous treatment of limits, sequences, and series. He formally stated that if a series converges, then its general term must tend to zero, providing the logical foundation for the nth Term Test.

1880s–1900s • Weierstrass and Formal Analysis

Karl Weierstrass and others completed the epsilon-delta formalization of limits. The nth Term Test became a standard first-step diagnostic taught alongside the integral test, comparison tests, and ratio test in modern analysis courses.

The critical insight that motivated the nth Term Test is deceptively simple: if you keep adding nonzero (or non-vanishing) quantities forever, the running total cannot stabilize at a finite number. This observation, once placed on rigorous footing by Cauchy, gave mathematicians a quick and powerful filter. Before investing effort in sophisticated convergence tests, one can check whether the individual terms of a series tend to zero. If they do not, the series must diverge, and no further analysis is needed.

SECTION 2

Core Principles & Definitions

The nth Term Test for Divergence is the most elementary test in the series-convergence toolkit, yet it is also one of the most frequently misapplied. To wield it correctly, you need to understand not only what it says but also — critically — what it does not say. The test is a one-directional logical gate: it can confirm divergence, but it can never confirm convergence. This asymmetry is the single most important conceptual point in the entire topic.

The Statement

If lim(n→∞) aₙ ≠ 0 or the limit does not exist, then the series Σaₙ diverges.

The Contrapositive

If Σaₙ converges, then lim(n→∞) aₙ = 0. This is logically equivalent to the test and is sometimes called the 'necessary condition for convergence.'

The Converse is FALSE

If lim(n→∞) aₙ = 0, the series may converge or diverge. The harmonic series Σ(1/n) is the classic counterexample: its terms tend to zero, yet it diverges.

The Test is Inconclusive When lim = 0

When lim(n→∞) aₙ = 0, the nth Term Test yields no conclusion. You must apply another test — comparison, integral, ratio, root, or alternating series — to determine convergence.

✦

KEY TAKEAWAY

Think of the nth Term Test like a bouncer at the door of a convergence club. If a series's terms do not approach zero, the bouncer rejects it immediately — the series diverges. But if the terms do approach zero, the bouncer merely lets the series through the door for further screening. Passing the bouncer does not guarantee admission; it only means the series has not been ruled out yet. In engineering and signal processing, this is analogous to a necessary-but-not-sufficient condition for a filter to be stable: the impulse response decaying to zero is required, but additional conditions determine actual stability.

SECTION 3

Visual Explanation

Visualizing the Necessary Condition

The diagram below contrasts two series: one whose general term tends to zero (the convergent geometric series Σ(1/2)ⁿ) and one whose general term does not tend to zero (the constant series Σ1). By plotting the partial sums, we see that the convergent series levels off toward a horizontal asymptote while the divergent series grows without bound. The nth Term Test captures precisely this intuition: if the individual terms being added are not shrinking to zero, the running total cannot stabilize.

The red curve shows the partial sums of Σ1, which diverges because each term equals 1 (the limit of aₙ is not zero). The cyan curve shows Σ(1/2)ⁿ, whose terms shrink to zero and whose partial sums approach the finite limit S = 1.

Notice the stark visual difference: the red partial sums accelerate away from any finite bound because each term contributes a full unit, while the cyan partial sums asymptotically flatten. The nth Term Test formalizes this observation — if the terms fail to vanish, the partial sums cannot level off, and the series diverges.

SECTION 4

Mathematical Framework

Formal Statement and Proof Sketch

NTH TERM TEST FOR DIVERGENCE

If lim(n→∞) aₙ ≠ 0 (or the limit does not exist), then Σ(n=1 to ∞) aₙ diverges.

Here aₙ denotes the general (nth) term of the series. The symbol Σ(n=1 to ∞) aₙ represents the infinite sum a₁ + a₂ + a₃ + ⋯.

CONTRAPOSITIVE (NECESSARY CONDITION)

If Σ(n=1 to ∞) aₙ converges, then lim(n→∞) aₙ = 0.

This is logically equivalent to the test above. It states that convergence of a series forces its terms to approach zero.

Proof of the Contrapositive

Suppose Σaₙ converges to a finite limit S. Define the partial sums Sₙ = a₁ + a₂ + ⋯ + aₙ. By hypothesis, lim(n→∞) Sₙ = S. Observe that aₙ = Sₙ − Sₙ₋₁ for n ≥ 2. Taking the limit of both sides yields lim(n→∞) aₙ = lim(n→∞) Sₙ − lim(n→∞) Sₙ₋₁ = S − S = 0. This elegant argument relies only on the algebraic limit laws, and it demonstrates that lim(n→∞) aₙ = 0 is necessary for convergence. Therefore, by contrapositive logic, if the limit of aₙ is not zero (or fails to exist), the series must diverge.

KEY RELATIONSHIP

aₙ = Sₙ − Sₙ₋₁ ⟹ lim(n→∞) aₙ = S − S = 0

Sₙ is the nth partial sum. If the series converges to S, both Sₙ and Sₙ₋₁ approach S, so their difference approaches 0.

⚠

⚠ Common Pitfall

Students frequently apply the nth Term Test in the wrong direction. If lim(n→∞) aₙ = 0, the test is inconclusive — it does NOT prove convergence. On the AP exam, this is one of the most common sources of lost points. The harmonic series Σ(1/n) satisfies lim(n→∞)(1/n) = 0, yet it diverges. Always write 'the nth Term Test is inconclusive' rather than 'the series converges by the nth Term Test.'

SECTION 5

Decision Flowchart & Common Patterns

When and How to Apply the Test

In practice, the nth Term Test should be the very first test you apply to any series. It is computationally inexpensive — you only need to evaluate a single limit — and when it gives a definitive answer (divergence), it saves you from applying more elaborate tests unnecessarily. The flowchart below illustrates where the nth Term Test sits within the broader series-testing strategy on the AP exam.

Flowchart showing the nth Term Test as the first step in series analysis. When the limit is nonzero or does not exist, the series diverges immediately. When the limit equals zero, the test is inconclusive and you must proceed to other convergence tests.

Common Patterns Where lim aₙ ≠ 0

Series Pattern	lim(n→∞) aₙ	Conclusion
Σ c (constant, c ≠ 0)	c ≠ 0	Diverges by NTT
Σ (−1)ⁿ	DNE (oscillates between −1 and 1)	Diverges by NTT
Σ n/(2n + 1)	1/2 ≠ 0	Diverges by NTT
Σ sin(n)	DNE (never settles)	Diverges by NTT
Σ eⁿ/(eⁿ + 1)	1 ≠ 0	Diverges by NTT

Common series types that fail the nth Term Test

SECTION 6

Worked Example

Applying the nth Term Test Step by Step

Determine whether the series Σ(n=1 to ∞) (3n² + 1)/(5n² − 2n) converges or diverges using the nth Term Test.

Example: Σ(n=1 to ∞) (3n² + 1)/(5n² − 2n)

1
Step 1 — Identify the General Term

The general term of the series is aₙ = (3n² + 1)/(5n² − 2n). Both the numerator and denominator are polynomials of degree 2 in n.

aₙ = (3n² + 1)/(5n² − 2n)
2
Step 2 — Compute the Limit of aₙ as n → ∞

Divide every term in the numerator and denominator by the highest power of n, which is n². This yields aₙ = (3 + 1/n²)/(5 − 2/n). As n → ∞, the terms 1/n² → 0 and 2/n → 0, so the limit becomes (3 + 0)/(5 − 0) = 3/5.

lim(n→∞) aₙ = 3/5
3
Step 3 — Apply the nth Term Test

Since lim(n→∞) aₙ = 3/5 ≠ 0, the nth Term Test for Divergence applies. The terms of this series do not approach zero; intuitively, adding 3/5 (approximately) at every step means the partial sums grow without bound.

The series DIVERGES by the nth Term Test.
4
Step 4 — Write a Conclusion (AP Style)

On a free-response question, always state the test used, compute the limit, compare the result to 0, and write a clear concluding sentence. For example: 'Because lim(n→∞)(3n² + 1)/(5n² − 2n) = 3/5 ≠ 0, the series Σ(3n² + 1)/(5n² − 2n) diverges by the nth Term Test for Divergence.'

SECTION 7

Strengths, Limitations & Common Errors

Strengths	Limitations
Extremely fast — requires only computing a single limit.	Can only prove divergence, never convergence.
Works on any series, not restricted by sign, monotonicity, or form of aₙ.	Inconclusive when lim aₙ = 0, which is the case for most interesting series on the AP exam.
Applies when the limit does not exist (e.g., oscillating sequences), not just when the limit is nonzero.	Cannot distinguish between conditional and absolute convergence.
Logically watertight — based on a direct consequence of the definition of convergence.	Provides no information about the rate of divergence or partial-sum behavior.

Strengths vs. Limitations of the nth Term Test

The Three Most Common Exam Errors

Error 1 — Claiming convergence: Writing 'lim aₙ = 0, so the series converges by the nth Term Test.' This is logically invalid. The test can only conclude divergence.
Error 2 — Forgetting the DNE case: If the limit of aₙ does not exist at all (e.g., aₙ = sin(n) or aₙ = (−1)ⁿ), the series still diverges by the nth Term Test. The test covers both 'limit ≠ 0' and 'limit DNE.'
Error 3 — Skipping the test entirely: Students sometimes jump directly to the ratio or comparison test without first checking the nth term limit. If the limit is nonzero, these more complex tests are unnecessary and may even yield misleading results.

💡

REMEMBER

The nth Term Test is a necessary condition checker, not a sufficient condition checker. In logic terms: convergence ⟹ lim aₙ = 0, but lim aₙ = 0 ⟹ convergence is false. The converse of a true statement is not necessarily true — the harmonic series is the definitive counterexample.

SECTION 8

Connection to Other Convergence Tests

The nth Term Test occupies a foundational position in the hierarchy of series tests, but it is only the beginning. Once the test is inconclusive (because lim aₙ = 0), you need to select an appropriate follow-up test. The table below situates the nth Term Test alongside the major convergence tests you will encounter on the AP Calculus BC exam, emphasizing what each test can and cannot determine.

Test	What It Determines	Requirements / Restrictions
nth Term Test	Divergence only	None — applies to all series
Geometric Series Test	Convergence and divergence (with exact sum)	Series must be of the form Σarⁿ
p-Series Test	Convergence (p > 1) and divergence (p ≤ 1)	Series must be Σ(1/nᵖ)
Integral Test	Convergence and divergence	aₙ = f(n) must be positive, continuous, decreasing
Comparison / Limit Comparison	Convergence and divergence	Positive terms; need a known comparison series
Ratio / Root Test	Absolute convergence and divergence	Inconclusive when limit = 1
Alternating Series Test	Convergence only	Terms must alternate sign; \|aₙ\| must be decreasing → 0

Comparison of major convergence tests on the AP Calculus BC exam

Looking ahead, the nth Term Test connects to deeper ideas in mathematical analysis. In a real analysis course, you will encounter the Cauchy criterion for convergence, which strengthens the necessary condition: not only must the individual terms tend to zero, but the partial sums over increasingly long stretches must also become arbitrarily small. For power series, the nth Term Test combined with the ratio or root test helps determine the radius of convergence, a concept central to Taylor and Maclaurin series on the BC exam. Mastering the nth Term Test builds the logical foundation for understanding these more advanced tools.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student claims: 'Since lim(n→∞)(1/n) = 0, the harmonic series Σ(n=1 to ∞)(1/n) converges by the nth Term Test.' Which of the following best describes the error in this reasoning?

A. The limit was computed incorrectly; lim(n→∞)(1/n) = 1, not 0.
B. The nth Term Test can only establish divergence, not convergence; when the limit equals zero, the test is inconclusive.
C. The nth Term Test only applies to series with positive terms, and 1/n can be negative for some n.
D. The harmonic series actually converges, so there is no error in the student's reasoning.
E. The nth Term Test requires computing lim(n→∞) Sₙ, not lim(n→∞) aₙ.

Show answer (B)

The correct answer is B. The nth Term Test for Divergence states that if lim(n→∞) aₙ ≠ 0 (or DNE), then Σaₙ diverges. However, if lim(n→∞) aₙ = 0, the test is inconclusive — it cannot prove convergence. The student has applied the converse of the test, which is logically invalid. Choice A is wrong because lim(n→∞)(1/n) is indeed 0. Choice C is wrong because the nth Term Test has no restriction to positive-term series. Choice D is wrong because the harmonic series is a famous divergent series (provable via the integral test or by Cauchy's condensation). Choice E is wrong because the nth Term Test specifically examines the limit of the general term aₙ, not the partial sums.

Question 2: BASIC CALCULATION

What can be concluded about the series Σ(n=1 to ∞) (2n)/(n + 3) using the nth Term Test for Divergence?

A. The series converges because lim(n→∞) 2n/(n + 3) = 0.
B. The series diverges because lim(n→∞) 2n/(n + 3) = 2 ≠ 0.
C. The series converges because lim(n→∞) 2n/(n + 3) = 2.
D. The test is inconclusive because lim(n→∞) 2n/(n + 3) = 0.
E. The series diverges because lim(n→∞) 2n/(n + 3) = ∞.

Show answer (B)

The correct answer is B. To compute the limit, divide numerator and denominator by n: 2n/(n + 3) = 2/(1 + 3/n). As n → ∞, 3/n → 0, so the limit equals 2/(1 + 0) = 2. Since 2 ≠ 0, the nth Term Test confirms divergence. Choice A is incorrect because the limit is 2, not 0. Choice C incorrectly claims convergence despite a nonzero limit. Choice D is wrong because the limit is 2, not 0. Choice E is wrong because the limit is the finite value 2, not infinity — both the numerator and denominator are degree 1, so the limit is the ratio of leading coefficients.

Question 3: INTERMEDIATE

Consider the series Σ(n=1 to ∞) cos(1/n). Which of the following correctly applies the nth Term Test?

A. lim(n→∞) cos(1/n) = cos(0) = 1 ≠ 0, so the series diverges.
B. lim(n→∞) cos(1/n) = 0, so the test is inconclusive.
C. lim(n→∞) cos(1/n) = 0, so the series converges.
D. The limit does not exist because cosine oscillates, so the series diverges.
E. lim(n→∞) cos(1/n) = cos(∞), which is undefined, so the test is inconclusive.

Show answer (A)

The correct answer is A. As n → ∞, 1/n → 0, and since cosine is continuous, lim(n→∞) cos(1/n) = cos(0) = 1. Because 1 ≠ 0, the nth Term Test guarantees divergence. This problem requires recognizing the composition of limits: lim cos(f(n)) = cos(lim f(n)) when cosine is continuous. Choice B and C incorrectly evaluate the limit as 0. Choice D incorrectly claims oscillation — while cos(n) oscillates, cos(1/n) does not, because 1/n approaches a single value. Choice E makes the error of substituting ∞ directly into cos rather than recognizing the limit of the argument.

Question 4: APPLIED

4 pts

A particle moves along a number line such that at the nth second (n = 1, 2, 3, …), it is displaced by dₙ = n²/(n² + 5n + 6) meters in the positive direction. (a) Determine lim(n→∞) dₙ and interpret the result in the context of the particle's motion. (2 points) (b) Using the nth Term Test, determine whether the total displacement Σ(n=1 to ∞) dₙ is finite. Justify your answer. (1 point) (c) Explain why the nth Term Test would be inconclusive if the displacement at the nth second were instead dₙ = 1/n², and name one test that could determine the convergence of Σ(1/n²). (1 point)

Show answer

Rubric (4 points total): Part (a) — 2 points: • (1 point) Correctly computes lim(n→∞) n²/(n² + 5n + 6). Dividing numerator and denominator by n² gives 1/(1 + 5/n + 6/n²). As n → ∞, this approaches 1/(1 + 0 + 0) = 1. • (1 point) Interprets: The individual displacements approach 1 meter per second, meaning the particle never slows down — each second it moves approximately 1 meter forward. Part (b) — 1 point: • (1 point) Since lim(n→∞) dₙ = 1 ≠ 0, the series Σdₙ diverges by the nth Term Test for Divergence. The total displacement is not finite; the particle travels infinitely far in the positive direction. Part (c) — 1 point: • (1 point) If dₙ = 1/n², then lim(n→∞)(1/n²) = 0, so the nth Term Test is inconclusive (it cannot confirm convergence or divergence). A test that works: the p-series test (p = 2 > 1 implies convergence), or the integral test (∫₁^∞ 1/x² dx converges).

Question 5: CRITICAL THINKING

3 pts

Let {aₙ} be a sequence such that lim(n→∞) aₙ = 0. (a) Give a specific example of such a sequence where Σaₙ converges, and state the sum if known. (1 point) (b) Give a specific example of such a sequence where Σaₙ diverges. (1 point) (c) Explain, using the logical structure of the nth Term Test, why these two examples are not contradictory. (1 point)

Show answer

Rubric (3 points total): Part (a) — 1 point: • (1 point) A valid convergent example: aₙ = (1/2)ⁿ, and Σ(n=1 to ∞)(1/2)ⁿ = 1 (geometric series with r = 1/2). Alternatively, aₙ = 1/n² and Σ(1/n²) = π²/6 (p-series with p = 2). Part (b) — 1 point: • (1 point) A valid divergent example: aₙ = 1/n, and Σ(1/n) is the harmonic series, which diverges. Alternatively, aₙ = 1/√n (p-series with p = 1/2 ≤ 1, diverges). Part (c) — 1 point: • (1 point) The nth Term Test states: if lim aₙ ≠ 0 then Σaₙ diverges. Its contrapositive is: if Σaₙ converges then lim aₙ = 0. Neither the test nor its contrapositive claims that lim aₙ = 0 implies convergence. That would be the converse, which is a logically independent statement (and is false). Therefore, having lim aₙ = 0 is consistent with both convergence and divergence — the two examples simply illustrate this logical gap.

SUMMARY

Lesson Summary

Summary

The nth Term Test for Divergence is the simplest and most widely applicable first step in analyzing an infinite series. It states that if lim(n→∞) aₙ ≠ 0 or the limit does not exist, then the series Σaₙ diverges. The proof follows from the observation that aₙ = Sₙ − Sₙ₋₁, so convergence of the partial sums forces the terms to vanish.

The test's greatest pitfall is the converse error: concluding convergence from lim aₙ = 0. The harmonic series Σ(1/n) is the classic counterexample, satisfying lim(1/n) = 0 yet diverging. When the nth Term Test is inconclusive, proceed to the integral, comparison, ratio, root, or alternating series tests to determine convergence. Always apply the nth Term Test first — it is quick, universal, and when it yields a verdict, it is definitive.