The nth term test for divergence — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The formal statement of the nth term test for divergence, underlying limit logic, step-by-step application, interpretation of results, and resolution of common misconceptions about test scope.

You should already know: How to evaluate limits of sequences at infinity; Definition of convergence for infinite series; Basic algebraic manipulation of general terms.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is The nth term test for divergence?

The nth term test for divergence (often shortened to the nth term test, or called the test for divergence) is the most fundamental first-step convergence test for infinite series, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections of the AP Calculus BC exam, accounting for roughly 1-3% of total exam weight per the official College Board CED. It leverages a core necessary condition for infinite series convergence to quickly eliminate obviously divergent series before you apply more complex, computationally intensive tests. Formally, for an infinite series ${\sum }_{n=1}^{\infty }{a}_n$ with general term $a_n$, the test states that 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. Critically, this test can only confirm divergence; it can never confirm convergence. This makes it the first test you should always apply after identifying the general term of any series you’re testing, as it saves you from wasting time on complicated tests for a series that is clearly divergent.

2. The Core Theorem and Underlying Logic

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: $$\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }(S_k-S_{k-1})=L-L=0$$ 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 ${\sum }_{n=1}^{\infty }\frac{2n^2+5n}{3n^2-1}$ diverges.

1. Identify the general term of the series: $a_n=\frac{2n^2+5n}{3n^2-1}$.
2. Evaluate ${\lim }_{n\to \infty }{a}_n$ by dividing numerator and denominator by the highest power of $n$, which is $n^2$: $$\underset{n\to \infty }{\lim }\frac{2n^2+5n}{3n^2-1}=\underset{n\to \infty }{\lim }\frac{2+\frac{5}{n}}{3-\frac{1}{n^2}}$$
3. As $n\to \infty$, terms with $\frac{1}{n}$ and $\frac{1}{n^2}$ approach 0, so the limit simplifies to $\frac{2}{3}\ne 0$.
4. Since ${\lim }_{n\to \infty }{a}_n\ne 0$, 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. A 10-second limit calculation can save you 5 minutes of unnecessary integration or ratio test computation on a divergent series.

3. Interpreting Inconclusive Test Results

The most confusing and 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. The same test result gives two different outcomes, so the nth term test cannot confirm anything when the limit is 0. You must always use a different test (like the integral test or comparison test) to resolve convergence in this case.

Worked Example

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

1. First, confirm the student’s limit calculation: ${\lim }_{n\to \infty }\frac{1}{\sqrt{n}}=0$, which is algebraically correct.
2. 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. While we need a p-series test to confirm, this particular series actually diverges (it is a p-series with $p=1/2<1$), even though its general term approaches 0.
4. Conclusion: The student is incorrect, because a limit of zero does not guarantee convergence via the nth term test.

Exam tip: On FRQ questions that ask you to justify convergence or divergence, never write "the series converges because ${\lim }_{n\to \infty }{a}_n=0$"—this will always lose points, because the nth term test cannot prove convergence.

4. Non-Existent Limits and Divergence

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 where the limit does not exist are: oscillating sequences that do not approach a single fixed value, and sequences that grow without bound (unbounded sequences). In both cases, the condition that ${\lim }_{n\to \infty }{a}_n=0$ fails, so the necessary condition for convergence is not met, and the series diverges. For example, a sequence $a_n=(-1{)}^n$ oscillates between $-1$ and $1$, so it never approaches 0. Each new term changes the partial sum by at least 1, so the partial sum can never settle to a finite limit, hence the series diverges. This is a straightforward application of the nth term test that is often missed by students who only look for finite non-zero limits.

Worked Example

Test the series ${\sum }_{n=1}^{\infty }\cos \left(\frac{n\pi }{2}\right)$ for divergence using the nth term test.

1. Write out the first few terms of $a_n=\cos \left(\frac{n\pi }{2}\right)$ 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. 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. 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. Conclusion: ${\sum }_{n=1}^{\infty }\cos \left(\frac{n\pi }{2}\right)$ diverges by the nth term test.

Exam tip: Always check if the limit exists before concluding anything about the nth term test. Oscillating or unbounded general terms always result in divergence, even if the terms equal zero infinitely often.

5. Common Pitfalls (and how to avoid them)

• 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 (sum of terms) vs partial sums, especially on exam questions that ask 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 (e.g., claiming ${\lim }_{n\to \infty }\frac{n}{n+1000}\approx 0$). 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 ${\lim }_{n\to \infty }\frac{a_{n+1}}{a_n}\ne 1$, 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.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following series diverge by the nth term test for divergence? I. ${\sum }_{n=1}^{\infty }\frac{5n}{n^2+1}$ II. ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n}3n}{2n+7}$ 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

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

Question 2 (Free Response)

Consider the infinite series ${\sum }_{n=1}^{\infty }{a}_n$, where $a_n=\frac{n^2-4n+10}{2n^2-n+3}$. (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.

Worked Solution: (a) Divide numerator and denominator by $n^2$ to evaluate the limit: $$\underset{n\to \infty }{\lim }\frac{n^2-4n+10}{2n^2-n+3}=\underset{n\to \infty }{\lim }\frac{1-\frac{4}{n}+\frac{10}{n^2}}{2-\frac{1}{n}+\frac{3}{n^2}}=\frac{1}{2}$$ So ${\lim }_{n\to \infty }{a}_n=\frac{1}{2}$.

(b) By the nth term test, since ${\lim }_{n\to \infty }{a}_n=\frac{1}{2}\ne 0$, the series ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

(c) Evaluate the limit of the new general term: ${\lim }_{n\to \infty }\frac{1}{a_n}=\frac{1}{1/2}=2\ne 0$. 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.

Question 3 (Application / Real-World Style)

An urban planner models the annual net change in the number of households in a growing city with an infinite series ${\sum }_{n=0}^{\infty }\Delta H_n$, where $\Delta H_n=1200\cdot \frac{0.8^n+1.2^n}{1.2^n}$ 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.

Worked Solution: First simplify the general term: $$\Delta H_n=1200\left(\frac{0.8^n}{1.2^n}+\frac{1.2^n}{1.2^n}\right)=1200\left({\left(\frac{2}{3}\right)}^n+1\right)$$ Evaluate the limit as $n\to \infty$: ${\left(\frac{2}{3}\right)}^n\to 0$, so ${\lim }_{n\to \infty }\Delta H_n=1200(0+1)=1200$, which is non-zero. 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.

7. Quick Reference Cheatsheet

8. What's Next

The nth term test is the foundational first test for infinite series, and it is 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 the nth term test to rule out obvious divergence before moving to more computationally intensive tests. Without mastering the logic of the nth term test and its limitations, you will almost certainly lose points on FRQ justifications, as AP readers commonly penalize incorrect justifications that claim convergence from a limit of zero. Next, you will apply this foundational knowledge to studying special series and more targeted convergence tests for complex series.

• Geometric series convergence
• p-series test for convergence
• Integral test for convergence
• Alternating series test