Defining convergent and divergent infinite series — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: infinite series sigma notation, partial sum sequences, formal definitions of convergent and divergent infinite series, the nth-Term Test for Divergence, and core properties of convergent series for the AP Calculus BC exam.

You should already know: How to evaluate limits of sequences as $n\to \infty$. How to compute finite sums and work with sigma notation. Basic properties of finite arithmetic and geometric series.

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 Defining convergent and divergent infinite series?

This topic is the foundational building block for all of Unit 10 (Infinite Sequences and Series), which makes up 17-18% of the total AP Calculus BC exam weight per the official College Board CED. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it is often tested as a standalone conceptual question, and it is the required prerequisite for every other convergence test you will learn in this unit.

An infinite series is the sum of the terms of an infinite sequence, written in standard sigma notation as ${\sum }_{n=1}^{\infty }{a}_n=a_1+a_2+a_3+...+a_n+...$, where $a_n$ is the nth term of the sequence. We cannot add infinitely many terms directly, so we instead build a sequence of partial sums: the nth partial sum $S_n$ is defined as the sum of the first n terms of the series. Convergence and divergence are defined entirely by the behavior of the limit of this sequence of partial sums. If the limit exists and is finite, the series converges; if not, it diverges. This core definition underpins all subsequent work in this unit, so mastery is non-negotiable.

2. Partial Sums and the Formal Definition of Convergence/Divergence

To understand convergence, you first must be able to write the nth partial sum $S_n$ for any given series. For an infinite series ${\sum }_{n=1}^{\infty }{a}_n$, the sequence of partial sums $\{S_n\}$ follows the definition: $$S_n=\sum _{k=1}^n{a}_k=a_1+a_2+...+a_n$$ The formal definition tested on the exam is:

An infinite series ${\sum }_{n=1}^{\infty }{a}_n$ converges if and only if ${\lim }_{n\to \infty }{S}_n=L$, where $L$ is a finite real number. The sum of the series is equal to $L$. An infinite series that does not converge is called divergent.

Divergence occurs in two common forms: either the limit of $S_n$ grows without bound to $\pm \infty$ (called diverging to infinity), or the sequence of partial sums oscillates between values and never settles to a single finite limit. For simple series like telescoping or geometric series, we can compute $S_n$ directly and evaluate the limit to determine convergence, no additional tests needed.

Worked Example

Determine whether the series ${\sum }_{n=1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right)$ converges or diverges. If it converges, find its sum.

1. First, write the general nth partial sum by expanding the first few and last terms: $$S_n=\sum _{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
2. Cancel all intermediate (telescoping) terms, leaving only the first term of the first expression and the last term of the last expression: $S_n=1-\frac{1}{n+1}$.
3. Evaluate the limit of $S_n$ as $n\to \infty$: ${\lim }_{n\to \infty }{S}_n={\lim }_{n\to \infty }\left(1-\frac{1}{n+1}\right)=1-0=1$, which is finite.
4. By the definition of convergence, the series converges, and its sum is 1.

Exam tip: When working with telescoping series, always expand at least the first 3 terms and the last 2 terms to confirm which terms do not cancel. Skipping this step is the most common cause of incorrect expressions for $S_n$.

3. The nth-Term Test for Divergence

One of the first core results derived from the definition of convergence is the nth-Term Test (also called the Test for Divergence), a quick check that lets you immediately rule out convergence for many series. If a series ${\sum }_{n=1}^{\infty }{a}_n$ converges, we know that ${\lim }_{n\to \infty }{S}_n=L$ (finite) and ${\lim }_{n\to \infty }{S}_{n-1}=L$. Since $a_n=S_n-S_{n-1}$, we can take the limit of both sides: $$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }(S_n-S_{n-1})=L-L=0$$ This gives us a necessary condition for convergence: for a series to converge, its nth term must approach 0. The contrapositive of this statement is the nth-Term Test for Divergence: If ${\lim }_{n\to \infty }{a}_n\ne 0$, then the series ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

A critical common misconception here: the converse of this statement is not true. ${\lim }_{n\to \infty }{a}_n=0$ does not guarantee that the series converges. We will see this with the harmonic series ${\sum }_{n=1}^{\infty }\frac{1}{n}$, which diverges even though its nth term approaches 0. The nth-Term Test can only prove divergence, it cannot prove convergence.

Worked Example

Determine whether ${\sum }_{n=1}^{\infty }\frac{3n^2-4n}{2n^2+1}$ converges or diverges.

1. First, compute the limit of the nth term as $n\to \infty$ to apply the nth-Term Test.
2. Divide the numerator and denominator by the highest power of $n$ ($n^2$) to evaluate the limit: $$\underset{n\to \infty }{\lim }\frac{3n^2-4n}{2n^2+1}=\underset{n\to \infty }{\lim }\frac{3-\frac{4}{n}}{2+\frac{1}{n^2}}=\frac{3-0}{2+0}=\frac{3}{2}$$
3. The limit of the nth term is $\frac{3}{2}\ne 0$, which violates the necessary condition for convergence.
4. By the nth-Term Test for Divergence, the series diverges.

Exam tip: Always apply the nth-Term Test first when approaching any convergence question. It only takes a few seconds, and if it tells you the series diverges, you can stop working and move on, saving valuable time on the exam.

4. Properties of Convergent Series

If we already know the convergence behavior of two separate series, we can use the following properties to determine the convergence of their combinations, which is a common topic for conceptual MCQ questions. Formally: if ${\sum }_{n=1}^{\infty }{a}_n=A$ (converges to finite $A$) and ${\sum }_{n=1}^{\infty }{b}_n=B$ (converges to finite $B$), and $c$ is any constant real number, then:

1. ${\sum }_{n=1}^{\infty }c{a}_n=cA$, so the scaled series also converges.
2. ${\sum }_{n=1}^{\infty }(a_n\pm b_n)=A\pm B$, so the combined series also converges.

Key consequences from these properties: multiplying a divergent series by a non-zero constant always results in a divergent series, and adding a convergent series to a divergent series always results in a divergent series. The only ambiguous case is the sum of two divergent series: that can be either convergent or divergent, so you must test it explicitly.

Worked Example

Given that ${\sum }_{n=1}^{\infty }{a}_n$ converges to 5 and ${\sum }_{n=1}^{\infty }{b}_n$ diverges, what can you conclude about the convergence of ${\sum }_{n=1}^{\infty }(3a_n-2b_n)$?

1. We know $\sum a_n$ converges, so by the scalar multiple property, $\sum 3a_n=3\cdot 5=15$, which is still convergent.
2. We know $\sum b_n$ diverges, and 2 is a non-zero constant, so $\sum 2b_n$ is also divergent, which means $\sum -2b_n$ is also divergent.
3. The sum of a convergent series ($\sum 3a_n$) and a divergent series ($\sum -2b_n$) must be divergent: if the result were convergent, we could rearrange the property to show $\sum b_n$ must also converge, which contradicts the given information.
4. Therefore, ${\sum }_{n=1}^{\infty }(3a_n-2b_n)$ diverges.

Exam tip: On conceptual MCQ questions about combined series, never assume the sum of two divergent series is automatically divergent. It can converge, e.g. $\sum 1$ and $\sum -1$ both diverge, but their sum $\sum 0$ converges to 0.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Concluding that 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 only proves divergence, not convergence. Correct move: Always remember that ${\lim }_{n\to \infty }{a}_n=0$ means "test further", it does not mean "converges".
• Wrong move: Canceling the wrong terms when finding the partial sum of a telescoping series. Why: Students skip expanding the first few and last few terms and incorrectly cancel the constant term or final term. Correct move: Always write out the first three terms and last two terms of the partial sum expansion to confirm which terms remain after canceling.
• Wrong move: Claiming that a series that diverges to $\infty$ converges because it "has a limit of $\infty$". Why: Students confuse "existing as a limit in the extended real numbers" with the definition of convergence, which requires a finite limit. Correct move: If ${\lim }_{n\to \infty }{S}_n=\pm \infty$, the limit does not exist as a finite number, so the series diverges.
• Wrong move: Confusing the limit of the nth term $a_n$ with the limit of the nth partial sum $S_n$. Why: Both are limits as $n\to \infty$, so students mix up which defines convergence. Correct move: Always remember: convergence of the series depends on ${\lim }_{n\to \infty }{S}_n$, not ${\lim }_{n\to \infty }{a}_n$.
• Wrong move: Claiming that the sum of two divergent series is always divergent. Why: Students overgeneralize the rule that convergent + divergent = divergent, and apply it to two divergent series. Correct move: Remember that the sum of two divergent series can be either convergent or divergent, so you must test it explicitly.
• Wrong move: Assuming that if the first 100 terms of a series get smaller, the series must converge. Why: Students confuse the behavior of early terms with the long-term behavior of partial sums. Correct move: Always base convergence on the limit of the nth partial sum as $n\to \infty$, not the behavior of early terms.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following statements about the infinite series ${\sum }_{n=1}^{\infty }{a}_n$ is guaranteed to be true if ${\lim }_{n\to \infty }{a}_n=0$? A) The series converges. B) The series diverges. C) The nth-Term Test for Divergence does not allow us to conclude divergence. D) The sum of the series is equal to the limit of $a_n$ as $n\to \infty$.

Worked Solution: This question tests correct understanding of the nth-Term Test for Divergence. The nth-Term Test only states that if ${\lim }_{n\to \infty }{a}_n\ne 0$, then the series diverges. If ${\lim }_{n\to \infty }{a}_n=0$, the test gives no information: the series could converge or diverge. Option A is incorrect because the harmonic series $\sum \frac{1}{n}$ has $\lim a_n=0$ but diverges. Option B is incorrect because $\sum \frac{1}{n^2}$ has $\lim a_n=0$ and converges. Option D is incorrect because the sum of the series is the limit of the partial sums $S_n$, not the limit of $a_n$. The only guaranteed true statement is C. Correct answer: $\boxed{C}$

Question 2 (Free Response)

Let ${\sum }_{n=1}^{\infty }\frac{1}{n(n+2)}$. (a) Write the first 4 terms of the sequence of partial sums $S_1,S_2,S_3,S_4$. (b) Use partial fractions to find a closed-form expression for the nth partial sum $S_n$. (c) Use the definition of convergence to determine whether the series converges or diverges. If it converges, find its sum.

Worked Solution: (a) Calculate each partial sum directly: $S_1=\frac{1}{1(3)}=\frac{1}{3}$ $S_2=\frac{1}{3}+\frac{1}{2(4)}=\frac{1}{3}+\frac{1}{8}=\frac{11}{24}$ $S_3=\frac{11}{24}+\frac{1}{3(5)}=\frac{21}{40}$ $S_4=\frac{21}{40}+\frac{1}{4(6)}=\frac{17}{30}$

(b) Decompose the nth term: $\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$. Expand the partial sum and cancel terms: $$S_n=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{n}-\frac{1}{n+2}\right)\right]$$ All intermediate terms cancel, leaving: $$S_n=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)=\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}$$

(c) Evaluate the limit of $S_n$ as $n\to \infty$: $$\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\left(\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\right)=\frac{3}{4}$$ Since the limit is finite, by definition the series converges to $\frac{3}{4}$.

Question 3 (Application / Real-World Style)

A patient receives a 120 mg dose of a medication every 12 hours. Before each new dose, only 25% of the previous amount of medication remains in the patient's body. Let $T$ be the total long-term (steady-state) amount of medication in the patient's body right after a dose is given. Write $T$ as an infinite series, and use the definition of convergence to determine if $T$ converges to a finite value, and if so, find it.

Worked Solution: Right after the first dose, the total medication is 120 mg. Right after the second dose, 25% of the first dose remains, so total is $120+120(0.25)$. Right after the third dose, 25% of the second dose's total remains, so total is $120+120(0.25)+120(0.25{)}^2$. The infinite series for steady-state $T$ is: $$T=\sum _{n=0}^{\infty }120(0.25{)}^n$$ The nth partial sum of this geometric series is: $$S_n=120\cdot \frac{1-(0.25{)}^{n+1}}{1-0.25}$$ Evaluate the limit as $n\to \infty$: ${\lim }_{n\to \infty }(0.25{)}^{n+1}=0$, so ${\lim }_{n\to \infty }{S}_n=\frac{120}{0.75}=160$ mg. In context, the patient's medication level will approach a steady state of 160 mg right after each repeated dose.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the absolute foundation for all remaining topics in Unit 10 (Infinite Sequences and Series). Every convergence test you will learn next relies on the core definition of convergence as the limit of partial sums, and the nth-Term Test you learned here is the first test you will apply to any series question on the exam. Without mastering this definition, you will not be able to correctly interpret the results of other convergence tests, or distinguish between the limit of terms and the limit of partial sums, which is a common source of lost points on the AP exam. Next you will apply this definition to geometric and telescoping series, then move on to specialized convergence tests for other series types. This topic also feeds into the final topic of the unit: finding Taylor series approximations for functions, which makes up a large portion of the exam’s FRQ section.

• The nth-Term Test for Divergence
• Geometric series
• Telescoping series
• Integral Test for convergence