# Defining convergent and divergent infinite series

> AP Calculus BC · Infinite Sequences and Series (Unit 10)
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-defining-convergent-and-divergent-infinite/

This subtopic introduces core definitions for infinite series, including partial sums, convergence, divergence, the nth-Term Test, and properties of convergent series, the foundational prerequisite for all subsequent convergence tests on the AP exam.

**Prerequisites:** Evaluating limits of sequences as $n \to \infty$; Working with finite sums and sigma notation; Basic properties of finite geometric series

## Learning objectives

- Define infinite series and sequences of partial sums
- State the formal definition of convergent and divergent infinite series
- Correctly apply the nth-Term Test for Divergence
- Use properties of convergent series to classify combined series

## Formal Definition of Convergence and Divergence

An infinite series is the sum of the terms of an infinite sequence, written in standard sigma notation as $\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + ...$, where $a_n$ is the nth term of the sequence. We cannot add infinitely many terms directly, so we analyze convergence via the sequence of partial sums.

**nth Partial Sum** — The sum of the first $n$ terms of the infinite series $\sum_{n=1}^{\infty} a_n$, given by $S_n = \sum_{k=1}^n a_k$. Convergence of the series depends entirely on the limit of $\{S_n\}$ as $n \to \infty$.

*Notation:* $S_n$

**Convergent and Divergent Series** — An infinite series converges if and only if $\lim_{n \to \infty} S_n = L$, where $L$ is a finite real number. The sum of the series equals $L$. A series that does not converge is called divergent; divergence occurs when the limit of $S_n$ is infinite or does not exist due to oscillation.

**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:
2. $$S_n = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + ... + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$
3. Cancel all intermediate (telescoping) terms, leaving only the first and last unpaired terms:
4. $$S_n = 1 - \frac{1}{n+1}$$
5. Evaluate the limit of $S_n$ as $n \to \infty$:
6. $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1$$
7. Since the limit is finite, the series converges to 1 by definition.

> **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 $S_n$.

## The nth-Term Test for Divergence

The nth-Term Test for Divergence is a core result derived directly from the definition of convergence. If a series $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} S_n = L$ (finite) and $\lim_{n \to \infty} S_{n-1} = L$. Since $a_n = S_n - S_{n-1}$, we take the limit of both sides:

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = L - L = 0$$

> **tip**
>
> This gives a necessary (but not sufficient) condition for convergence: for a series to converge, its nth term must approach 0. The nth-Term Test for Divergence (the contrapositive) states: *If $\lim_{n \to \infty} a_n \neq 0$, then the series diverges*. This 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 $n \to \infty$:
2. Divide numerator and denominator by the highest power of $n$ ($n^2$):
3. $$\lim_{n \to \infty} \frac{3n^2 - 4n}{2n^2 + 1} = \lim_{n \to \infty} \frac{3 - \frac{4}{n}}{2 + \frac{1}{n^2}} = \frac{3}{2}$$
4. The limit of the nth term is $\frac{3}{2} \neq 0$, which violates the necessary condition for convergence.
5. 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 exam time.

## Properties of Convergent Series

If we know the convergence behavior of two separate series, we can use core properties to classify their combinations, a common topic for conceptual multiple-choice questions. 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 real constant, then:

1. $\sum_{n=1}^{\infty} c a_n = c A$, 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: multiplying a divergent series by a non-zero constant always gives a divergent series, and adding a convergent series to a divergent series always gives a divergent series. The sum of two divergent series 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 $\sum_{n=1}^{\infty} (3a_n - 2b_n)$?

1. Since $\sum a_n$ converges, $\sum 3a_n = 3 \cdot 5 = 15$, which is still convergent.
2. Since $\sum b_n$ diverges and 2 is non-zero, $\sum 2b_n$ diverges, so $\sum -2b_n$ also diverges.
3. The sum of a convergent and a divergent series must be divergent: if the result were convergent, $\sum b_n$ would have to converge, contradicting the given information.
4. Therefore, $\sum_{n=1}^{\infty} (3a_n - 2b_n)$ diverges.

> **Exam tip:** Never assume the sum of two divergent series is automatically divergent. For example, $\sum 1$ and $\sum -1$ both diverge, but their sum $\sum 0$ converges to 0.

## Concept Check

**Check your understanding**

Test your understanding of core concepts:

1. Which of the following statements about the infinite series $\sum_{n=1}^{\infty} a_n$ is guaranteed to be true if $\lim_{n \to \infty} a_n = 0$?

   - The series converges
   - The series diverges
   - The nth-Term Test does not allow us to conclude divergence
   - The sum of the series equals $\lim_{n \to \infty} a_n$

   *Answer:* The nth-Term Test does not allow us to conclude divergence

   *Why:* The nth-Term Test only concludes divergence if $\lim_{n \to \infty} a_n \neq 0$. If the limit is 0, the test gives no information: the series could converge or diverge. The sum of the series is the limit of partial sums $S_n$, not the limit of $a_n$, so only the third option is correct.

## Common pitfalls

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

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| Infinite Series | $\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + ...$ | Sum of terms of an infinite sequence; cannot be evaluated directly by addition |
| nth Partial Sum | $S_n = \sum_{k=1}^{n} a_k$ | Sum of the first n terms; convergence depends on $\lim_{n \to \infty} \{S_n\}$ |
| Convergent Series Definition | $\sum a_n$ converges $\iff \lim_{n \to \infty} S_n = L$, $L$ finite | The sum of the series equals the finite limit $L$ |
| Divergent Series Definition | $\sum a_n$ diverges if $\lim_{n \to \infty} S_n$ does not exist (finite) | Divergence includes diverging to $\pm \infty$ and oscillation |
| nth-Term Test for Divergence | If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges | $\lim_{n \to \infty} a_n = 0$ does NOT prove convergence; this only proves divergence |
| Necessary Condition for Convergence | If $\sum a_n$ converges, then $\lim_{n \to \infty} a_n = 0$ | Core result that the nth-Term Test is built from |
| Properties of Convergent Series | If $\sum a_n = A$, $\sum b_n = B$, then $\sum c a_n = cA$, $\sum (a_n \pm b_n) = A \pm B$ | Convergent + Divergent = Divergent; sum of two divergents is ambiguous |

## What's next

This subtopic is the absolute foundation for all remaining topics in Unit 10 (Infinite Sequences and Series), which makes up 17-18% of the total AP Calculus BC exam score. 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 always 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 results of other convergence tests or distinguish between the limit of terms and the limit of partial sums, a common source of lost points. Next you will apply this definition to specific series types, then move on to specialized convergence tests, and finally to Taylor series approximations, a major component of the exam's FRQ section.

- [Working with Geometric Series](https://www.owlsprep.com/study/ap-calculus-bc-u10-working-with-geometric-series/)
- [The nth term test for divergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-the-nth-term-test-for/)
- [Integral Test for Convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-integral-test-for-convergence/)

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