Defining convergent and divergent infinite series

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

1. Formal Definition of Convergence and Divergence★★☆☆☆⏱ 3 min

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.

📘 Definition

nth Partial Sum

$S_{n}$

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 $n \to \infty$ .

📘 Definition

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 $\sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{1}{n + 1})$ 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 $S_{n}$ .

2. The nth-Term Test for Divergence★★★☆☆⏱ 4 min

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

📐 Worked Example

Determine whether $\sum_{n = 1}^{\infty} \frac{3 n ^{2} - 4 n}{2 n ^{2} + 1}$ 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.

3. Properties of Convergent Series★★★☆☆⏱ 3 min

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:

$\sum_{n = 1}^{\infty} c a_{n} = c A$ , so the scaled series also converges
$\sum_{n = 1}^{\infty} (a_{n} \pm b_{n}) = A \pm B$ , 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 $\sum_{n = 1}^{\infty} (3 a_{n} - 2 b_{n})$ ?

1
Since $\sum a_{n}$ converges, $\sum 3 a_{n} = 3 \cdot 5 = 15$ , which is still convergent.
2
Since $\sum b_{n}$ diverges and 2 is non-zero, $\sum 2 b_{n}$ diverges, so $\sum - 2 b_{n}$ 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, $\sum_{n = 1}^{\infty} (3 a_{n} - 2 b_{n})$ 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.

4. Concept Check★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding of core concepts:

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 $lim_{n \to \infty} a_{n}$
Reveal answer
2 —
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.

5. Common Pitfalls

Wrong move:

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

Why:

Students confuse a necessary condition for convergence with a sufficient condition; the nth-Term Test only proves divergence, not convergence

Correct move:

Remember that $lim_{n \to \infty} a_{n} = 0$ means you need to test further, it does not guarantee convergence

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 or final term

Correct move:

Always write out the first three terms and last two terms to confirm which terms remain after canceling

Wrong move:

Claiming that a series diverging to $\infty$ converges because it "has a limit of $\infty$ "

Why:

Students confuse extended real limits 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 $lim_{n \to \infty} a_{n}$ with $lim_{n \to \infty} S_{n}$

Why:

Both are limits as $n \to \infty$ , so students mix up which defines series 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 the sum of two divergent series is always divergent

Why:

Students overgeneralize the convergent + divergent = divergent rule 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 get smaller, the series must converge

Why:

Students confuse early term behavior with the long-term behavior of partial sums as $n \to \infty$

Correct move:

Always base convergence on the limit of the nth partial sum, not the behavior of early terms

6. Quick Reference 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 $⟺ lim_{n \to \infty} S_{n} = L$ , $L$ 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 $\sum c a_{n} = c A$ , $\sum (a_{n} \pm b_{n}) = A \pm B$	Convergent + Divergent = Divergent; sum of two divergents is ambiguous

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.

2023 · MCQ
Convergence conceptual question
2022 · FRQ
Partial sum calculation

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.