AP Calculus BC

Series Convergence Tests: The Complete Table (BC)

Eight tests cover AP Calculus BC series (CED 10.1 to 10.9): nth-term, geometric, p-series, integral, direct comparison, limit comparison, alternating, and ratio. Match the shape of the terms to the test, run the nth-term check first, and remember it proves divergence only, never convergence.

The eight tests at a glance

These eight are the only convergence tools the AP Calculus BC exam assesses. The CED (the College Board's Course and Exam Description) Topic 10.8 exclusion statement names the six formal tests it assesses (nth-term, integral, comparison, limit comparison, alternating series, and ratio); geometric series (10.2) and p-series (10.5) are the two known series you add to complete the toolkit. Read the table left to right: match the shape of the terms in $\sum a_n$ to a test, confirm what that test can actually conclude, then read the trap that costs the most points. Series live in Unit 10, worth 15 to 20 percent of the exam.

Test	Applies to	Concludes	Watch out
nth-Term Test (10.3)	Any series, as the opening check	If $\lim_{n \to \infty} a_n \ne 0$ (or the limit does not exist), diverges. If the limit is $0$ , inconclusive.	Never proves convergence. $a_n \to 0$ is necessary, not sufficient: the harmonic series obeys it and still diverges.
Geometric Series (10.2)	A constant ratio $r$ between successive terms, $\sum a r^n$	Converges to $\frac{a}{1-r}$ if $\|r\| < 1$ ; diverges if $\|r\| \ge 1$ .	Before using $\frac{a}{1-r}$ , check the starting index: $a$ is the first term actually written out.
p-Series (10.5)	Terms of the form $\frac{1}{n^p}$	Converges if $p > 1$ ; diverges if $p \le 1$ .	$p = 1$ is the harmonic series and diverges; $p = \frac{1}{2}$ (a $\frac{1}{\sqrt{n}}$ term) also diverges.
Integral Test (10.4)	$a_n = f(n)$ with $f$ positive, continuous, and decreasing for $x \ge k$	$\sum a_n$ and $\int_k^\infty f(x)\,dx$ converge or diverge together.	The integral's value is not the sum. You must be able to antidifferentiate $f$ for this to be usable.
Direct Comparison (10.6)	Positive terms you can bound by a known series	If $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $0 \le b_n \le a_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.	Direction is everything: bound above by a convergent series to prove convergence, below by a divergent one to prove divergence. Any other pairing proves nothing.
Limit Comparison (10.6)	Rational or algebraic terms close to a p-series	With $b_n > 0$ , if $L = \lim_{n \to \infty} \frac{a_n}{b_n}$ and $0 < L < \infty$ , both series do the same thing.	The limit must be finite and positive. Pick $b_n$ from the leading powers of $a_n$ .
Alternating Series Test (10.7)	Alternating signs, $\sum (-1)^n b_n$ with $b_n > 0$	Converges if $b_n$ is decreasing and $\lim_{n \to \infty} b_n = 0$ .	Both conditions are required, and the rubric expects you to state that $b_n$ decreases. Proves convergence only, never divergence.
Ratio Test (10.8)	Factorials, exponentials, or $n$ th powers	$L = \lim_{n \to \infty} \left\| \frac{a_{n+1}}{a_n} \right\|$ : $L < 1$ converges absolutely, $L > 1$ diverges, $L = 1$ inconclusive.	$L = 1$ on every p-series and rational term. When it happens, switch to a comparison test.

Run the nth-term test first

It is one limit and it can finish the problem: if $\lim_{n \to \infty} a_n \ne 0$ , the series diverges immediately. If the limit is $0$ , you have learned nothing yet, so keep going. Writing "the terms go to zero, so it converges" is the single most common lost point on series.

The two series you memorize

Two series are pure recall on the exam: you never derive them, you recognize them. Every comparison you set up eventually leans on one of these, so memorize the cutoffs cold.

Series	Form	Converges when	Key fact
Geometric Series (10.2)	$\sum_{n=0}^{\infty} a r^n$	$\|r\| < 1$	Sum equals $\frac{a}{1-r}$ , the only series in this list with a general closed-form sum formula you can evaluate for any $a$ and $r$ .
p-Series (10.5)	$\sum_{n=1}^{\infty} \frac{1}{n^p}$	$p > 1$	No exam-required closed sum; it is the benchmark almost every comparison targets.
Harmonic Series (10.5)	$\sum_{n=1}^{\infty} \frac{1}{n}$	Never (diverges)	The $p = 1$ case: terms shrink to $0$ , yet the series diverges. The canonical counterexample.
Alternating Harmonic Series (10.5)	$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$	Always (converges)	Conditionally convergent: it converges to $\ln 2$ , but drop the signs and it diverges.

\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}, \quad |r| < 1

Recognition cues: reading the terms

Method selection is the whole game: most lost time on series comes from starting the wrong test. Each pattern in the terms points to one test. Find the row that matches what $a_n$ looks like, then run it.

What the terms look like	Reach for	Why it works
A constant multiplier between successive terms	Geometric Series (10.2)	Divide any term by the previous one; if it is always the same $r$ , convergence reads straight off $\|r\|$ .
$\frac{1}{n^p}$	p-Series (10.5)	Convergence is decided by a single number, $p$ , with the cutoff at $1$ .
A factorial $n!$ , an exponential $c^n$ , or an $n$ th power	Ratio Test (10.8)	Forming $\frac{a_{n+1}}{a_n}$ cancels the growing factor and leaves a clean limit.
A rational or algebraic function of $n$	Limit Comparison (10.6)	Strip to the leading powers to find the p-series it behaves like, then confirm with the limit.
Signs that flip through $(-1)^n$ or $(-1)^{n-1}$	Alternating Series Test (10.7)	Only two conditions to check; a series that fails other tests may still converge here.
$a_n = f(n)$ with $f$ positive, decreasing, and easy to antidifferentiate	Integral Test (10.4)	The improper integral settles it. Natural for $\frac{1}{n \ln n}$ -type terms with a logarithm.
Terms trapped between a known series and $0$	Direct Comparison (10.6)	When the bounding inequality is obvious, no limit is needed.
Anything, as the opening move	nth-Term Test (10.3)	One limit can end the problem before any real work begins.

Absolute versus conditional convergence (10.9)

Once a series converges, Topic 10.9 asks a second question: is the convergence absolute or conditional? Test the series of absolute values $\sum |a_n|$ and match your result to the table.

Situation	Classification	What it means
$\sum \|a_n\|$ converges	Absolutely convergent	Convergence is guaranteed (absolute convergence implies convergence), and rearranging the terms keeps the same sum.
$\sum a_n$ converges but $\sum \|a_n\|$ diverges	Conditionally convergent	The series survives only through sign cancellation; the alternating harmonic series is the model case.
$\sum a_n$ diverges	Divergent	Neither classification applies.

A shortcut worth memorizing

Before running the alternating series test, try the ratio test on $|a_n|$ . If it returns $L < 1$ , you already have absolute convergence, which implies convergence, and you can skip the alternating series test entirely. If it returns $L > 1$ , the terms do not go to zero, so the series diverges by the nth-term test. If it returns $L = 1$ , fall back to the alternating series test.