AP Calculus BC

Which Series Convergence Test Should I Use?

Q: Which convergence test should I try first?

The nth-term test, every time. It is one limit and it can end the problem immediately: if \lim_{n \to \infty} a_n \ne 0, the series diverges. If the limit is zero, move on, because the test cannot confirm convergence.

Q: Can the nth-term test prove a series converges?

No. It only proves divergence. Terms shrinking to zero is required for convergence but does not guarantee it, as the diverging harmonic series shows. To prove convergence you need a different test.

Q: What do I do when the ratio test gives 1?

L = 1 is inconclusive, so the ratio test tells you nothing and you switch tests. This happens on p-series and rational terms; those are best handled by the comparison or limit comparison test instead.

Q: Which convergence tests are on the AP Calculus BC exam?

The CED (Topics 10.2 through 10.9) lists exactly these: the nth-term test for divergence, and the integral, comparison, limit comparison, alternating series, and ratio tests, plus the geometric-series and p-series facts. Other tests such as the root test are not assessed.

Run a fixed checklist, starting with the nth-term test: if the terms do not shrink to zero, the series diverges. Otherwise match the shape: a geometric ratio, a p-series, alternating signs (alternating series test), factorials or exponentials (ratio test), or a rational function of n (limit or direct comparison).

Start with the shape, not the test

The skill AP Calculus BC actually rewards is recognition: reading the terms of $\sum a_n$ and knowing which test the College Board expects. Series live in Unit 10, worth 15 to 20 percent of the BC exam, and the convergence tests sit in Topics 10.1 through 10.9. There is no single test that works on everything, so trained students run a fixed checklist instead of guessing.

Work the list top to bottom and stop at the first test whose conditions the series clearly meets. The order is deliberate: the cheapest checks (a limit, a ratio you can eyeball) come first, and the tests that need setup (comparison, integral) come last.

nth-term test first. Compute $\lim_{n \to \infty} a_n$ . If it is not zero (or does not exist), the series diverges and you are finished. If the limit is zero, the test is inconclusive, so keep reading. This is Topic 10.3.
Geometric? If each term is a fixed multiple of the one before it, so $a_n = a r^n$ , the series diverges when $|r| \ge 1$ and converges when $|r| < 1$ . The closed form $\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}$ uses the first term as the numerator, so a sum that starts at $n = 1$ converges to $\frac{ar}{1-r}$ (Topic 10.2).
p-series? If the terms look like $\frac{1}{n^p}$ , the series converges exactly when $p > 1$ and diverges when $p \le 1$ . The harmonic series is the $p = 1$ case, and it diverges (Topic 10.5).
Alternating? If the signs flip through $(-1)^n$ or $(-1)^{n-1}$ , try the alternating series test, then decide whether the convergence is absolute or conditional (Topics 10.7 and 10.9).
Factorials, or $n$ sitting in an exponent? Reach for the ratio test (Topic 10.8). It is built for terms that multiply or divide by a growing factor each step.
A messy rational or algebraic function of $n$ ? Match it to a p-series using the direct comparison or limit comparison test (Topic 10.6).
Do the terms come from a positive, decreasing function you can integrate? Use the integral test (Topic 10.4).

The one-way trap

The nth-term test can only prove divergence. If $\lim_{n \to \infty} a_n = 0$ , you have learned nothing yet: the harmonic series has terms going to zero and still diverges. Never write "the terms go to zero, so it converges." That sentence loses points.

The eight tests at a glance

This table is the checklist compressed into a reference. The middle column is the recognition cue (what makes you pick that test), and the last column is the exact condition plus the verdict. These eight are the only convergence tools assessed on the BC exam. The CED covers the six named tests (nth-term, integral, comparison, limit comparison, alternating series, ratio) across Topics 10.3 through 10.9, and the geometric-series (Topic 10.2) and p-series (Topic 10.5) facts round out the eight tools you actually use.

Test	Reach for it when	Condition and verdict
nth-term test (10.3)	Always, as the first check	If $\lim_{n \to \infty} a_n \ne 0$ , diverges. If the limit is $0$ , inconclusive.
geometric series (10.2)	Constant ratio $r$ between terms	Converges to $\frac{a}{1-r}$ (first term $a$ , sum from $n = 0$ ) if $\|r\| < 1$ ; diverges if $\|r\| \ge 1$ .
p-series (10.5)	Terms are $\frac{1}{n^p}$	Converges if $p > 1$ ; diverges if $p \le 1$ (harmonic is $p = 1$ ).
alternating series (10.7)	Signs alternate: $\sum (-1)^n b_n$ , $b_n > 0$	Converges if $b_n$ decreases and $\lim_{n \to \infty} b_n = 0$ .
ratio test (10.8)	Factorials, exponentials, $n$ th powers	$L = \lim_{n \to \infty} \left\| \frac{a_{n+1}}{a_n} \right\|$ : $L < 1$ converges, $L > 1$ diverges, $L = 1$ inconclusive.
comparison test (10.6)	Terms bounded above or below by a known series	If $0 \le a_n \le b_n$ and $\sum b_n$ converges, so does $\sum a_n$ ; if $\sum a_n$ diverges, so does $\sum b_n$ .
limit comparison test (10.6)	Rational or algebraic in $n$ , close to a p-series	$L = \lim_{n \to \infty} \frac{a_n}{b_n}$ with $0 < L < \infty$ : both series do the same thing.
integral test (10.4)	$a_n = f(n)$ , with $f$ positive and decreasing	$\sum a_n$ and $\int_k^\infty f(x)\,dx$ converge or diverge together.

Reading the terms: recognition triggers

Most lost time on series problems comes from starting the wrong test, so the cue matters more than the mechanics. Here is what each pattern in the terms is telling you.

Factorials or exponentials point to the ratio test. When $a_n$ contains $n!$ , $2^n$ , or $n^n$ , forming $\frac{a_{n+1}}{a_n}$ cancels most of the mess and leaves a clean limit. Factorials almost always win the growth race, so $\sum \frac{3^n}{n!}$ converges while $\sum \frac{n!}{3^n}$ diverges.

A rational function of $n$ points to comparison. When $a_n$ is a ratio of polynomials, its long-run size is set by the leading powers. Strip everything but the highest power on top and bottom to get the p-series it behaves like, then confirm with limit comparison. For example $\frac{2n+1}{n^3+5}$ behaves like $\frac{2n}{n^3} = \frac{2}{n^2}$ , a convergent p-series.

A constant multiplier between terms points to geometric. If dividing any term by the previous one gives the same number every time, it is geometric and you can read off convergence from $|r|$ instantly, no other test needed.

Alternating signs let you use the gentlest test. The alternating series test only asks two things, that $b_n$ decreases and that $b_n \to 0$ , so a series that fails the ratio or comparison tests may still converge as an alternating series. That is exactly how conditional convergence happens.

Terms that match a clean function point to the integral test. If $a_n = f(n)$ for an $f$ you can actually antidifferentiate, and $f$ is positive and decreasing, the improper integral $\int_k^\infty f(x)\,dx$ settles the series. It is the natural choice for $\sum \frac{1}{n \ln n}$ and similar terms with a logarithm.

Alternating series: absolute versus conditional

Once a series converges by the alternating series test, Topic 10.9 asks a follow-up: is the convergence absolute or conditional? Test the series of absolute values $\sum |a_n|$ .

If $\sum |a_n|$ converges, the original series is absolutely convergent. Absolute convergence is the stronger statement: if a series converges absolutely, it converges.
If $\sum |a_n|$ diverges but $\sum a_n$ still converges (by the alternating series test), the series is conditionally convergent.

The alternating harmonic series $\sum \frac{(-1)^{n-1}}{n}$ is the textbook conditional case: it converges, but stripping the signs gives the harmonic series, which diverges. A useful shortcut on the exam is to run the ratio test on $|a_n|$ first. If the ratio test returns $L < 1$ , you already have absolute convergence and can skip the alternating series test entirely.

Where students lose points

Using the nth-term test to claim convergence. It never proves convergence, only divergence. Terms going to zero is necessary, not sufficient.
Stopping when the ratio test gives $L = 1$ . That outcome is inconclusive, and it always happens on p-series and rational terms, which is your signal to switch to comparison instead.
Getting the comparison direction backward. To prove convergence you need a larger convergent series above your terms; to prove divergence you need a smaller divergent series below them. Bounding a series above by a divergent one proves nothing.
Forgetting to check that $b_n$ decreases in the alternating series test. Both conditions are required, and the free-response rubric expects you to state that $b_n$ is decreasing and that $b_n \to 0$ .
Misremembering the p-series cutoff. Convergence needs $p > 1$ strictly; $p = 1$ is the harmonic series and diverges. On the exam $p = \frac{1}{2}$ (a common $\frac{1}{\sqrt{n}}$ term) diverges.

Worked examples

Worked example

nth-term test settles it in one line

Determine whether $\sum_{n=1}^{\infty} \frac{3n^2}{5n^2 + 2}$ converges or diverges.

The terms are a ratio of polynomials of equal degree, so the very first checklist item, the nth-term test, is worth trying: does $a_n \to 0$ ?
Divide numerator and denominator by $n^2$ to rewrite the terms: $\lim_{n \to \infty} \frac{3n^2}{5n^2 + 2} = \lim_{n \to \infty} \frac{3}{5 + \frac{2}{n^2}}$ .
As $n \to \infty$ the $\frac{2}{n^2}$ term drops out, so the limit is $\frac{3}{5}$ .
The limit is $\frac{3}{5}$ , which is not zero. By the nth-term test for divergence (Topic 10.3), the series diverges. No further test is needed.

Diverges by the nth-term test, since $\lim_{n \to \infty} a_n = \frac{3}{5} \ne 0$ .

Worked example

Limit comparison for a rational term

Determine whether $\sum_{n=1}^{\infty} \frac{2n+1}{n^3 + 5}$ converges or diverges.

First run the nth-term test: $\lim_{n \to \infty} \frac{2n+1}{n^3+5} = 0$ , so that test is inconclusive and we continue down the checklist.
The term is rational in $n$ , so compare it to a p-series. The leading behavior is $\frac{2n}{n^3} = \frac{2}{n^2}$ , which points to $b_n = \frac{1}{n^2}$ , a convergent p-series with $p = 2 > 1$ .
Apply the limit comparison test by forming the ratio: $\frac{a_n}{b_n} = \frac{2n+1}{n^3+5} \cdot \frac{n^2}{1}$ .
Multiply out the numerator: $\frac{(2n+1)n^2}{n^3+5} = \frac{2n^3 + n^2}{n^3 + 5}$ .
Take the limit by dividing top and bottom by $n^3$ : $L = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{1 + \frac{5}{n^3}} = \frac{2}{1} = 2$ .
Since $L = 2$ is finite and positive, $0 < L < \infty$ , the two series share a fate. Because $\sum \frac{1}{n^2}$ converges, the original series converges too.

Converges by limit comparison with the convergent p-series $\sum \frac{1}{n^2}$ (limit $L = 2$ ).

Worked example

Ratio test for a factorial term

Determine whether $\sum_{n=1}^{\infty} \frac{3^n}{n!}$ converges or diverges.

The presence of a factorial is the recognition cue for the ratio test (Topic 10.8): forming $\frac{a_{n+1}}{a_n}$ will cancel most of the factorial.
Write the next term: $a_{n+1} = \frac{3^{n+1}}{(n+1)!}$ . Then $\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n}$ .
Rewrite the pieces using $3^{n+1} = 3 \cdot 3^n$ and $(n+1)! = (n+1) \cdot n!$ .
Substitute and cancel the common $3^n$ and $n!$ : $\frac{3 \cdot 3^n}{(n+1) \cdot n!} \cdot \frac{n!}{3^n} = \frac{3}{n+1}$ .
The terms are positive, so the absolute values change nothing. Take the limit: $L = \lim_{n \to \infty} \frac{3}{n+1} = 0$ .
Since $L = 0 < 1$ , the ratio test gives absolute convergence, and absolute convergence implies convergence.

Converges (absolutely) by the ratio test, since $L = \lim_{n \to \infty} \frac{3}{n+1} = 0 < 1$ .

Frequently asked questions

Which convergence test should I try first?

The nth-term test, every time. It is one limit and it can end the problem immediately: if $\lim_{n \to \infty} a_n \ne 0$ , the series diverges. If the limit is zero, move on, because the test cannot confirm convergence.

Can the nth-term test prove a series converges?

No. It only proves divergence. Terms shrinking to zero is required for convergence but does not guarantee it, as the diverging harmonic series shows. To prove convergence you need a different test.

What do I do when the ratio test gives 1?

$L = 1$ is inconclusive, so the ratio test tells you nothing and you switch tests. This happens on p-series and rational terms; those are best handled by the comparison or limit comparison test instead.

Which convergence tests are on the AP Calculus BC exam?

The CED (Topics 10.2 through 10.9) lists exactly these: the nth-term test for divergence, and the integral, comparison, limit comparison, alternating series, and ratio tests, plus the geometric-series and p-series facts. Other tests such as the root test are not assessed.