AP Calculus BC

Series Convergence Test Chooser: Which Test and Why (BC)

Q: What is the difference between the comparison test and the limit comparison test?

The direct comparison test needs an inequality: if 0 \le a_n \le b_n and \sum b_n converges, then \sum a_n converges, and if a_n \ge b_n \ge 0 and \sum b_n diverges, then \sum a_n diverges. The limit comparison test drops the inequality and only asks for \lim_{n \to \infty} \frac{a_n}{b_n}: if that limit is a finite positive number, both series share the same fate. Use limit comparison when the term-by-term inequality is awkward to prove, which is most rational-function cases (Topic 10.6).

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

No. It is a one-directional test. If \lim_{n \to \infty} a_n \ne 0 the series diverges, but \lim_{n \to \infty} a_n = 0 tells you nothing by itself. The harmonic series \sum \frac{1}{n} has terms going to 0 yet diverges, so a zero limit only earns the series a real convergence test (Topic 10.3).

Which convergence test to use is set by the shape of the terms. Start with the nth-term test: if the terms do not approach 0, the series diverges. If they do, match the shape: geometric and p-series by sight, alternating signs to the alternating series test, factorials to the ratio test, rational terms to comparison.

BCUnit 10: series convergence (BC only)

Pick a series, then answer one question at a time. The chooser walks the standard decision order: nth term test, geometric and p-series pattern match, alternating structure, ratio test, then the comparison family and integral test.

Choose one of the ten series above to start the decision tree.

This tool walks the convergence decision tree one question at a time. You pick or enter a series and it asks the same questions a strong BC student asks, in order, then explains why each test either applies to your series or fails on it. That reasoning is the whole skill of Unit 10 (Topics 10.1 through 10.9): a solver can tell you a series converges, but the exam rewards choosing the right test and justifying it. Only six named tests are assessed on the BC exam, and the tool uses exactly those, plus the geometric and p-series facts you recognize on sight.

There is no single convergence test. Instead you run a short list of them in a fixed order, and the shape of the general term $a_n$ decides which one settles the question. The order matters because the cheapest check comes first. Every series starts on the same rung, the nth-term test, and you only move deeper into the tree when a test comes back inconclusive.

The nth-term test only proves divergence

If $\lim_{n \to \infty} a_n \ne 0$ , the series diverges, full stop (Topic 10.3). But $\lim_{n \to \infty} a_n = 0$ proves nothing on its own: the harmonic series $\sum \frac{1}{n}$ has terms going to $0$ and still diverges. A zero limit just clears the series to move on to a real test, so never write 'the terms go to $0$ , therefore it converges.'

Once the nth-term test is inconclusive, the shape of $a_n$ points to the test that will finish the job. Read the left column as what your term looks like, and the tool selects the row that matches.

When the term looks like	The tool reaches for	Because
$\lim_{n \to \infty} a_n \ne 0$	nth-term test (Topic 10.3)	The series diverges at once; no further test is needed.
$a r^n$ , a constant ratio between terms	Geometric series (Topic 10.2)	It converges to $\frac{a}{1-r}$ exactly when $\|r\| < 1$ .
$\frac{1}{n^p}$ , a single power of $n$	p-series (Topic 10.5)	It converges exactly when $p > 1$ ; the harmonic case $p = 1$ diverges.
Signs alternate, $(-1)^n b_n$	Alternating series test (Topic 10.7)	It converges if the positive terms $b_n$ decrease monotonically to $0$ .
Factorials or $n$ -th powers	Ratio test (Topic 10.8)	The ratio $\frac{a_{n+1}}{a_n}$ makes those pieces cancel.
A rational or algebraic function of $n$	Comparison or limit comparison (Topic 10.6)	Match it to the p-series its leading powers resemble.
A positive, continuous, decreasing term you can integrate	Integral test (Topic 10.4)	The series and $\int_1^\infty f(x)\,dx$ converge together.

Three of those rows deserve a sharper rule. The ratio test is the specialist for factorials and $n$ -th powers, because $\frac{a_{n+1}}{a_n}$ makes those pieces cancel; on a plain rational function of $n$ its limit is $1$ and tells you nothing, so do not reach for it there. Comparison and limit comparison are the tools for rational and algebraic terms: match the term to the p-series its leading powers resemble, then compare. The integral test is the fallback when a term is positive, continuous, decreasing, and something you can actually antidifferentiate, like $\frac{1}{n \ln n}$ . Two shapes never need a test at all, because their rules are exact:

\sum_{n=0}^{\infty} a r^n \text{ converges} \iff |r| < 1 \qquad \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges} \iff p > 1

Once a series passes the alternating series test, one more question follows: is it absolutely or conditionally convergent (Topic 10.9)? Take the series of absolute values and run a positive-term test on it. If that converges, the original converges absolutely, and absolute convergence guarantees convergence. If the absolute series diverges but the alternating series test still passes, the convergence is only conditional, like the alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$ .

Worked example

Choosing a test for a rational-term series

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 + 2}{n^4 + n + 1}$ converges or diverges, and name the test that decides it.

Start on the bottom rung, the nth-term test. The denominator has degree $4$ and the numerator degree $2$ , so $\lim_{n \to \infty} \frac{3n^2 + 2}{n^4 + n + 1} = 0$ . A zero limit never proves convergence, so this test is inconclusive and you keep going (Topic 10.3).
Rule out the by-sight forms. Every term is positive, so it is not alternating (Topic 10.7 is off the table). There is no constant ratio between terms, so it is not geometric (Topic 10.2), and it is not a single power $\frac{1}{n^p}$ , so it is not a bare p-series (Topic 10.5). No factorials or $n$ -th powers appear, and on a rational function of $n$ the ratio test's limit is $1$ , which is inconclusive, so skip it (Topic 10.8).
What is left is a rational function of $n$ , which points to comparison (Topic 10.6). Pick the partner from the leading powers: for large $n$ the term behaves like $\frac{3n^2}{n^4} = \frac{3}{n^2}$ , so compare against $b_n = \frac{1}{n^2}$ , a p-series with $p = 2$ . Because a clean term-by-term inequality is awkward here, use the limit comparison test, which only needs the limit of the ratio.
Set up the limit comparison ratio, a fraction divided by a fraction: $\frac{a_n}{b_n} = \frac{\frac{3n^2 + 2}{n^4 + n + 1}}{\frac{1}{n^2}}$ .
Multiply numerator and denominator through by $n^2$ to clear the complex fraction: $\frac{n^2(3n^2 + 2)}{n^4 + n + 1} = \frac{3n^4 + 2n^2}{n^4 + n + 1}$ .
Evaluate the limit by dividing top and bottom by $n^4$ : $\lim_{n \to \infty} \frac{3 + \frac{2}{n^2}}{1 + \frac{1}{n^3} + \frac{1}{n^4}}$ .
Every term with $n$ in a denominator vanishes, leaving $\frac{3}{1} = 3$ , a finite, positive number.
Conclude. The comparison series $\sum \frac{1}{n^2}$ is a convergent p-series ( $p = 2 > 1$ ). When the limit comparison ratio is finite and positive, both series share the same fate, so the given series converges (Topic 10.6).

The series converges, by limit comparison with the convergent p-series sum of 1/n^2 (the ratio has finite positive limit 3).

Frequently asked questions

What is the difference between the comparison test and the limit comparison test?

The direct comparison test needs an inequality: if $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges, and if $a_n \ge b_n \ge 0$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. The limit comparison test drops the inequality and only asks for $\lim_{n \to \infty} \frac{a_n}{b_n}$ : if that limit is a finite positive number, both series share the same fate. Use limit comparison when the term-by-term inequality is awkward to prove, which is most rational-function cases (Topic 10.6).

Does the nth-term test prove a series converges?

No. It is a one-directional test. If $\lim_{n \to \infty} a_n \ne 0$ the series diverges, but $\lim_{n \to \infty} a_n = 0$ tells you nothing by itself. The harmonic series $\sum \frac{1}{n}$ has terms going to $0$ yet diverges, so a zero limit only earns the series a real convergence test (Topic 10.3).

When should I use the ratio test?

Reach for the ratio test when the term contains factorials or $n$ -th powers such as $r^n$ or $n^n$ , because $\frac{a_{n+1}}{a_n}$ cancels those pieces cleanly. If the limit $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$ the series converges absolutely, if $L > 1$ it diverges, and if $L = 1$ the test is inconclusive and you switch to another test. On a plain rational function of $n$ the ratio limit is always $1$ , so use comparison there instead (Topic 10.8).

Which convergence tests are on the AP Calculus BC exam?

The CED assesses exactly six: the nth-term test for divergence, the integral test, the comparison test, the limit comparison test, the alternating series test, and the ratio test (the Topic 10.8 exclusion statement). You also need the geometric series and p-series convergence rules from Topics 10.2 and 10.5. The root test is not assessed, so you will never need it on the exam.