The alternating series test tells you an alternating series converges if the terms shrink toward zero and decrease in size. You check two things: that the limit of the positive part is 0, and that those terms are eventually decreasing. For AP Calculus BC, separate the alternating sign from the positive term before checking the test conditions.

How Does the Alternating Series Test Work?

The alternating series test applies to series whose signs switch back and forth. Write the series as $\sum (-1)^n a_n$ or $\sum (-1)^{n+1}a_n$ with $a_n>0$ , then check that $a_n\to0$ and that $a_n$ is eventually decreasing.

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Why This Matters for the AP Calculus Exam

This is a BC-only topic in the infinite series unit. Knowing the alternating series test gives you a quick way to handle series that flip signs, which show up often when you choose a convergence test on multiple-choice questions and when you justify convergence on free-response work. It also sets up later topics like conditional versus absolute convergence and the alternating series error bound, so getting comfortable here pays off across the rest of the unit.

Key Takeaways

An alternating series looks like $\sum(-1)^n a_n$ or $\sum(-1)^{n+1} a_n$ , where $a_n > 0$ .
The test needs two conditions: $\lim_{n\to\infty} a_n = 0$ and $a_n$ is (eventually) decreasing.
If both conditions hold, the series converges. If $\lim_{n\to\infty} a_n \neq 0$ , it diverges by the nth term test.
Always factor out the $(-1)^n$ part first so you can see $a_n$ clearly.
$\cos(n\pi)$ equals $(-1)^n$ , so series with $\cos(n\pi)$ are alternating in disguise.
Convergence here can be conditional, which matters when you compare it to absolute convergence later.

Alternating Series Test Theorem

For an alternating series $\sum(-1)^n\cdot a_n$ with $a_n > 0$ , if

1. \lim_{n\to \infty} a_n=0 \ \text{and}

2. \ a_n \ \text{decreases,}

then the series converges. If the limit of $a_n$ is not 0, the series diverges.

Breaking Down the Theorem

A famous example is the alternating harmonic series:

\sum_{n=1}^\infty\frac{(-1)^n}{n}

Start with the first condition: the limit of $a_n$ must equal zero. To find $a_n$ , factor out the alternating part $(-1)^n$ . That leaves

a_n=\frac{1}{n}

Now take the limit.

\lim_{n\to \infty}\frac{1}{n}=0

The first condition is satisfied. Next, check whether $a_n$ decreases by showing $a_n>a_{n+1}$ .

a_n=\frac{1}{n}>\frac{1}{n+1}=a_{n+1}

Plug in a number for $n$ to confirm.

a_2=\frac{1}{2}>\frac{1}{3}=a_{2+1}

Both conditions are met, so the series converges.

How to Use This on the AP Calculus Exam

MCQ

When a series has $(-1)^n$ , $(-1)^{n+1}$ , or $\cos(n\pi)$ , reach for the alternating series test. Check the limit of $a_n$ first. If it is not 0, you are done: the series diverges, and you do not need the decreasing check. If the limit is 0, confirm the terms decrease before saying it converges.

Free Response

When you justify convergence, state both conditions clearly. Write that $\lim_{n\to\infty} a_n = 0$ and that $a_n$ is decreasing, then conclude convergence. Showing both conditions makes your reasoning complete and is important for clear exam work.

Common Trap

The test only proves convergence. Failing the decreasing condition does not automatically mean the series diverges. The only divergence conclusion you get directly is when $\lim_{n\to\infty} a_n \neq 0$ .

Practice with Alternating Series Test

For each series, state whether it converges or diverges.

1. \sum_{n=1}^\infty \frac{(-1)^{n+2}\cdot n^5}{n^5+3}

2. \sum_{n=2}^\infty\frac{\cos(n\pi)}{n}

3. \sum_{n=2}^\infty (-1)^n \ln(n)

Solutions

Question 1

First, identify $a_n$ .

a_n=\frac{n^5}{n^5+3}

Take the limit.

\lim_{n\to \infty}\frac{n^5}{n^5+3}=1\neq0

The first condition fails, so you do not check the second. This series is divergent.

Question 2

This one requires recognizing another type of alternating series, $\cos(n\pi)$ . Plug in some values and you will see that $\cos(n\pi)=(-1)^n$ . So you can treat this like the alternating harmonic series from earlier. That series met both conditions, so this one does too. This series is convergent.

Question 3

First, find $a_n=\ln(n)$ . Then take the limit:

\lim_{n\to \infty}\ln(n)=\infty

Like Question 1, the first condition fails, so the series is divergent without checking the other condition.

Common Misconceptions

The test does not prove divergence on its own. If $a_n$ fails to decrease but still goes to 0, the test is simply inconclusive, not a guarantee of divergence.
The condition is about $a_n$ , the positive part, not the whole term with the sign. Factor out $(-1)^n$ before checking the limit and the decreasing condition.
Convergence from this test can be conditional. Just because an alternating series converges does not mean it converges absolutely.
"Decreasing" can mean eventually decreasing. The terms only need to decrease past some starting point, not from the very first term.
A series with $\cos(n\pi)$ or $\sin$ terms that produce alternating signs still counts as alternating, even if the $(-1)^n$ is not written out directly.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
alternating series	A series whose terms alternate in sign, typically written in the form Σ(-1)^n * a_n where a_n > 0.
alternating series test	A convergence test that determines whether an alternating series converges based on whether its terms decrease in absolute value and approach zero.
converges	A series converges when the sequence of partial sums approaches a finite limit as n approaches infinity.
diverges	A series diverges when the sequence of partial sums does not approach a finite limit as the number of terms increases indefinitely.

Frequently Asked Questions

What are the conditions for the alternating series test?

For an alternating series written with positive part a_n, the alternating series test needs two conditions: a_n approaches 0 and a_n is eventually decreasing. If both are true, the alternating series converges.

How do you identify an alternating series?

Look for signs that switch every term, often shown by (-1)^n, (-1)^{n+1}, or cos(nπ). After identifying the sign pattern, separate the positive part a_n from the alternating factor.

Can the alternating series test prove divergence?

The alternating series test itself proves convergence when its conditions hold. If a_n does not approach 0, the series diverges by the nth term test, not by the alternating series test alone.

What if the terms are not decreasing?

If a_n approaches 0 but is not eventually decreasing, the alternating series test is inconclusive. That does not automatically mean the series diverges; you need another valid test.

Is the alternating harmonic series convergent?

Yes. The alternating harmonic series converges because 1/n approaches 0 and decreases. It does not converge absolutely, which is why it becomes a key example of conditional convergence.

How is the alternating series test used on AP Calculus BC?

AP Calculus BC questions often ask you to justify convergence for an alternating series. A complete response names the positive part a_n, shows a_n approaches 0, states it decreases eventually, and concludes convergence.