10.7 Alternating Series Test for Convergence

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.

The alternating series error bound, which comes later in the unit, builds on this test, so the two conditions you check here carry forward.

Two Conditions That Make an Alternating Series Converge

• 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.

Choosing and Justifying the Alternating Series Test

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

• Misconception: If $a_n$ fails to decrease, the series diverges.
  Actually: 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.
• Misconception: The condition is about the whole signed term, not just $a_n$.
  Actually: 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.
• Misconception: Just because an alternating series converges means it converges absolutely.
  Actually: Convergence from this test can be conditional. Just because an alternating series converges does not mean it converges absolutely.
• Misconception: The terms must decrease from the very first term.
  Actually: "Decreasing" can mean eventually decreasing. The terms only need to decrease past some starting point, not from the very first term.
• Misconception: Only series with an explicitly written $(-1)^n$ are alternating.
  Actually: 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.

Related AP Calculus Guides

• Unit 10 Review: Infinite Series and Sequences
• 10.1 Defining Convergent and Divergent Infinite Series
• 10.3 The nth Term Test for Divergence
• 10.5 Harmonic Series and p-Series
• 10.2 Working with Geometric Series
• 10.4 Integral Test for Convergence
• 10.9 Determining Absolute or Conditional Convergence