#10.7 Alternating Series Test for Convergence

Welcome to AP Calc 10.7! In this lesson, you’ll how to test for convergence when dealing with an alternating series.

#➕ Alternating Series Test Theorem

The alternating series test for convergence states that for an alternating series $\sum(-1)^n\cdot a_n$ , if

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

$2. \ a_n \ \text{decreases,}$

then the series converges. Otherwise, it diverges.

#🧱 Breaking Down the Theorem

To illustrate this theorem, let’s look at a famous example—the alternating harmonic sequence:

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

Let’s look at our first criteria: the limit of $a_n$ must be equal to zero. To figure this out, we first must figure out what $a_n$ is! To do this, all we have to do is factor out the alternating part of the sequence, $(-1)^n$ . Then, we get

$a_n=\frac{1}{n}$

Now, let’s take the limit.

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

Well, that’s our first criteria satisfied! Now we need to know whether $a_n$ decreases. To show this, we must show that $a_n>a_{n+1}$ .

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

Try plugging in a random number for $n$ to see that this is true!

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

Therefore, both of our conditions for convergence are met, and our series converges!

#📝 Practice with Alternating Series Test

Now it’s your turn to apply what you’ve learned!

#❓Alternating Series Test Problems

For each of the following series, state whether they converge or diverge.

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

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

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

#💡 Alternating Series Test Solutions

#Question 1 Solution

First, identify $a_n$ .

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

Now, take the limit.

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

Since our first condition isn’t met, we don’t need to check the second condition. This series is divergent.

#Question 2 Solution

This one is a little tricky—it requires you to recognize another type of alternating series, $\text{cos}(n\pi)$ . If you plug some examples into your calculator, you’ll see that $\text{cos}(n\pi)=(-1)^n$ . Therefore, we can treat this equation just like the harmonic series in our first example. We showed that the harmonic series met the conditions for convergence, so this one does too! This series is convergent.

#Question 3 Solution

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

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

Like our first problem, since the first condition isn’t met, we can say that this series is divergent without checking the other condition.

#💫 Closing

Great work! With this test mastered, you’re well equipped to take on all sorts of convergence problems. Make sure you recognize both types of alternating series so that you know when to apply this test! 🧠