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10.9 Determining Absolute or Conditional Convergence

1 min readjune 18, 2024

10.9 Determining Absolute or Conditional Convergence

Welcome to the ninth topic in the final unit! In this topic, we’ll talk about the different types of convergence you can have and how to determine if something is absolutely convergent, conditionally convergent, or divergent! ✨


📈 Different Types of Convergence

There are two types of convergence that you’ll go over in AP Calculus BC: absolute convergence and conditional convergence.

💭 Absolute vs. Conditional Convergence

A series an\sum a_n is called absolutely convergent if and only if the absolute value of ana_n is convergent. If an\sum a_n is convergent but an\sum |a_n| is divergent, then the series is considered conditionally convergent.

📝 Example 1: Conditional Convergence

Let’s take a look at the following series and show that it is conditionally convergent!

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

This series converges due to the alternating series test! Take a look at topic 10.7 if you want a more in-depth review of the alternating series test!

Now to prove that it is conditionally convergent, we have to follow the following steps:

🟰 Take the absolute value of the sequence that is inside the infinite series.

In this case, this would be the new series:

n=11n\sum_{n=1}^{\infty} \frac{1}{n}

This is the new series because when you take the absolute value of 1-1, no matter what the exponent was, it will always be equal to 11 in the end. 🧮

🤔 Use tests that you already know in order to determine the convergence of the absolute value of your series!

In this case, we can recognize that the series is a harmonic series and is therefore divergent. On the AP exam, your final answer would be something like, “The series is conditionally convergent due to the alternating series test and due to the absolute value of the series being a harmonic series!”

📝 Example 2: Absolute Convergence

Let’s take a look at this example, where you do not have an alternating series, but you still want to determine whether it is conditionally or absolutely convergent.

n=1sin(n)n3\sum_{n=1}^{\infty} \frac{\sin(n)}{n^3}

As you can tell by inspection 🔎, this series is clearly not an alternating series! So how do we determine the type of convergence it displays? Let’s follow the same steps as before.

🟰 Take the absolute value of the sequence that is inside the infinite series.

n=1sin(n)n3\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n^3}

Since n3n^3 is always positive, we only need absolute value brackets around the sine term. Since sine oscillates between 1-1 and 11, taking the absolute value means that sinn1|\sin n| \leq 1.

🤔 Use tests that you already know in order to determine the convergence of the absolute value of your series!

In this case, you can use the direct comparison test to determine the convergence of the series! For a review on comparison tests, check out topic 10.6!

sin(n)n31n3\frac{|\sin(n)|}{n^3} \leq \frac{1}{n^3}

Since we know that 1n3\frac{1}{n^3} converges because it is a p-series, we also know that the other series converges due to the direct comparison test.

Therefore, this series is absolutely convergent!


📕 Closing

Now you know the difference between absolutely and conditionally convergent series! Using this knowledge and practice, you’ve got this! 🌟