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The nth Term Test for Divergence

Benjamin Wright

Benjamin Wright

4 min read

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Study Guide Overview

This study guide covers the nth Term Test for Divergence for AP Calculus BC. It explains how to apply the test to determine if a series diverges by evaluating the limit of the nth term as n approaches infinity. It includes a walkthrough example and practice problems with solutions, focusing on converting to limit notation, evaluating the limit, and drawing conclusions based on the test results.

10.3 The nth Term Test for Divergence

Welcome back to Unit 10 of AP Calculus BC! Today, we’re going to discuss the nth-term test for divergence with series. Let’s get started!

πŸ€·β€β™€οΈΒ What is the nth Term Test for Divergence?

As the name suggests the nth Divergence test tells us if a series will diverge! (mind-blowing stuff guys, I know 🀯). The Divergence test states that:

ifΒ lim⁑nβ†’βˆžan=ΜΈ0,βˆ‘anΒ diverges\text{if} \ \displaystyle{\lim_{n \to \infty}} a_n \not = 0, \sum a_n \ \text{diverges}

As we can see, if the nth term doesn't approach 0, the series diverges. On the other hand, if the nth term approaches 0, it creates a situation where the series might converge or still diverge. The crucial point here is that the fate of the series hinges on whether the nth term tends towards zero or not.

πŸ€“ Divergence Test Walkthrough

Let’s try a practice problem together! There are really only 3 steps involved with this:

  1. ✏️ Convert to limit notation.
  2. πŸ“ Evaluate the limit.
  3. πŸ€” Make your conclusion based on the nth-term test.

Determine if the series ana_n diverges.

βˆ‘n=1∞5nβˆ’1nβˆ’2\sum_{n=1}^∞ \dfrac{5n -1}{n-2}

✏️ Step 1: Convert to limit notation.

lim⁑nβ†’βˆž5nβˆ’1nβˆ’2\lim_{n \to \infty} \dfrac{5n -1}{n-2}

πŸ“ Step 2: Evaluate the limit.

lim⁑nβ†’βˆžn(5βˆ’1n)n(1βˆ’2n)\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})}

5βˆ’1∞1βˆ’2∞=51\dfrac{5- \frac{1}{\infty} }{1-\frac{2}{\infty}} = \frac{5}{1}

Recall that any number divided by ∞\infty is 0. lim⁑nβ†’βˆžn(5βˆ’1n)n(1βˆ’2n)=51β‰ 0\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})} = \frac{5}{1} \neq 0

βˆ΄Β βˆ‘n=1∞5nβˆ’1nβˆ’2Β Β diverges\therefore \ \sum_{n=1}^∞ \dfrac{5n -1}{n-2} \ \ \text{diverges}

Not too bad, right? We’re mainly just applying a new test to the mathematics that we are already familiar with!

πŸ“ Divergence Test Practice Problems

Try the following two practice questions yourself!

1.Β βˆ‘n=1∞n3+3n2n3βˆ’51. \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5}

2.Β βˆ‘n=1∞arctan⁑(n)2. \ \sum_{n=1}^∞ \arctan(n)

βœ… Divergence Test: Solution 1

Remember the three steps involved and the nth-term test itself.

lim⁑nβ†’βˆžn3+3n2n3βˆ’5\lim_{n \to \infty} \dfrac{n^3+3n}{2n^3-5}

lim⁑nβ†’βˆžn3(1+3n2)n3(2βˆ’5n3)\lim_{n \to \infty} \dfrac{n^3(1+\dfrac{3}{n^2})}{n^3(2-\dfrac{5}{n^3})}

1+3∞2βˆ’5∞=12\dfrac{1+\dfrac{3}{\infty}}{2-\dfrac{5}{\infty}} = \dfrac{1}{2}

βˆ΄Β βˆ‘n=1∞n3+3n2n3βˆ’5Β Β diverges\therefore \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5} \ \ \text{diverges}

Great work!

βœ… Divergence Test: Solution 2

Last question πŸŽ‰

lim⁑nβ†’βˆžarctan⁑(n)\lim_{n \to \infty} \arctan(n)

!Untitled

Graph displaying the function y = arctan(x)

Image courtesy of Math.net

As arctan goes to ∞\infty, it stays at Ο€2\dfrac{\pi}{2}.

Β lim⁑nβ†’βˆžarctan⁑(n)=Ο€2\ \lim_{n \to \infty} \arctan(n) = \dfrac{\pi}{2}

βˆ΄Β βˆ‘n=1∞arctan⁑(n)Β diverges\therefore \ \sum_{n=1}^∞ \arctan(n) \ \text{diverges}

πŸ•ΊΒ Closing

In conclusion, the nth Term Test for Divergence is a powerful tool for determining whether a series diverges. Remember, if the limit of the nth term does not approach zero, the series diverges. However, passing the divergence test doesn't provide information about convergence. Good luck! πŸ€