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10.3 The nth Term Test for Divergence

1 min read•april 5, 2021

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    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!

    🤷‍♀️ 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:

    if lim⁡n→∞an≠0,∑an 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.

    🤓 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. 📏 Evaluate the limit.
    3. 🤔 Make your conclusion based on the nth-term test.

    Determine if the series ana_n diverges.

    ∑n=1∞5n−1n−2\sum_{n=1}^∞ \dfrac{5n -1}{n-2}

    ✏️ Step 1: Convert to limit notation.

    lim⁡n→∞5n−1n−2\lim_{n \to \infty} \dfrac{5n -1}{n-2}

    📏 Step 2: Evaluate the limit.

    lim⁡n→∞n(5−1n)n(1−2n)\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})}
    5−1∞1−2∞=51\dfrac{5- \frac{1}{\infty} }{1-\frac{2}{\infty}} = \frac{5}{1}

    Recall that any number divided by ∞\infty is 0.

    lim⁡n→∞n(5−1n)n(1−2n)=51≠0\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})} = \frac{5}{1} \neq 0
    ∴ ∑n=1∞5n−1n−2  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!

    📝 Divergence Test Practice Problems

    Try the following two practice questions yourself!

    1. ∑n=1∞n3+3n2n3−51. \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5}
    2. ∑n=1∞arctan⁡(n)2. \ \sum_{n=1}^∞ \arctan(n)

    ✅ Divergence Test: Solution 1

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

    lim⁡n→∞n3+3n2n3−5\lim_{n \to \infty} \dfrac{n^3+3n}{2n^3-5}
    lim⁡n→∞n3(1+3n2)n3(2−5n3)\lim_{n \to \infty} \dfrac{n^3(1+\dfrac{3}{n^2})}{n^3(2-\dfrac{5}{n^3})}
    1+3∞2−5∞=12\dfrac{1+\dfrac{3}{\infty}}{2-\dfrac{5}{\infty}} = \dfrac{1}{2}
    ∴ ∑n=1∞n3+3n2n3−5  diverges\therefore \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5} \ \ \text{diverges}

    Great work!

    ✅ Divergence Test: Solution 2

    Last question 🎉

    lim⁡n→∞arctan⁡(n)\lim_{n \to \infty} \arctan(n)

    Untitled

    Image courtesy of Math.net

    As arctan goes to ∞\infty, it stays at π2\dfrac{\pi}{2}.

     lim⁡n→∞arctan⁡(n)=π2 \ \lim_{n \to \infty} \arctan(n) = \dfrac{\pi}{2}
    ∴ ∑n=1∞arctan⁡(n) diverges\therefore \ \sum_{n=1}^∞ \arctan(n) \ \text{diverges}

    🕺 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! 🍀

    Key Terms to Review (1)

    nth Term Test

    : The nth Term Test is a method used to determine whether an infinite series converges or diverges by examining the behavior of its individual terms. If the limit of the sequence of terms as n approaches infinity is not zero, then the series diverges. If the limit is zero, further tests are needed to determine convergence.

    10.3 The nth Term Test for Divergence

    1 min read•april 5, 2021

    Attend a live cram event

    Review all units live with expert teachers & students

      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!

      🤷‍♀️ 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:

      if lim⁡n→∞an≠0,∑an 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.

      🤓 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. 📏 Evaluate the limit.
      3. 🤔 Make your conclusion based on the nth-term test.

      Determine if the series ana_n diverges.

      ∑n=1∞5n−1n−2\sum_{n=1}^∞ \dfrac{5n -1}{n-2}

      ✏️ Step 1: Convert to limit notation.

      lim⁡n→∞5n−1n−2\lim_{n \to \infty} \dfrac{5n -1}{n-2}

      📏 Step 2: Evaluate the limit.

      lim⁡n→∞n(5−1n)n(1−2n)\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})}
      5−1∞1−2∞=51\dfrac{5- \frac{1}{\infty} }{1-\frac{2}{\infty}} = \frac{5}{1}

      Recall that any number divided by ∞\infty is 0.

      lim⁡n→∞n(5−1n)n(1−2n)=51≠0\lim_{n \to \infty} \dfrac{n(5- \frac{1}{n} )}{n(1-\frac{2}{n})} = \frac{5}{1} \neq 0
      ∴ ∑n=1∞5n−1n−2  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!

      📝 Divergence Test Practice Problems

      Try the following two practice questions yourself!

      1. ∑n=1∞n3+3n2n3−51. \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5}
      2. ∑n=1∞arctan⁡(n)2. \ \sum_{n=1}^∞ \arctan(n)

      ✅ Divergence Test: Solution 1

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

      lim⁡n→∞n3+3n2n3−5\lim_{n \to \infty} \dfrac{n^3+3n}{2n^3-5}
      lim⁡n→∞n3(1+3n2)n3(2−5n3)\lim_{n \to \infty} \dfrac{n^3(1+\dfrac{3}{n^2})}{n^3(2-\dfrac{5}{n^3})}
      1+3∞2−5∞=12\dfrac{1+\dfrac{3}{\infty}}{2-\dfrac{5}{\infty}} = \dfrac{1}{2}
      ∴ ∑n=1∞n3+3n2n3−5  diverges\therefore \ \sum_{n=1}^∞ \dfrac{n^3+3n}{2n^3-5} \ \ \text{diverges}

      Great work!

      ✅ Divergence Test: Solution 2

      Last question 🎉

      lim⁡n→∞arctan⁡(n)\lim_{n \to \infty} \arctan(n)

      Untitled

      Image courtesy of Math.net

      As arctan goes to ∞\infty, it stays at π2\dfrac{\pi}{2}.

       lim⁡n→∞arctan⁡(n)=π2 \ \lim_{n \to \infty} \arctan(n) = \dfrac{\pi}{2}
      ∴ ∑n=1∞arctan⁡(n) diverges\therefore \ \sum_{n=1}^∞ \arctan(n) \ \text{diverges}

      🕺 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! 🍀

      Key Terms to Review (1)

      nth Term Test

      : The nth Term Test is a method used to determine whether an infinite series converges or diverges by examining the behavior of its individual terms. If the limit of the sequence of terms as n approaches infinity is not zero, then the series diverges. If the limit is zero, further tests are needed to determine convergence.
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      About Us
      About FiveableBlogCareersCode of ConductTerms of UsePrivacy PolicyCCPA Privacy Policy
      Resources
      Cram ModeAP Score CalculatorsStudy GuidesPractice QuizzesGlossaryCram EventsMerch ShopCrisis Text LineHelp Center

      © 2024 Fiveable Inc. All rights reserved.

      AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

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