The $n^{th}$ Term Test for Divergence - AP Calculus BC Study Guide

The Core Idea: The $n^{th}$ Term Test for Divergence

The $n^{th}$ Term Test for Divergence serves as a fundamental, preliminary check for the behavior of an infinite series. The core concept is based on a simple, logical requirement for convergence. For an infinite sum of terms, $\sum a_n$, to possibly add up to a finite number (converge), the terms being added must eventually become infinitesimally small. In other words, the sequence of terms $\{a_n\}$ must approach zero as $n$ approaches infinity.

This test formalizes that idea. If the terms of a series do not approach zero, it is impossible for the sum to converge, as we would be adding non-trivial numbers an infinite number of times. Therefore, if the limit of the $n^{th}$ term is anything other than zero, the series is guaranteed to diverge. It is critical to understand that this is a test for divergence only. If the terms do approach zero, it does not guarantee convergence; the series might still diverge for other reasons. In such cases, the test is inconclusive, and other, more powerful tests are required.

The $n^{th}$ Term Test for Divergence

The test is based on the behavior of the limit of the general term of the series. Given an infinite series ${\sum }_{n=1}^{\infty }{a}_n$:

1. Evaluate the limit of the general term:

   $$L=\underset{n\to \infty }{\lim }{a}_n$$

2. Apply the test conditions:

   • If $L\ne 0$ (this includes cases where the limit is a non-zero number or the limit does not exist), then the series ${\sum }_{n=1}^{\infty }{a}_n$diverges.

   • If $L=0$, the $n^{th}$ Term Test is inconclusive. The series may converge or it may diverge. This test provides no information, and another convergence test must be used.

Understanding the Inconclusive Case

The most critical nuance of the $n^{th}$ Term Test is understanding what happens when ${\lim }_{n\to \infty }{a}_n=0$. This result does not mean the series converges. It simply means the series might converge. The condition that the terms approach zero is a necessary prerequisite for convergence, but it is not a sufficient condition to guarantee it.

Consider two series where the limit of the terms is zero:

• The harmonic series, ${\sum }_{n=1}^{\infty }\frac{1}{n}$. Here, ${\lim }_{n\to \infty }\frac{1}{n}=0$. However, this series famously diverges. The terms do not get small "fast enough" for the sum to be finite.

• The series ${\sum }_{n=1}^{\infty }\frac{1}{n^2}$. Here, ${\lim }_{n\to \infty }\frac{1}{n^2}=0$. This series converges.

In both cases, the $n^{th}$ Term Test yields a limit of 0 and is therefore inconclusive. It cannot distinguish between the divergent harmonic series and the convergent series $\sum \frac{1}{n^2}$. This illustrates why you can never use the $n^{th}$ Term Test to conclude that a series converges. Its only power is to confirm divergence.

Core Concepts & Rules

• The $n^{th}$ Term Test is a test for divergence only. It can never be used to prove that a series converges.

• To apply the test to a series $\sum a_n$, you must first calculate the limit of the sequence of terms, ${\lim }_{n\to \infty }{a}_n$.

• If the limit of the terms is not zero or does not exist, the series $\sum a_n$ is guaranteed to diverge.

• If the limit of the terms is zero, the test is inconclusive. The series could either converge or diverge, and a different test must be applied to make a determination.

Step-by-Step Example 1: A Clear Case of Divergence

Problem: Determine if the series ${\sum }_{n=1}^{\infty }\frac{4n^3-5n}{2n^3+1}$ converges or diverges.

Step 1: Identify the general term, $a_n$.

The general term of the series is the expression being summed:

$$a_n=\frac{4n^3-5n}{2n^3+1}$$

Step 2: Set up the limit for the $n^{th}$ Term Test.

We need to evaluate the limit of $a_n$ as $n$ approaches infinity.

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{4n^3-5n}{2n^3+1}$$

Step 3: Evaluate the limit.

For a rational function of polynomials of the same degree, the limit is the ratio of the leading coefficients. Alternatively, we can divide the numerator and denominator by the highest power of $n$, which is $n^3$.

$$\underset{n\to \infty }{\lim }\frac{\frac{4n^3}{n^3}-\frac{5n}{n^3}}{\frac{2n^3}{n^3}+\frac{1}{n^3}}=\underset{n\to \infty }{\lim }\frac{4-\frac{5}{n^2}}{2+\frac{1}{n^3}}$$

As $n\to \infty$, the terms $\frac{5}{n^2}$ and $\frac{1}{n^3}$ approach 0.

$$\frac{4-0}{2+0}=2$$

Step 4: State the conclusion with justification.

The limit of the terms is ${\lim }_{n\to \infty }{a}_n=2$. Since this limit is not equal to 0, the series ${\sum }_{n=1}^{\infty }\frac{4n^3-5n}{2n^3+1}$diverges by the $n^{th}$ Term Test for Divergence.

Step-by-Step Example 2: The Inconclusive Case

Problem: A student claims the series ${\sum }_{n=1}^{\infty }\sin \left(\frac{1}{n}\right)$ converges because ${\lim }_{n\to \infty }\sin \left(\frac{1}{n}\right)=0$. Is the student's reasoning correct? Justify your answer.

Step 1: Identify the general term, $a_n$.

The general term of the series is:

$$a_n=\sin \left(\frac{1}{n}\right)$$

Step 2: Evaluate the limit of the general term as the student did.

We need to find the limit of $a_n$ as $n$ approaches infinity.

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\sin \left(\frac{1}{n}\right)$$

As $n\to \infty$, the argument $\frac{1}{n}\to 0$. Since the sine function is continuous at 0, we can evaluate the limit:

$$\sin \left(\underset{n\to \infty }{\lim }\frac{1}{n}\right)=\sin (0)=0$$

Step 3: Analyze the student's reasoning based on the rules of the $n^{th}$ Term Test.

The student correctly found that the limit of the terms is 0. However, they incorrectly concluded that this implies convergence.

Step 4: State the correct conclusion with justification.

The student's reasoning is incorrect. According to the $n^{th}$ Term Test for Divergence, if the limit of the terms of a series is 0, the test is inconclusive. It does not provide any information about whether the series converges or diverges. Therefore, one cannot conclude that the series ${\sum }_{n=1}^{\infty }\sin \left(\frac{1}{n}\right)$ converges based on this test alone. Another test, such as the Limit Comparison Test, would be needed to determine the behavior of this series.

Using Your Calculator

The $n^{th}$ Term Test is a purely analytical test; the conclusion must be reached by evaluating a limit and applying the rule. A graphing calculator cannot perform the test for you, but it can be a powerful tool for investigating the limit of the terms, ${\lim }_{n\to \infty }{a}_n$, especially for more complex functions.

To investigate ${\lim }_{n\to \infty }{a}_n$ for the series ${\sum }_{n=1}^{\infty }n{e}^{-n}$:

1. Function Mode: Enter the expression for $a_n$ as a function in the Y= editor. Use $X$ in place of $n$.

   • Y1 = X*e^(-X)

2. Investigate Numerically using a Table:

   • Go to TBLSET (2nd + WINDOW).

   • Set TblStart to a large number, for example, $100$.

   • Set ΔTbl to a large step, for example, $10$.

   • View the TABLE (2nd + GRAPH). You will see that the Y1 values are extremely small and clearly approaching 0. This provides strong evidence that ${\lim }_{n\to \infty }n{e}^{-n}=0$.

3. Investigate Graphically:

   • Graph the function Y1. You may need to adjust the WINDOW settings (e.g., $Xmax=10$, $Ymax=1$) to see the behavior.

   • Observe the graph. You will see that the function approaches the x-axis as $x$ increases, indicating a horizontal asymptote at $y=0$. This provides graphical evidence that the limit is 0.

Conclusion from Calculator Evidence:

The numerical and graphical evidence strongly suggests that ${\lim }_{n\to \infty }n{e}^{-n}=0$. Based on this, you would conclude that the $n^{th}$ Term Test is inconclusive for the series ${\sum }_{n=1}^{\infty }n{e}^{-n}$. The calculator helps find the limit, but the final conclusion is based on the rules of the test.

AP Exam Quick Hit

Common Question Types

• Direct Application (Multiple Choice or FRQ): You will be given a series and asked to determine if it converges or diverges. The $n^{th}$ Term Test is often the quickest way to identify a divergent series.

  • Example: Does the series ${\sum }_{n=1}^{\infty }{\left(1+\frac{1}{n}\right)}^n$ converge or diverge?

  • Solution:${\lim }_{n\to \infty }{\left(1+\frac{1}{n}\right)}^n=e$. Since $e\ne 0$, the series diverges by the $n^{th}$ Term Test.

• Identifying Inconclusive Tests (Multiple Choice): You may be asked to identify which series from a list results in an inconclusive $n^{th}$ Term Test. This requires you to find the one series where the limit of the terms is zero.

  • Example: For which of the following series is the $n^{th}$ Term Test inconclusive?

    (A) $\sum \frac{n^2}{5n^2+3}$ (B) $\sum \frac{1}{n}$ (C) $\sum \cos (n)$

  • Solution: (B), because ${\lim }_{n\to \infty }\frac{1}{n}=0$. For (A), the limit is $\frac{1}{5}\ne 0$. For (C), the limit does not exist.

• Conceptual Justification (Multiple Choice or FRQ): Questions may test your understanding of the conditions of the test itself.

  • Example: If $\{a_n\}$ is a sequence of positive terms and ${\lim }_{n\to \infty }{a}_n=0.001$, what can be concluded about the series ${\sum }_{n=1}^{\infty }{a}_n$?

  • Solution: The series diverges. The limit of the terms is a non-zero number (0.001), so by the $n^{th}$ Term Test, the series must diverge.

Common Mistakes

• The Converse Fallacy: The most common mistake is to conclude that a series converges just because the limit of its terms is zero. Remember: If ${\lim }_{n\to \infty }{a}_n=0$, the test is INCONCLUSIVE. You cannot use this test to prove convergence.

• Limit Calculation Errors: Simple algebraic or calculus errors when finding the limit of the sequence of terms. This is especially common with indeterminate forms requiring L'Hôpital's Rule or with oscillating sequences like $a_n=(-1{)}^n$.

• Confusing Sequence Convergence with Series Convergence: For a series like ${\sum }_{n=1}^{\infty }\frac{2n}{n+1}$, the sequence of terms $a_n=\frac{2n}{n+1}$ converges to 2. However, the series (the sum) diverges by the $n^{th}$ Term Test because that limit is not 0. Be precise with your language.

• Poor or Incomplete Justification: On an FRQ, simply stating "The series diverges" is not enough. You must state the name of the test used, show the evaluation of the limit, and explicitly state why the result of that limit leads to your conclusion. For example: "Since ${\lim }_{n\to \infty }{a}_n=L\ne 0$, the series diverges by the $n^{th}$ Term Test."