Alternating Series Test for Convergence - AP Calculus BC Study Guide

The Core Idea: Alternating Series Test for Convergence

An alternating series is a series in which the terms alternate between positive and negative signs. These series appear in the form ${\sum }_{n=1}^{\infty }(-1{)}^n{a}_n$ or ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$, where the $a_n$ part represents the positive magnitude of each term.

The fundamental challenge is to determine if such a series converges to a finite sum. The Alternating Series Test provides a straightforward, two-step method to determine if an alternating series converges. The test formalizes the intuitive idea that if the terms are consistently shrinking in magnitude and approaching zero, the back-and-forth "steps" of the partial sums will eventually hone in on a specific value.

The Alternating Series Test

For an alternating series of the form ${\sum }_{n=1}^{\infty }(-1{)}^n{a}_n$ or ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$, where $a_n>0$ for all $n$, the series converges if both of the following conditions are satisfied:

1. The limit of the terms' magnitudes is zero:

   $$\underset{n\to \infty }{\lim }{a}_n=0$$

2. The terms' magnitudes are non-increasing:

   $$a_{n+1}\le a_n$$

   This condition must hold for all $n$, or at least for all $n$ greater than some integer $N$.

Understanding the Conditions

The two conditions of the Alternating Series Test are both essential for convergence, and each serves a distinct purpose.

The first condition, ${\lim }_{n\to \infty }{a}_n=0$, ensures that the individual terms of the series eventually become infinitesimally small. If the terms did not approach zero, the partial sums would continue to oscillate by a significant amount, preventing them from settling on a finite limit. This condition is a necessary prerequisite for the convergence of any series.

The second condition, $a_{n+1}\le a_n$, ensures that the magnitude of the terms is consistently decreasing (or at least not increasing). This guarantees that each successive term brings the partial sum closer to the final limit than the previous term did. The "for all $n$ greater than some integer $N$" clause is an important nuance; the terms do not need to be decreasing from the very beginning. As long as the sequence of terms $a_n$ is eventually decreasing, the condition is met.

If both conditions are met, the test guarantees convergence. The test itself does not provide the sum of the series, only that a finite sum exists.

Core Concepts & Rules

• Definition: An alternating series is one whose terms alternate in sign, such as $\sum (-1{)}^n{a}_n$ or $\sum (-1{)}^{n+1}{a}_n$, where $a_n$ is always positive.

• The Test: The Alternating Series Test is a tool used specifically to determine if an alternating series converges.

• Condition 1 (Limit): The limit of the positive part of the term, $a_n$, must be zero as $n$ approaches infinity. That is, ${\lim }_{n\to \infty }{a}_n=0$.

• Condition 2 (Decreasing): The sequence of positive terms, $\{a_n\}$, must be non-increasing. This means $a_{n+1}\le a_n$ for all terms eventually (i.e., for all $n>N$ for some integer $N$).

• Conclusion: If both of the above conditions are satisfied, you can conclude that the alternating series converges. The test only provides a conclusion of convergence.

Step-by-Step Example 1: Basic Application

Problem: Determine if the alternating harmonic series, ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}$, converges.

Step 1: Identify the series type and $a_n$.

The series is $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$. This is an alternating series.

The form is ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$, where the positive part of the term is $a_n=\frac{1}{n}$.

Step 2: Check the first condition of the Alternating Series Test.

We must check if ${\lim }_{n\to \infty }{a}_n=0$.

$$\underset{n\to \infty }{\lim }\frac{1}{n}=0$$

The first condition is met.

Step 3: Check the second condition of the Alternating Series Test.

We must check if $a_{n+1}\le a_n$ for all $n$ (or eventually).

$$a_{n+1}=\frac{1}{n+1}\text{and}{a}_n=\frac{1}{n}$$

Is $\frac{1}{n+1}\le \frac{1}{n}$?

Since $n+1>n$ for all $n\ge 1$, taking the reciprocal reverses the inequality. Thus, $\frac{1}{n+1}<\frac{1}{n}$ is true for all $n\ge 1$.

The second condition is met.

Step 4: State the conclusion.

Since both conditions of the Alternating Series Test are met (${\lim }_{n\to \infty }{a}_n=0$ and $\{a_n\}$ is a decreasing sequence), the series ${\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}$ converges.

Step-by-Step Example 2: Exam-Style Application

Problem: Determine if the series ${\sum }_{n=1}^{\infty }(-1{)}^n\frac{n^2}{e^n}$ converges.

Step 1: Identify the series type and $a_n$.

The series is alternating. The positive part of the term is $a_n=\frac{n^2}{e^n}$.

Step 2: Check the first condition of the Alternating Series Test.

We must check if ${\lim }_{n\to \infty }{a}_n=0$.

$$\underset{n\to \infty }{\lim }\frac{n^2}{e^n}$$

This limit is of the indeterminate form $\frac{\infty }{\infty }$. We can use L'Hôpital's Rule.

$$\underset{n\to \infty }{\lim }\frac{n^2}{e^n}\stackrel{L^{\prime }H}{=}\underset{n\to \infty }{\lim }\frac{2n}{e^n}$$

This is still $\frac{\infty }{\infty }$, so we apply L'Hôpital's Rule again.

$$\stackrel{L^{\prime }H}{=}\underset{n\to \infty }{\lim }\frac{2}{e^n}=0$$

The first condition is met.

Step 3: Check the second condition of the Alternating Series Test.

We must check if $a_{n+1}\le a_n$ eventually. This is not immediately obvious, so we can analyze the related function $f(x)=\frac{x^2}{e^x}$ and determine where it is decreasing by checking its derivative.

$$f^{\prime }(x)=\frac{e^x(2x)-x^2(e^x)}{(e^x{)}^2}=\frac{x{e}^x(2-x)}{e^{2x}}=\frac{x(2-x)}{e^x}$$

We want to find where $f^{\prime }(x)\le 0$. Since $x\ge 1$ and $e^x>0$, the sign of $f^{\prime }(x)$ is determined by the term $(2-x)$.

$$2-x\le 0⟹x\ge 2$$

The derivative $f^{\prime }(x)$ is negative for all $x\ge 2$. This means the function $f(x)$ is decreasing for $x\ge 2$. Therefore, the sequence of terms $a_n=\frac{n^2}{e^n}$ is decreasing for all $n\ge 2$. The "eventually decreasing" condition is met.

Step 4: State the conclusion.

Since ${\lim }_{n\to \infty }\frac{n^2}{e^n}=0$ and the sequence $\{a_n\}$ is decreasing for $n\ge 2$, both conditions of the Alternating Series Test are met. Therefore, the series ${\sum }_{n=1}^{\infty }(-1{)}^n\frac{n^2}{e^n}$ converges.

Using Your Calculator

The Alternating Series Test is an analytical test and cannot be performed directly by a calculator. However, a graphing calculator can be a useful tool for verifying the two conditions, especially when they are not obvious.

To verify Condition 1: ${\lim }_{n\to \infty }{a}_n=0$

1. Define the function corresponding to $a_n$. For the series $\sum (-1{)}^n\frac{\ln (n)}{n}$, you would enter Y1 = ln(X)/X.

2. Use the TABLE feature. Set TblStart to a large number (e.g., 1000) and ΔTbl to a large step (e.g., 100). Observe the Y1 values. They should be getting progressively closer to 0.

3. Alternatively, GRAPH the function and use TRACE for large values of $X$ to visually confirm that the graph is approaching the x-axis ($y=0$).

To verify Condition 2: $a_{n+1}\le a_n$ (eventually decreasing)

1. Define the function corresponding to $a_n$ in Y1.

2. GRAPH the function. Visually inspect the graph for large values of $X$. The graph should be consistently decreasing (moving downward as you move to the right) after some point X=N.

3. For a more rigorous check, you can analyze the derivative.

   • In Y2, enter the derivative of Y1 using the calculator's numerical derivative function (e.g., nDeriv(Y1, X, X)).

   • Graph Y2. For the decreasing condition to hold, the graph of the derivative (Y2) must be at or below the x-axis (i.e., be negative or zero) for all $X$ greater than some value $N$.

AP Exam Quick Hit

Common Question Types

• Direct Application: You will be given an alternating series and asked to determine if it converges. You must state the test you are using and show that both conditions are met.

  • Example: "Does the series ${\sum }_{n=2}^{\infty }\frac{(-1{)}^n}{\ln n}$ converge? Justify your answer."

• Justifying with Conditions: A multiple-choice question might ask which statement must be true for a series that is known to converge by the Alternating Series Test.

  • Example: "The series ${\sum }_{n=1}^{\infty }(-1{)}^n{a}_n$ converges by the Alternating Series Test. Which of the following must be true?"

    (A) ${\lim }_{n\to \infty }{a}_n=L$, where $L\ne 0$

    (B) $a_{n+1}>a_n$ for all $n$

    (C) ${\lim }_{n\to \infty }{a}_n=0$ and $a_{n+1}\le a_n$ for all $n$ sufficiently large.

Common Mistakes

• Forgetting a Condition: A very common error is to only check that ${\lim }_{n\to \infty }{a}_n=0$ and immediately conclude convergence, completely forgetting to verify that the terms are decreasing. Both conditions are required.

• Incorrectly Defining $a_n$: The term $a_n$ is the positive part of the series term. Students sometimes incorrectly include the $(-1{)}^n$ or $(-1{)}^{n+1}$ when defining and checking the conditions for $a_n$. Remember, the condition is that $a_n>0$.

• Giving Up on the Decreasing Condition: If the first few terms are not decreasing (e.g., $a_2>a_1$), students may incorrectly conclude the test fails. Remember to check if the terms are eventually decreasing, which may require analyzing the derivative of the related function.

• Incorrect Conclusion from a Failed Test: The Alternating Series Test can only be used to prove convergence. If one of the conditions fails, you cannot conclude that the series diverges by the Alternating Series Test. If ${\lim }_{n\to \infty }{a}_n\ne 0$, the series diverges by the nth-Term Test for Divergence, not the AST. If ${\lim }_{n\to \infty }{a}_n=0$ but the terms are not decreasing, the AST is inconclusive.