Comparison Tests for Convergence - AP Calculus BC Study Guide

The Core Idea: Comparison Tests for Convergence

When faced with a complicated series, determining its convergence or divergence directly can be challenging. The core idea behind the comparison tests is to determine the convergence of a given series by comparing its terms to the terms of another, simpler series whose convergence behavior is already known (such as a p-series or a geometric series). This method provides a powerful tool for deducing the behavior of a complex series based on its relationship to a familiar one.

There are two primary methods for this comparison. The Direct Comparison Test examines the relative size of the terms of the two series; if the terms of the unknown series are smaller than those of a known convergent series, it too must converge. The Limit Comparison Test examines the long-term behavior of the ratio of the terms of the two series; if this ratio approaches a finite, positive number, the two series share the same fate—they either both converge or both diverge.

Key Theorems: The Comparison Tests

The convergence of a series with positive terms can be determined by comparing it to a series of known convergence or divergence.

The Direct Comparison Test (DCT)

Let $\sum a_n$ and $\sum b_n$ be series with positive terms. The test is based on the following conditions, which must hold for all $n$ greater than or equal to some integer $N$:

1. For Convergence: If $0<a_n\le b_n$ for all $n\ge N$, and the series $\sum b_n$ converges, then the series $\sum a_n$ also converges.

   • Intuition: If the terms of your series ($a_n$) are smaller than the terms of a known convergent series ($b_n$), your series is "trapped" and cannot grow to infinity.

2. For Divergence: If $0<b_n\le a_n$ for all $n\ge N$, and the series $\sum b_n$ diverges, then the series $\sum a_n$ also diverges.

   • Intuition: If the terms of your series ($a_n$) are larger than the terms of a known divergent series ($b_n$), your series is "pushed" to infinity.

The Limit Comparison Test (LCT)

Let $\sum a_n$ and $\sum b_n$ be series with positive terms ($a_n>0$ and $b_n>0$).

Calculate the limit of the ratio of the corresponding terms:

$$L=\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$$

If $L$ is a finite and positive number ($0<L<\infty$), then the two series $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

• Intuition: If the limit $L$ is finite and positive, it means that for large $n$, the terms $a_n$ and $b_n$ are roughly proportional to each other ($a_n\approx L\cdot b_n$). Therefore, their corresponding series must behave in the same way.

Understanding the Conditions

The conditions for applying the comparison tests are strict and essential for a valid conclusion.

1. Positive Terms: Both the Direct Comparison Test and the Limit Comparison Test require that the terms of the series being compared ($a_n$ and $b_n$) are positive. These tests are designed for series whose partial sums are non-decreasing. If a series has negative terms, these tests cannot be directly applied.

2. The Direct Comparison Inequality: The direction of the inequality in the Direct Comparison Test is critical and is a common source of error.

• To prove convergence, you must show your series is less than or equal to a known convergent series.

• To prove divergence, you must show your series is greater than or equal to a known divergent series.

• Inconclusive Cases: The test provides no information if your series is greater than a convergent series or less than a divergent series. In these situations, you must choose a different comparison series or use a different test, like the Limit Comparison Test.

3. The Limit Comparison Value: For the Limit Comparison Test, the conclusion that both series "share the same fate" is only valid if the resulting limit, $L$, is both finite and positive.

• Finite:$L\ne \infty$.

• Positive:$L\ne 0$.

If the limit is 0 or $\infty$, the specific conclusion stated in the Essential Knowledge does not apply, and another test or a more nuanced version of the LCT would be needed. For the AP exam, focus on cases where $L$ is a finite, positive number.

4. Choosing the Comparison Series $\sum b_n$: The success of both tests hinges on choosing an appropriate series $\sum b_n$. A good strategy is to look at the dominant terms in the numerator and denominator of $a_n$ as $n\to \infty$. For example, if $a_n=\frac{3n^2+5}{n^4+2n-1}$, the dominant terms are $3n^2$ in the numerator and $n^4$ in the denominator. This suggests a comparison with $b_n=\frac{n^2}{n^4}=\frac{1}{n^2}$, which is a known convergent p-series.

Core Concepts & Rules

• Purpose: Comparison tests determine the convergence or divergence of a series $\sum a_n$ by comparing it to a known series $\sum b_n$.

• Prerequisite: The terms of both series, $a_n$ and $b_n$, must be positive.

• Direct Comparison Test (DCT):

  • To prove convergence: Find a known convergent series $\sum b_n$ such that $a_n\le b_n$ for all sufficiently large $n$.

  • To prove divergence: Find a known divergent series $\sum b_n$ such that $a_n\ge b_n$ for all sufficiently large $n$.

• Limit Comparison Test (LCT):

  • Choose a known series $\sum b_n$ that you believe behaves like $\sum a_n$.

  • Calculate $L={\lim }_{n\to \infty }\frac{a_n}{b_n}$.

  • If $L$ is a finite, positive number, then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Step-by-Step Example 1: Direct Comparison Test

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

Step 1: Identify the series and check conditions.

The series is $\sum a_n$ where $a_n=\frac{1}{n^3+5}$. For $n\ge 1$, the terms $a_n$ are positive.

Step 2: Choose a comparison series $\sum b_n$.

The dominant term in the denominator is $n^3$. This suggests we compare our series to the p-series $\sum b_n=\sum \frac{1}{n^3}$.

Step 3: Analyze the comparison series.

The series ${\sum }_{n=1}^{\infty }\frac{1}{n^3}$ is a p-series with $p=3$. Since $p>1$, the series $\sum \frac{1}{n^3}$ converges.

Step 4: Establish the inequality for the Direct Comparison Test.

We want to show that our series is "smaller" than the known convergent series.

For $n\ge 1$:

$$n^3+5>n^3$$

Taking the reciprocal of both sides reverses the inequality:

$$\frac{1}{n^3+5}<\frac{1}{n^3}$$

So, we have established that $a_n<b_n$ for all $n\ge 1$.

Step 5: State the conclusion.

Because $0<\frac{1}{n^3+5}<\frac{1}{n^3}$ and the series ${\sum }_{n=1}^{\infty }\frac{1}{n^3}$ converges, the series ${\sum }_{n=1}^{\infty }\frac{1}{n^3+5}$ must also converge by the Direct Comparison Test.

Step-by-Step Example 2: Limit Comparison Test

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

Step 1: Identify the series and check conditions.

The series is $\sum a_n$ where $a_n=\frac{\sqrt{n}}{n^2-3}$. For $n\ge 2$, the terms $a_n$ are positive.

Note: Direct comparison with $\frac{\sqrt{n}}{n^2}$ is difficult because $n^2-3<n^2$, which means $\frac{\sqrt{n}}{n^2-3}>\frac{\sqrt{n}}{n^2}$. This is the "greater than a convergent series" case, which is inconclusive for the DCT.

Step 2: Choose a comparison series $\sum b_n$.

Consider the dominant terms in $a_n$. The numerator is $\sqrt{n}=n^{1/2}$ and the dominant term in the denominator is $n^2$.

This suggests a comparison with the series $\sum b_n$ where $b_n=\frac{n^{1/2}}{n^2}=\frac{1}{n^{3/2}}$.

Step 3: Analyze the comparison series.

The series ${\sum }_{n=2}^{\infty }\frac{1}{n^{3/2}}$ is a p-series with $p=\frac{3}{2}$. Since $p>1$, the series $\sum b_n$ converges.

Step 4: Set up and evaluate the limit for the Limit Comparison Test.

We need to calculate $L={\lim }_{n\to \infty }\frac{a_n}{b_n}$.

$$L=\underset{n\to \infty }{\lim }\frac{\frac{\sqrt{n}}{n^2-3}}{\frac{1}{n^{3/2}}}$$

Step 5: Simplify the limit expression.

$$L=\underset{n\to \infty }{\lim }\left(\frac{\sqrt{n}}{n^2-3}\cdot \frac{n^{3/2}}{1}\right)=\underset{n\to \infty }{\lim }\frac{n^2}{n^2-3}$$

Step 6: Evaluate the limit.

To evaluate this limit, we can divide the numerator and denominator by the highest power of $n$, which is $n^2$.

$$L=\underset{n\to \infty }{\lim }\frac{\frac{n^2}{n^2}}{\frac{n^2}{n^2}-\frac{3}{n^2}}=\underset{n\to \infty }{\lim }\frac{1}{1-\frac{3}{n^2}}=\frac{1}{1-0}=1$$

Step 7: State the conclusion.

The limit $L=1$ is a finite and positive number. Because the comparison series $\sum \frac{1}{n^{3/2}}$ converges, the original series ${\sum }_{n=2}^{\infty }\frac{\sqrt{n}}{n^2-3}$ must also converge by the Limit Comparison Test.

Using Your Calculator

The Comparison Tests are fundamentally analytical and logical tests; they cannot be "solved" on a calculator. A calculator cannot perform the symbolic manipulation or logical deduction required. However, it can be a useful tool for building intuition or checking your work.

1. Verifying the Limit in LCT:

For the LCT example above, you can investigate the limit $L={\lim }_{n\to \infty }\frac{n^2}{n^2-3}$.

• Graphically: Graph the function $Y_1=\frac{x^2}{x^2-3}$. Use the TRACE function for large values of $x$ or observe the graph's horizontal asymptote. You will see the y-values approach 1.

• Numerically: Use the table feature. Set TblStart to a large number (e.g., 1000) and ΔTbl to a large step (e.g., 100). Observe that the values in the $Y_1$ column get closer and closer to 1.

2. Investigating Series Behavior:

You can calculate partial sums of a series to guess whether it might converge or diverge. This can help guide your choice of a comparison series, but it is not a proof.

• To investigate $\sum \frac{\sqrt{n}}{n^2-3}$, you can use the $sum(seq(...))$ command: $sum(seq(\surd (X)/(X^2-3),X,2,100))$. Calculate this for a large upper bound (e.g., 100, then 1000). If the partial sums appear to level off at a specific value, it suggests convergence. If they continue to grow without bound, it suggests divergence.

AP Exam Quick Hit

Common Question Types

• Direct Application (Free Response): "Determine whether the series $\sum _{n=1}^{\infty }\frac{n}{n^3+{\sin }^2(n)}$ converges or diverges. State the conditions of the test used and show your work."

• Choosing the Right Comparison (Multiple Choice): "The convergence of which of the following series can be determined by the Limit Comparison Test with the series ${\sum }_{n=1}^{\infty }\frac{1}{n^2}$?"

  (A) $\sum \frac{1}{n}$ (B) $\sum \frac{n}{n^2+1}$ (C) $\sum \frac{n^2}{n^3+1}$ (D) $\sum \frac{n^3}{n^4+1}$

  The answer would be (D), as the ratio of dominant terms is $\frac{n^3}{n^4}=\frac{1}{n}$, not $\frac{1}{n^2}$. The correct answer would be a series whose dominant terms simplify to $\frac{1}{n^2}$, such as $\sum \frac{n}{n^3+1}$.

• Justifying a Conclusion (Multiple Choice): "The series $\sum a_n$ converges by direct comparison with $\sum \frac{1}{n^4}$. Which of the following could be $a_n$?" The student must find an option where $a_n\le \frac{1}{n^4}$.

Common Mistakes

• Incorrect Inequality in DCT: To prove convergence, students show $a_n\ge b_n$ (where $\sum b_n$ converges) or to prove divergence, they show $a_n\le b_n$ (where $\sum b_n$ diverges). The inequality must be in the correct direction for the test to be conclusive.

• Drawing a Conclusion from an Inconclusive DCT: Stating that $\sum a_n$ converges because $a_n\ge b_n$ and $\sum b_n$ converges. This is invalid; being larger than a finite sum does not guarantee your own sum is finite. Similarly, being smaller than a divergent series (which sums to infinity) provides no information.

• Forgetting Conditions: Failing to state that the terms of both series are positive, which is a necessary condition for both tests. On a free-response question, this can lead to a loss of points.

• LCT Limit Conclusion Error: Concluding that if $L=1$ in the LCT, the series converges to 1. The LCT only determines if the series converges, not the value to which it converges.

• Poor Choice of Comparison Series: Choosing a series $\sum b_n$ that does not have the same end behavior as $\sum a_n$. For example, trying to use LCT on $\sum \frac{1}{n^2+1}$ with $\sum \frac{1}{n}$. The limit would be 0, and the test would not apply based on the provided EK. The correct choice is $\sum \frac{1}{n^2}$.