Ratio Test for Convergence - AP Calculus BC Study Guide

The Core Idea: Ratio Test for Convergence

The Ratio Test is a powerful tool used to determine whether an infinite series converges or diverges. The fundamental idea behind this test is to examine the behavior of the ratio of consecutive terms in the series. By taking the limit of the absolute value of this ratio as $n$ approaches infinity, we can understand how the terms are growing or shrinking.

If the terms are shrinking sufficiently fast, meaning the limit of the ratio is less than one, the series is guaranteed to converge. Conversely, if the terms are growing or not shrinking fast enough, indicated by a limit greater than one, the series will diverge. The test provides a clear conclusion in these cases, making it particularly effective for series that involve expressions like factorials or n$th powers, where the growth rate of terms is pronounced. ## The Ratio Test For a given infinite series\sum a_n$, the Ratio Test is defined by the following limit and conditions:

First, calculate the limit $L$:

$$L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$$

Then, apply the following three conditions based on the value of $L$:

1. If $L<1$, the series $\sum a_n$ converges absolutely.

2. If $L>1$, the series $\sum a_n$ diverges.

3. If $L=1$, the Ratio Test is inconclusive. This means the test provides no information about the convergence or divergence of the series, and another method must be used.

Understanding the Conditions

The power of the Ratio Test lies in its comparison of the series to a geometric series. The limit $L$ can be thought of as the eventual common ratio of the series. If $L<1$, the series behaves like a convergent geometric series, where each term is a fraction of the previous one, causing the terms to approach zero quickly enough for the sum to be finite. The "converges absolutely" conclusion is a strong one; it means that the series formed by the absolute values of the terms, $\sum |a_n|$, also converges.

If $L>1$, the terms of the series are eventually increasing in magnitude. Since the terms do not approach zero, the series must diverge by the n$th Term Test for Divergence. The most critical nuance is the inconclusive case where $L=1. This result does not mean the series converges or diverges; it means the Ratio Test is not sensitive enough to make a determination. The terms are not shrinking fast enough for a clear conclusion of convergence, but they are not growing either. For example, both the convergent p-series $\sum \frac{1}{n^2}$ and the divergent harmonic series $\sum \frac{1}{n}$ yield a limit of $L=1$ with the Ratio Test. Therefore, when $L=1$, you must state that the test is inconclusive and another convergence test is required.

Core Concepts & Rules

• Purpose: The Ratio Test is a method used to determine if an infinite series converges or diverges.

• Procedure: The test involves calculating the limit of the absolute value of the ratio of the (n+1)$th term to the $n$th term. * **The Limit Formula:** The core calculation is $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.

• Convergence Condition: If the limit $L$ is less than 1 ($L<1$), the series converges absolutely.

• Divergence Condition: If the limit $L$ is greater than 1 ($L>1$), the series diverges.

• Inconclusive Condition: If the limit $L$ is equal to 1 ($L=1$), the test fails to provide a conclusion.

• Best Use Cases: The Ratio Test is most effective and should be considered first for series containing factorials (e.g., $n!$) or n$th powers (e.g., $c^n).

Step-by-Step Example 1: Series with n$th Powers **Problem:** Determine if the series\sum_{n=1}^{\infty} \frac{n^2}{3^n}converges or diverges. **Step 1: Identify $a_n and $a_{n+1}$.**

For this series, the general term is $a_n=\frac{n^2}{3^n}$.

To find $a_{n+1}$, we replace every $n$ in $a_n$ with $(n+1)$:

$$a_{n+1}=\frac{(n+1{)}^2}{3^{n+1}}$$

Step 2: Set up the ratio $\left|\frac{a_{n+1}}{a_n}\right|$.

Since all terms are positive, the absolute value is not strictly necessary but is good practice to include.

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{(n+1{)}^2}{3^{n+1}}}{\frac{n^2}{3^n}}\right|$$

Step 3: Simplify the ratio.

To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator.

$$\frac{(n+1{)}^2}{3^{n+1}}\cdot \frac{3^n}{n^2}=\frac{(n+1{)}^2}{n^2}\cdot \frac{3^n}{3^{n+1}}$$

Using exponent rules ($3^{n+1}=3\cdot 3^n$), we simplify the second fraction:

$${\left(\frac{n+1}{n}\right)}^2\cdot \frac{1}{3}$$

Step 4: Evaluate the limit $L$.

Now, we take the limit as $n\to \infty$.

$$L=\underset{n\to \infty }{\lim }\left[{\left(\frac{n+1}{n}\right)}^2\cdot \frac{1}{3}\right]$$

We can rewrite the fraction inside the parenthesis:

$$L=\underset{n\to \infty }{\lim }\left[{\left(1+\frac{1}{n}\right)}^2\cdot \frac{1}{3}\right]$$

As $n\to \infty$, the term $\frac{1}{n}\to 0$.

$$L=(1+0{)}^2\cdot \frac{1}{3}=1^2\cdot \frac{1}{3}=\frac{1}{3}$$

Step 5: Draw a conclusion based on the value of $L$.

The limit is $L=\frac{1}{3}$. Since $L<1$, the series converges absolutely by the Ratio Test.

Step-by-Step Example 2: Series with Factorials

Problem: Determine if the series ${\sum }_{n=0}^{\infty }\frac{(-2{)}^n}{n!}$ converges or diverges.

Step 1: Identify $a_n$ and $a_{n+1}$.

The general term is $a_n=\frac{(-2{)}^n}{n!}$.

The next term is $a_{n+1}=\frac{(-2{)}^{n+1}}{(n+1)!}$.

Step 2: Set up the ratio $\left|\frac{a_{n+1}}{a_n}\right|$.

The absolute value is critical here due to the alternating term $(-2{)}^n$.

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{(-2{)}^{n+1}}{(n+1)!}}{\frac{(-2{)}^n}{n!}}\right|$$

Step 3: Simplify the ratio.

First, simplify the complex fraction and handle the absolute value.

$$\left|\frac{(-2{)}^{n+1}}{(n+1)!}\cdot \frac{n!}{(-2{)}^n}\right|=\left|\frac{(-2{)}^{n+1}}{(-2{)}^n}\cdot \frac{n!}{(n+1)!}\right|$$

The absolute value makes the negative base positive: $|-2|=2$.

$$\frac{2^{n+1}}{2^n}\cdot \frac{n!}{(n+1)!}$$

Now, simplify the exponential and factorial parts.

Recall that $(n+1)!=(n+1)\cdot n!$.

$$\frac{2^{n+1}}{2^n}=2^{n+1-n}=2^1=2$$

$$\frac{n!}{(n+1)!}=\frac{n!}{(n+1)\cdot n!}=\frac{1}{n+1}$$

Combining these gives the simplified ratio:

$$2\cdot \frac{1}{n+1}=\frac{2}{n+1}$$

Step 4: Evaluate the limit $L$.

$$L=\underset{n\to \infty }{\lim }\frac{2}{n+1}$$

As the denominator $n+1$ grows infinitely large, the fraction approaches zero.

$$L=0$$

Step 5: Draw a conclusion based on the value of $L$.

The limit is $L=0$. Since $L<1$, the series converges absolutely by the Ratio Test.

Using Your Calculator

The Ratio Test is an analytical test, meaning its steps (setting up the ratio, simplifying algebra, and finding the limit) must be done by hand and shown as work on the AP Exam. A calculator cannot perform the symbolic algebra or limiting process required.

However, a calculator can be a useful tool for verifying your result. After you have algebraically simplified the expression for $\left|\frac{a_{n+1}}{a_n}\right|$, you can numerically estimate the limit by substituting a very large value for $n$.

Example: For the series ${\sum }_{n=1}^{\infty }\frac{n^2}{3^n}$ from Example 1, we simplified the ratio to ${\left(\frac{n+1}{n}\right)}^2\cdot \frac{1}{3}$.

Verification Steps (TI-84 Style):

1. Enter the simplified ratio into the Y= editor: Y1 = ((X+1)/X)^2 * (1/3)

2. Go to the table (2nd + GRAPH).

3. Set the table to start at a large number (e.g., TblStart = 1000) or scroll down to large values of $X$.

4. Observe the Y1 values. You will see that they approach $\mathrm{0.3333...}$, which supports the analytical conclusion that $L=\frac{1}{3}$.

This method does not replace the analytical work but can provide confidence that your algebraic simplification and limit evaluation are correct.

AP Exam Quick Hit

Common Question Types

• Direct Application with Factorials and Powers: You will be given a series with a mix of factorials, constants raised to the n$th power, and polynomials in $n, and asked to determine convergence or divergence.

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

• Showing the Test is Inconclusive: You may be given a series for which the Ratio Test fails (like a p-series) and be asked to apply the test and state the conclusion.

  • Example: "Show that the Ratio Test is inconclusive for the series ${\sum }_{n=1}^{\infty }\frac{n+1}{n^2+3}$."

Common Mistakes

• Algebraic Errors with Factorials: Incorrectly simplifying the ratio of factorials. A common mistake is writing $\frac{n!}{(n+1)!}=\frac{1}{n+1}$, which is correct, but fumbling more complex cases like $\frac{(n-1)!}{(n+1)!}=\frac{1}{(n+1)n}$.

• Incorrectly Setting Up the Ratio: Swapping the positions of $a_n$ and $a_{n+1}$ in the fraction, leading to the reciprocal of the correct limit.

• Forgetting the Absolute Value: Forgetting to apply the absolute value when the series has negative terms (e.g., a term like $(-1{)}^n$ or $(-5{)}^n$). This can lead to an incorrect limit or a limit that does not exist.

• Misinterpreting $L=1$: Stating that if $L=1$, the series diverges (often by confusion with the n$th Term Test). The correct conclusion is that the test is inconclusive. - **Errors in Limit Evaluation:** Incorrectly finding the limit of the simplified ratio, especially when it involves rational functions of $n where one must compare the degrees of the numerator and denominator.