Working with Geometric Series - AP Calculus BC Study Guide

The Core Idea: Working with Geometric Series

A geometric series is a specific type of infinite series where each term is generated by multiplying the preceding term by a fixed, non-zero constant known as the common ratio. The fundamental concept is to determine the behavior of the sum of these infinite terms. Specifically, we are concerned with whether this infinite sum approaches a single, finite value—a property called convergence—or if it grows without bound or oscillates, in which case it is said to diverge.

The entire analysis of a geometric series hinges on its common ratio, denoted by $r$. The magnitude of this ratio dictates whether the terms shrink towards zero fast enough for their sum to be finite. This topic provides a definitive test and a precise formula for the sum of a convergent geometric series, establishing a foundational tool for the study of infinite series. The primary tasks are to identify a series as geometric, determine its first term $a$ and common ratio $r$, and then apply the rules of convergence to find its sum if it exists.

Key Formulas

The behavior and sum of a geometric series are governed by a set of precise formulas derived directly from its definition. All analysis begins with identifying the series' components.

1. The Standard Form of a Geometric Series

A geometric series is defined by the following summation notation:

$$\sum _{n=0}^{\infty }a{r}^n=a+ar+a{r}^2+a{r}^3+\cdots +a{r}^n+\dots$$

• $a$ is the first term of the series ($a\ne 0$).

• $r$ is the common ratio, the constant value multiplied to get from one term to the next.

2. The Condition for Convergence

A geometric series converges to a finite sum if and only if the absolute value of its common ratio is less than 1.

$$\text{The series converges if }|r|<1$$

3. The Formula for the Sum of a Convergent Geometric Series

If the condition for convergence is met ($|r|<1$), the sum $S$ of the infinite series is given by:

$$S=\frac{a}{1-r}$$

where $a$ is the first term of the series.

4. The Condition for Divergence

A geometric series diverges (its sum is not a finite value) if the absolute value of its common ratio is greater than or equal to 1.

$$\text{The series diverges if }|r|\ge 1$$

Understanding the First Term and the Common Ratio

The formulas for geometric series are powerful, but their correct application depends entirely on accurately identifying the first term, $a$, and the common ratio, $r$. While the standard form ${\sum }_{n=0}^{\infty }a{r}^n$ is a useful template, many series on the AP Exam are presented in slightly different forms.

The common ratio, $r$, is the base that is being raised to the power of the index variable (e.g., $n$). For a series like ${\sum }_{n=0}^{\infty }5{\left(\frac{2}{3}\right)}^n$, the ratio $r$ is clearly $\frac{2}{3}$.

The first term, $a$, is not simply the coefficient in front of the ratio. The term $a$ is always the value of the first term that is actually being summed in the series. To find $a$, you must substitute the starting value of the index into the expression for the terms of the series.

• For a series starting at $n=0$, like ${\sum }_{n=0}^{\infty }5{\left(\frac{2}{3}\right)}^n$, the first term is found by plugging in $n=0$: $a=5{\left(\frac{2}{3}\right)}^0=5(1)=5$. In this case, $a$ matches the coefficient.

• For a series starting at a different index, like ${\sum }_{n=2}^{\infty }5{\left(\frac{2}{3}\right)}^n$, the first term is found by plugging in the starting index $n=2$: $a=5{\left(\frac{2}{3}\right)}^2=5\left(\frac{4}{9}\right)=\frac{20}{9}$.

Failing to correctly identify the first term $a$ based on the series' starting index is a frequent source of error. Always find $r$ first to check for convergence, and then find $a$ by evaluating the first term before applying the sum formula.

Core Concepts & Rules

• A series is geometric if it can be written in the form $\sum a{r}^n$, where $a$ is the first term and $r$ is the common ratio.

• The convergence or divergence of a geometric series is determined solely by its common ratio $r$.

• A geometric series converges if and only if the absolute value of its common ratio is strictly less than 1 (i.e., $|r|<1$).

• If a geometric series converges, its sum can be calculated using the formula $S=\frac{a}{1-r}$, where $a$ is the first term of the series.

• A geometric series diverges if the absolute value of its common ratio is greater than or equal to 1 (i.e., $|r|\ge 1$). A divergent geometric series does not have a finite sum.

Step-by-Step Example 1: Basic Application

Problem: Determine whether the series ${\sum }_{n=0}^{\infty }8{\left(\frac{1}{3}\right)}^n$ converges or diverges. If it converges, find its sum.

Step 1: Identify the series type and its components.

The series is in the form ${\sum }_{n=0}^{\infty }a{r}^n$. This is a geometric series.

• The common ratio $r$ is the base being raised to the power of $n$, so $r=\frac{1}{3}$.

• The series starts at $n=0$. The first term $a$ is found by substituting $n=0$ into the expression:

$$a=8{\left(\frac{1}{3}\right)}^0=8(1)=8$$

Step 2: Check the condition for convergence.

The test for convergence depends on the absolute value of the common ratio, $|r|$.

$$|r|=\left|\frac{1}{3}\right|=\frac{1}{3}$$

Since $|r|=\frac{1}{3}<1$, the series converges.

Step 3: Calculate the sum.

Because the series converges, we can use the sum formula $S=\frac{a}{1-r}$.

$$S=\frac{8}{1-\frac{1}{3}}$$

Simplify the denominator:

$$S=\frac{8}{\frac{2}{3}}$$

Calculate the final sum:

$$S=8\cdot \frac{3}{2}=12$$

Conclusion: The series converges, and its sum is 12.

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

Problem: Find the sum of the series ${\sum }_{n=1}^{\infty }3{\left(-\frac{3}{4}\right)}^{n+1}$.

Step 1: Identify the series type and its common ratio.

The expression has a constant base raised to a power involving the index $n$, so it is a geometric series. To identify $r$ clearly, we can use exponent rules to isolate the $n$ in the exponent.

$$3{\left(-\frac{3}{4}\right)}^{n+1}=3{\left(-\frac{3}{4}\right)}^n{\left(-\frac{3}{4}\right)}^1$$

The common ratio $r$ is the base being raised to the power of $n$.

$$r=-\frac{3}{4}$$

Step 2: Check the condition for convergence.

We evaluate the absolute value of the common ratio.

$$|r|=\left|-\frac{3}{4}\right|=\frac{3}{4}$$

Since $|r|=\frac{3}{4}<1$, the series converges and has a finite sum.

Step 3: Determine the first term, $a$.

The series starts at the index $n=1$. To find the first term $a$, we must substitute $n=1$ into the original expression for the terms of the series.

$$a=3{\left(-\frac{3}{4}\right)}^{1+1}=3{\left(-\frac{3}{4}\right)}^2$$

$$a=3\left(\frac{9}{16}\right)=\frac{27}{16}$$

This is the value of $a$ we must use in the sum formula.

Step 4: Calculate the sum.

Using the sum formula $S=\frac{a}{1-r}$ with our calculated $a$ and $r$:

$$S=\frac{\frac{27}{16}}{1-\left(-\frac{3}{4}\right)}$$

Simplify the denominator:

$$S=\frac{\frac{27}{16}}{1+\frac{3}{4}}=\frac{\frac{27}{16}}{\frac{4}{4}+\frac{3}{4}}=\frac{\frac{27}{16}}{\frac{7}{4}}$$

Calculate the final sum by multiplying by the reciprocal of the denominator:

$$S=\frac{27}{16}\cdot \frac{4}{7}=\frac{27}{4\cdot 7}=\frac{27}{28}$$

Conclusion: The series converges, and its sum is $\frac{27}{28}$.

Using Your Calculator

The determination of convergence and the calculation of the sum for a geometric series are purely analytical processes based on the formulas for $|r|$ and $S=\frac{a}{1-r}$. A calculator is not used to determine convergence or find the exact sum.

However, a graphing calculator can be a useful tool for verifying your answer. You can approximate the sum of a convergent infinite series by calculating a partial sum with a large number of terms.

Example: To verify the result of Example 1, $S=12$ for ${\sum }_{n=0}^{\infty }8{\left(\frac{1}{3}\right)}^n$.

On a TI-84 style calculator:

1. Press MATH and find the $summation$ function (often 0:summation Σ(` or found in the `MATH` menu). 2. Enter the expression to match the series notation. You will set a large upper limit to approximate the infinite sum. For example, summing to $n=50 is usually sufficient for a good approximation.

   • The input would look like: $sum(8\ast (1/3{)}^X,X,0,50)$

   • This command calculates ${\sum }_{X=0}^{50}8{\left(\frac{1}{3}\right)}^X$.

2. Press ENTER. The calculator will return a value extremely close to 12. This numerical result supports your analytical conclusion that the sum is exactly 12.

This method does not prove the sum is 12, but it provides strong evidence that your analytical calculation is correct. For a divergent series, this process would yield a very large number (or an error), supporting the conclusion of divergence.

AP Exam Quick Hit

Common Question Types

• Direct Convergence and Sum: You will be given a geometric series in standard form and asked to determine if it converges and, if so, to find its sum.

  • Example: Find the sum of the series $18-12+8-\frac{16}{3}+\dots$. (Here you must first find $r=-12/18=-2/3$ and $a=18$).

• Series with an Altered Index or Form: You will be given a geometric series that does not start at $n=0$ or requires algebraic manipulation to identify its components.

  • Example: Does the series ${\sum }_{n=2}^{\infty }\frac{4^n}{5^{n-1}}$ converge or diverge? If it converges, what is its sum? (You must rewrite the term as $5\cdot {\left(\frac{4}{5}\right)}^n$ to identify $r=4/5$ and then find the first term by plugging in $n=2$).

• Finding the Interval of Convergence: You will be given a geometric series where the common ratio is an expression involving a variable, and you must find the values of the variable for which the series converges.

  • Example: For what values of $x$ does the series ${\sum }_{n=0}^{\infty }{\left(\frac{x-1}{2}\right)}^n$ converge? (This requires solving the inequality $|r|<1$, which becomes $|\frac{x-1}{2}|<1$).

Common Mistakes

• Incorrectly Identifying $a$: Automatically using the coefficient as the first term $a$ without checking the starting index of the summation. Remember to always find $a$ by plugging the starting value of $n$ into the series expression.

• Ignoring the Absolute Value in the Convergence Condition: Checking if $r<1$ instead of $|r|<1$. A series with a ratio of $r=-2$ diverges because $|-2|=2\ge 1$, even though $-2<1$.

• Arithmetic Errors with the Sum Formula: Making simple calculation mistakes with fractions or negative signs in the formula $S=\frac{a}{1-r}$. For example, if $r=-1/2$, incorrectly calculating the denominator as $1-1/2$ instead of the correct $1-(-1/2)=3/2$.

• Attempting to Sum a Divergent Series: Forgetting to check for convergence first. If you are asked for the sum of a series, your first step should always be to find $r$ and check if $|r|<1$. If it is not, the series diverges and has no sum.