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Chapter 11: Problem 50

Determine whether each infinite geometric series has a limit.If a limit exists, find it. $$-6+3-\frac{3}{2}+\frac{3}{4}-\cdots$$

Short Answer

Expert verified
-4

Step by step solution

01

Identify the first term and common ratio

The first term of the series is the first number: a = -6The common ratio can be found by dividing the second term by the first term: \( r = \frac{3}{-6} = -\frac{1}{2} \)
02

Check the common ratio鈥檚 absolute value

For an infinite geometric series to have a limit, the absolute value of the common ratio must be less than 1: \( |r| = | -\frac{1}{2} | = \frac{1}{2} < 1 \)Since \( |r| < 1 \), the series has a limit.
03

Calculate the limit

To find the limit of the geometric series, use the formula: \[ S = \frac{a}{1-r} \]Substitute the values we know: \[ S = \frac{-6}{1-(-\frac{1}{2})} = \frac{-6}{1 + \frac{1}{2}} = \frac{-6}{\frac{3}{2}} = -6 \times \frac{2}{3} = -4 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio (r). An example is the series -6, 3, -1.5, 0.75, ..., where each term is multiplied by -1/2 to get the next. Understanding geometric series is crucial because they appear frequently in different areas of mathematics and applied sciences. Recognizing if you're dealing with a geometric series helps in knowing which formulas and techniques to apply. To sum up:
  • The first term of the series is often denoted by 'a'.
  • The common ratio between terms is denoted by 'r'.
Common Ratio
The common ratio (r) in a geometric series is the factor by which we multiply one term to get the next. To find it, divide any term by the previous term. For instance, in the series -6, 3, -1.5, 0.75, ..., we get:
  • The first term (a) is -6.
  • The second term is 3, so r = 3 / -6 = -1/2.
The common ratio can be positive or negative. The pattern of the numbers changes depending on whether 'r' is positive or negative. Additionally, the magnitude of 'r' (its absolute value) will influence whether the series converges to a limit or diverges (more on that in the next sections).
Limit of a Series
The limit of a series is the value that the sum of the terms approaches as more and more terms are added. Not all series have limits. For a geometric series, the limit exists only if the absolute value of the common ratio is less than 1. Mathematically, we check if \( |r| < 1 \). If this condition holds, the series will converge to a limit.

Otherwise, if \( |r| \) is 1 or greater, the series does not have a finite limit and instead diverges (grows infinitely large or does not settle to a single value). For example, in the series -6, 3, -1.5, 0.75, ..., \( |r| = |-1/2| = 1/2 < 1 \) so this series converges.
Infinite Series
An infinite series sums an infinite number of terms together. In a geometric infinite series, each term is determined by multiplying the previous term by the common ratio. It's essential to determine if the series has a limit to understand whether the sum converges to a specific value.

For our series -6, 3, -1.5, 0.75, ..., because \( |r| < 1 \), the series is convergent. This means that as you add more and more terms, the sum gets closer and closer to a certain number. In this case, we can calculate this limit using a specific formula, which we will discuss next.
Sum of Series
To find the sum of an infinite geometric series that converges, we use the formula: \[ S = \frac{a}{1-r} \]

Here, 'a' is the first term of the series, and 'r' is the common ratio. If \( |r| < 1 \), this formula gives us the sum to which the series converges. For the series -6, 3, -1.5, 0.75, ..., we've found \( a = -6 \) and \( r = -1/2 \). Using the formula:

\[ S = \frac{-6}{1 - (-1/2)} = \frac{-6}{1 + 1/2} = \frac{-6}{3/2} = -6 \times \frac{2}{3} = -4 \]

So, the sum, or limit, of the infinite geometric series is -4. Thus, as we add more terms of this series, the total will approach -4.

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Most popular questions from this chapter

use the formula for \(S_{n}\) to find the indicated sum for each geometric series. $$S_{7} \text { for } \frac{1}{18}-\frac{1}{6}+\frac{1}{2}-\cdots$$

Write out and evaluate each sum. $$ \sum_{k=2}^{6} \sqrt{5 k-1} $$

Perform the indicated operation and, if possible, simplify. $$ \frac{t}{t-1}-\frac{t}{1-t} $$

Rewrite each sum using sigma notation. Answers may vary. $$ \frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7} $$

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