Chapter 9: Problem 24
Use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{(-2)^{n}}{3^{n}}$$
Short Answer
Expert verified
The series converges because it is a geometric series with \(|r| = \frac{2}{3} < 1\).
Step by step solution
01
Identify the Given Series
The given series is \( \sum_{n=1}^{\infty} \frac{(-2)^{n}}{3^{n}} \). This is an infinite series where the general term \( a_n = \frac{(-2)^{n}}{3^{n}} \).
02
Notice the Structure of the Series
Recognize that the general term \( a_n = \left( \frac{-2}{3} \right)^n \) is in the form of a geometric series, where \( r = \frac{-2}{3} \).
03
Determine the Type of Series
Because the series is geometric, we use properties of geometric series. A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if \( |r| < 1 \) and diverges if \( |r| \geq 1 \).
04
Check the Convergence Criteria
Calculate \( |r| = \left| \frac{-2}{3} \right| = \frac{2}{3} \). Since \( \frac{2}{3} < 1 \), the series converges.
05
Conclude Based on the Tests
Since \( |r| = \frac{2}{3} < 1 \), the series \( \sum_{n=1}^{\infty} \frac{(-2)^{n}}{3^{n}} \) is convergent by the geometric series test.
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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 special type of infinite series where each term after the first is found by multiplying the previous one by a constant. This constant is known as the common ratio, denoted as \( r \). Geometric series can be written in the form:
Geometric series have interesting properties. If \( |r| < 1 \), the series \( \sum_{n=0}^{\infty} ar^n \) converges to a sum \( \frac{a}{1-r} \). If \( |r| \geq 1 \), the series diverges. Hence, the behavior of the series is determined mostly by the absolute value of the common ratio \( r \). This makes determining convergence straightforward in such series.
- \( a, ar, ar^2, ar^3, ... \)
Geometric series have interesting properties. If \( |r| < 1 \), the series \( \sum_{n=0}^{\infty} ar^n \) converges to a sum \( \frac{a}{1-r} \). If \( |r| \geq 1 \), the series diverges. Hence, the behavior of the series is determined mostly by the absolute value of the common ratio \( r \). This makes determining convergence straightforward in such series.
Infinite Series
An infinite series is a sum involving an infinite number of terms. It is written as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the general term of the series. Unlike a finite series that has a limited number of terms, an infinite series extends infinitely. The challenge with infinite series is deciding whether their sum reaches a finite limit or not as more terms are added.
In the context of math, two primary behaviors are associated with infinite series:
In the context of math, two primary behaviors are associated with infinite series:
- Convergence: An infinite series converges if the sum of its terms approaches a specific finite value as more and more terms are added.
- Divergence: If the sum does not approach a specific finite limit, the series is said to diverge.
Convergence Criteria
Convergence criteria are rules and tests used to determine if an infinite series converges. For a series to converge, its terms must decrease in such a manner that their sum reaches a specific value as the number of terms increases.
Geometric series have straightforward convergence criteria. Specifically, a geometric series \( \sum_{n=0}^{\infty} ar^n \) converges under the condition that \( |r| < 1 \). This is because, intuitively, the terms of the series get smaller and smaller, ensuring that their total forms a finite value. On the other hand, if \( |r| \geq 1 \), the terms either stay large or oscillate, meaning their total does not settle on a particular value, leading to divergence.
In the exercise example, the series \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \) is assessed using the geometric series test. We detect the common ratio \( r = \frac{-2}{3} \) and determine its absolute value \( |r| = \frac{2}{3} \). Since \( \frac{2}{3} < 1 \), the series is convergent based on this convergence criterion. Understanding and applying the right test provides a structured way to analyze the behavior of infinite series.
Geometric series have straightforward convergence criteria. Specifically, a geometric series \( \sum_{n=0}^{\infty} ar^n \) converges under the condition that \( |r| < 1 \). This is because, intuitively, the terms of the series get smaller and smaller, ensuring that their total forms a finite value. On the other hand, if \( |r| \geq 1 \), the terms either stay large or oscillate, meaning their total does not settle on a particular value, leading to divergence.
In the exercise example, the series \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \) is assessed using the geometric series test. We detect the common ratio \( r = \frac{-2}{3} \) and determine its absolute value \( |r| = \frac{2}{3} \). Since \( \frac{2}{3} < 1 \), the series is convergent based on this convergence criterion. Understanding and applying the right test provides a structured way to analyze the behavior of infinite series.