Problem 24 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{(-2)^{n}}{3^{n}} $$

Step-by-Step Solution

Answer

The series converges because the common ratio \( |-2/3| < 1 \).

Step 1: Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{(-2)^n}{3^n} \). This series can be rewritten as \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \). It resembles a geometric series where the general term is \( ar^n \). Here, \( a = -\frac{2}{3} \) and \( r = -\frac{2}{3} \).

Step 2: Determine the Convergence Criteria for Geometric Series

A geometric series \( \sum_{n=1}^{\infty} ar^n \) converges if the absolute value of the common ratio \( |r| < 1 \). In our series, \( r = -\frac{2}{3} \), so we need to check if \( |-2/3| < 1 \).

Step 3: Evaluate the Convergence Criterion

Calculate the absolute value of the common ratio: \( |-2/3| = 2/3 \). Since \( 2/3 < 1 \), the convergence criterion is satisfied.

Step 4: Conclude the Convergence of the Series

Since \( |-2/3| < 1 \), the series \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \) converges by the geometric series test.

Key Concepts

Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous one by a constant, known as the common ratio. The general form of a geometric series is \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.
In the provided exercise, the series \( \sum_{n=1}^{\infty} \frac{(-2)^n}{3^n} \) can be rewritten as \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \). Here, the first term \( a \) is \(-\frac{2}{3} \), and it's progressive based on multiplying by \( r = -\frac{2}{3}\) multiple times.

Convergence Criteria

Convergence criteria for a series determine whether the series adds up to a finite value or not. For geometric series, convergence is straightforward. The geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio \( |r| < 1 \).
This rule serves as a quick and efficient way to check if a series converges or diverges. In the given problem, the series converges because the absolute value of the common ratio \( |\frac{-2}{3}| = \frac{2}{3} \) is less than 1.

Absolute Value

The absolute value of a number is a way of describing its distance from zero on the number line, without considering its direction (positive or negative).
In the context of evaluating the convergence of a series, especially a geometric series, absolute value plays a critical role. It ensures that even if the terms have alternating signs, what ultimately matters for convergence is the magnitude or size of the common ratio.
In our geometric series, the common ratio is \( -\frac{2}{3} \). The absolute value \( |-\frac{2}{3}| = \frac{2}{3} \) focuses purely on its size, confirming the conditions for convergence.

Common Ratio

The common ratio in a geometric series is the factor by which we multiply each term to get the next term. This ratio is pivotal in understanding how the series behaves.
For the series \( \sum_{n=1}^{\infty} \left( \frac{-2}{3} \right)^n \), the common ratio \( r = -\frac{2}{3} \) dictates the relationship between successive terms.

• Each term is obtained by multiplying the previous term by \(-\frac{2}{3}\).
• Since the common ratio \(-\frac{2}{3}\) has an absolute value less than 1, it ensures that the terms of the series get smaller in magnitude as \( n \) increases, leading the series to converge.

Understanding the common ratio helps us predict the behavior of the series as it progresses.