Problem 18 - Thomas Calculus

Question

In Exercises \(15-22\) , determine if the geometric series converges or diverges. If a series converges, find its sum. $$\left(\frac{-2}{3}\right)^{2}+\left(\frac{-2}{3}\right)^{3}+\left(\frac{-2}{3}\right)^{4}+\left(\frac{-2}{3}\right)^{5}+\left(\frac{-2}{3}\right)^{6}+\cdots$$

Step-by-Step Solution

Answer

The series converges, and its sum is \(\frac{4}{15}\).

Step 1: Identify the First Term and Common Ratio

A geometric series is defined by its first term and common ratio. From the given series \( \left(\frac{-2}{3}\right)^2 + \left(\frac{-2}{3}\right)^3 + \left(\frac{-2}{3}\right)^4 + \cdots \), the first term \(a\) can be identified as \( a = \left(\frac{-2}{3}\right)^2 \) and the common ratio \(r\) as \( r = \frac{-2}{3} \), since the series goes from one term to the next by multiplying by \( \frac{-2}{3} \).

Step 2: Evaluate Convergence of the Series

A geometric series converges if the absolute value of the common ratio is less than 1, that is, \(|r| < 1\). Here, \(r = \frac{-2}{3}\), and \(|r| = \frac{2}{3}\). Because \(\frac{2}{3} < 1\), the series converges.

Step 3: Find the Sum of the Converging Series

The sum \(S\) of an infinite converging geometric series is given by the formula: \[ S = \frac{a}{1 - r} \] where \(a = \left(\frac{-2}{3}\right)^2 = \frac{4}{9}\) and \(r = \frac{-2}{3}\). Thus, the sum is: \[ S = \frac{\frac{4}{9}}{1 - \left(\frac{-2}{3}\right)} = \frac{\frac{4}{9}}{\frac{5}{3}} = \frac{4}{9} \times \frac{3}{5} = \frac{4 \times 3}{9 \times 5} = \frac{12}{45} = \frac{4}{15} \] Simplifying \(\frac{12}{45}\) gives \(\frac{4}{15}\).

Key Concepts

Convergence

Convergence is an important concept when dealing with infinite series. It tells us whether the sum of the series approaches a finite limit as the number of terms goes to infinity. For geometric series, convergence depends on the common ratio, denoted as \( r \). The rule of thumb is that a geometric series will converge if the absolute value of its common ratio is less than 1, that is, if \( |r| < 1 \).
This means the values of the terms in the series keep getting smaller and smaller, eventually becoming negligibly small.
In our exercise, the common ratio is \( \frac{-2}{3} \), which has an absolute value of \( \frac{2}{3} \). Since \( \frac{2}{3} < 1 \), our series converges.
This convergence allows us to find the sum of the infinite series.

Infinite Series

An infinite series is simply a sum of an infinite sequence of numbers. In mathematics, infinite series are denoted with an ellipsis (...), showing that the sequence continues indefinitely.
Geometric series are a kind of infinite series that grow or shrink by a constant factor, the common ratio.
The idea of adding up an infinite number of terms might sound perplexing, but with convergence, it is possible to end up with a finite sum for an infinite series.

• It's crucial to determine convergence when working with infinite series to ensure the sum is not infinite.
• Only convergent series allow for calculation of a sum.

This process is central to the exercises you see in many math courses because it combines understanding of sequences, limits, and convergence.

Common Ratio

The common ratio is a defining characteristic of a geometric series. It is the factor by which each term in the series is multiplied to get the next term.
In other words, if you know one term of the series and the common ratio, you can predict all subsequent terms.
In our geometric series, the expression is continuously multiplied by \( \frac{-2}{3} \). This means, for any term \( a_n \), the next term \( a_{n+1} \) is found by \( a_n \times \left( \frac{-2}{3} \right) \).
The common ratio's absolute value determines if the infinite series will converge. For \( |r| < 1 \), the series will converge, while a series with \( |r| \geq 1 \) will not. Thus, the common ratio is the key to predicting the series' behavior.

Sum of Series

When a geometric series converges, it is possible to find its sum using a specific formula. For a series with the first term \( a \) and a common ratio \( r \), the sum \( S \) of the series is given by the formula:\[S = \frac{a}{1 - r}\]
This formula works because as terms get progressively smaller, their contribution to the sum becomes negligible, allowing the series to approach a finite value.
In the exercise given, where \( a = \left( \frac{-2}{3} \right)^2 = \frac{4}{9} \) and \( r = \frac{-2}{3} \), you insert these into the sum formula:
\[S = \frac{\frac{4}{9}}{1 - \left( \frac{-2}{3} \right)} = \frac{4}{15}\]
The final result, \( \frac{4}{15} \), means that even though you are adding an infinite number of terms, the series sums up to this specific value. Understanding this process provides insight into the magic of geometric series in mathematics.