Problem 18 - Thomas Calculus

Question

In Exercises \(15-18,\) 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 with a sum of \( \frac{4}{15} \).

Step 1: Identify the First Term and Common Ratio

To analyze the geometric series, first identify the first term (\( a \)) and the common ratio (\( r \)). In this series, the first term is given by \( a = \left(\frac{-2}{3}\right)^2 \) and the common ratio \( r \) is also \( \frac{-2}{3} \). Thus, \( a = \frac{4}{9} \) and \( r = \frac{-2}{3} \).

Step 2: Check the Condition for Convergence

For a geometric series to converge, \( |r| < 1 \) must be satisfied. Here, \( r = \frac{-2}{3} \), and \( |\frac{-2}{3}| = \frac{2}{3} < 1 \). Therefore, the series converges.

Step 3: Use the Formula to Find the Sum

Since the series converges, use the sum formula for an infinite geometric series: \[ S = \frac{a}{1 - r} \]. Substituting the values, \( a = \frac{4}{9} \) and \( r = \frac{-2}{3} \), we have: \[ S = \frac{\frac{4}{9}}{1 - \left(-\frac{2}{3}\right)} = \frac{\frac{4}{9}}{\frac{5}{3}} = \frac{4}{9} \times \frac{3}{5} = \frac{4 \times 3}{9 \times 5} = \frac{12}{45} = \frac{4}{15} \].

Step 4: Conclusion

The series \( \left(\frac{-2}{3}\right)^{2} + \left(\frac{-2}{3}\right)^{3} + \left(\frac{-2}{3}\right)^{4} + \cdots \) is a convergent geometric series with the sum \( \frac{4}{15} \).

Key Concepts

Convergence of Series

Understanding when a series converges is crucial in mathematics, especially for geometric series. A series is a sum of terms of a sequence. Convergence means that as we sum more and more terms of the series, the value approaches a particular finite value. For geometric series, this behavior is all about the common ratio, denoted as \( r \). The series will converge only if the absolute value of \( r \) is less than 1, i.e., \( |r| < 1 \). This is because if \( |r| \) is 1 or more, the terms of the series do not get smaller with each step and instead grow or oscillate indefinitely.When the series converges, it becomes possible to find its sum, which is quite useful in various applications, from engineering to natural sciences. The sum formula, therefore, relies on this property of \( r \), ensuring that the infinite terms can still add up to a meaningful number.

Infinite Series

Infinite series are sequences where we add up terms forever and ever. Although it might seem impossible to find a sum without ever stopping, under certain conditions, it’s completely doable. This is especially true for geometric series that meet the criteria for convergence.An infinite series is represented as \( a + ar + ar^2 + ar^3 + \cdots \), with \( a \) being the first term and \( r \) the common ratio. When \( |r| < 1 \), the series converges, and the terms start getting closer to zero.Even if the series has an infinite number of terms, the sum remains fixed at a certain value if it converges. This fascinating property helps mathematicians and scientists make predictions and solve complex problems efficiently, transforming an endless sum into a single, finite number.

Sum of Series

Knowing whether a geometric series converges allows us to use a simple formula to find its sum. For an infinite geometric series, the formula to calculate the sum, when \(|r| < 1\), is: \[S = \frac{a}{1 - r}\]Here, \( S \) represents the sum, \( a \) is the first term, and \( r \) is the common ratio. The beauty of this formula is its simplicity despite the series having infinitely many terms. As each subsequent term's contribution diminishes, the formula gives us a clear and concise result.For example, in the series with \( a = \frac{4}{9} \) and \( r = \frac{-2}{3} \), the convergence condition \(|r| < 1\) holds, allowing us to compute the sum using the formula. Substituting the values in, we find that the sum is \( \frac{4}{15} \).This direct approach simplifies problem-solving, making it easier to evaluate complex series in mathematics and its numerous applications.