Problem 24 - Algebra 2

Question

Find the sum of each infinite geometric series, if it exists. $$\sum_{n=1}^{\infty}\left(\frac{3}{8}\right)\left(\frac{3}{4}\right)^{n-1}$$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{3}{2} \).

Step 1: Identify the First Term and Common Ratio

The given series is \( \sum_{n=1}^{\infty}\left(\frac{3}{8}\right)\left(\frac{3}{4}\right)^{n-1} \). The first term \( a \) is what we get when \( n = 1 \). Thus, \( a = \frac{3}{8} \cdot \left(\frac{3}{4}\right)^{0} = \frac{3}{8} \). The common ratio \( r \) is the constant factor between consecutive terms which is \( \frac{3}{4} \).

Step 2: Check the Conditions for Convergence

For an infinite geometric series \( \sum a r^{n-1} \) to converge, the common ratio \( r \) must satisfy \( |r| < 1 \). Here, \( r = \frac{3}{4} \), and since \( \left| \frac{3}{4} \right| < 1 \), the series converges.

Step 3: Apply the Formula for Sum of Infinite Geometric Series

The sum \( S \) of an infinite geometric series is given by the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Substituting \( a = \frac{3}{8} \) and \( r = \frac{3}{4} \) into the formula gives \( S = \frac{\frac{3}{8}}{1 - \frac{3}{4}} = \frac{\frac{3}{8}}{\frac{1}{4}} \).

Step 4: Simplify the Expression for the Sum

The expression \( \frac{\frac{3}{8}}{\frac{1}{4}} \) simplifies by multiplying both the numerator and the denominator by 4 (the reciprocal of \( \frac{1}{4} \)). Thus, \( \frac{3}{8} \times 4 = \frac{3 \times 4}{8} = \frac{12}{8} = \frac{3}{2} \).

Key Concepts

First Term

In the context of infinite geometric series, the first term, often denoted by \( a \), is the initial value of the series when the index \( n \) is equal to 1. It's essentially where the series starts as it builds consecutive terms. For the given series \( \sum_{n=1}^{\infty}\left(\frac{3}{8}\right)\left(\frac{3}{4}\right)^{n-1} \), the first term \( a \) can be determined by substituting \( n = 1 \). Hence, we find:

• First Term \( a = \frac{3}{8} \cdot \left(\frac{3}{4}\right)^{0} \)
• Since any number raised to the power of zero is 1, it simplifies to: \( a = \frac{3}{8} \)

This illustrates that, even if the series extends into infinity, it begins with a very distinct starting point, which is crucial for further calculations.

Common Ratio

The common ratio, represented as \( r \), is a crucial part of understanding geometric series. It describes how each term in the series relates to the previous one. In geometric series, each term is a product of the previous term and the common ratio. Identifying \( r \) helps in determining the nature of the series and whether it will converge or diverge.For the series \( \sum_{n=1}^{\infty}\left(\frac{3}{8}\right)\left(\frac{3}{4}\right)^{n-1} \), the common ratio \( r \) is given directly as:

• Common Ratio \( r = \frac{3}{4} \)

If \( r \) is a fraction between -1 and 1 (\( |r|<1 \)), it indicates that the series terms are progressively shrinking, an important aspect that leads to a convergent series.

Convergence Conditions

Convergence is a key concept when working with infinite series, particularly geometric ones. A series converges if it approaches a finite sum as more and more terms are added. For an infinite geometric series \( \sum a r^{n-1} \) to be convergent, the magnitude of the common ratio must be less than one, noted as \( |r|<1 \).In our specific example, where \( r = \frac{3}{4} \), we check this condition:

• \( |r| = |\frac{3}{4}| = \frac{3}{4} < 1 \)

The condition is satisfied, confirming that our series will converge. This means as you sum the infinite terms of the series, they will approach a specific finite value, which is essential when calculating the total or sum of the series.

Sum of Infinite Series

Finding the sum of an infinite geometric series is made possible through a simple yet powerful formula. Once a series is confirmed to converge, the sum \( S \) can be calculated. The formula used is:\[S = \frac{a}{1 - r}\]Here, \( a \) is the first term and \( r \) is the common ratio. This formula assumes the series starts with the term \( a \) and continues infinitely as the common ratio \( r \) is applied.For our series:

• Substitute \( a = \frac{3}{8} \) and \( r = \frac{3}{4} \) into the formula
• \( S = \frac{\frac{3}{8}}{1 - \frac{3}{4}} = \frac{\frac{3}{8}}{\frac{1}{4}} \)
• Simplifying this yields the sum: \( S = \frac{3}{2} \)

This demonstrates that, despite having an infinite number of terms, the sum is a finite value due to convergent nature of the series.