Problem 51 - Precalculus: Mathematical for Calculus

Question

Find the sum of the infinite geometric series. $$\frac{1}{3^{6}}+\frac{1}{3^{8}}+\frac{1}{3^{10}}+\frac{1}{3^{12}}+\cdots$$

Step-by-Step Solution

Answer

The sum of the infinite series is \( \frac{1}{648} \).

Step 1: Identify the First Term

The first term \( a \) of the series is \( \frac{1}{3^6} \).

Step 2: Identify the Common Ratio

To identify the common ratio \( r \), divide the second term by the first term. The second term is \( \frac{1}{3^8} \) and the first term is \( \frac{1}{3^6} \). Thus, \( r = \frac{1}{3^8} \div \frac{1}{3^6} = \frac{1}{3^2} = \frac{1}{9} \).

Step 3: Determine if the Series Converges

A geometric series converges if the absolute value of the common ratio \( |r| \) is less than 1. Here, \( r = \frac{1}{9} \) and \( |r| = \frac{1}{9} < 1 \), so the series converges.

Step 4: Apply the Infinite Geometric Series Sum Formula

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. Substitute \( a = \frac{1}{3^6} \) and \( r = \frac{1}{9} \) into the formula: \[ S = \frac{\frac{1}{3^6}}{1 - \frac{1}{9}} = \frac{\frac{1}{3^6}}{\frac{8}{9}} = \frac{1}{3^6} \times \frac{9}{8} \]}},{

Step 5: Simplify the Expression for the Sum

Simplify the expression from Step 4: \( S = \frac{9}{8 \times 3^6} = \frac{9}{8 \times 729} \). Calculate \( 8 \times 729 = 5832 \), so \( S = \frac{9}{5832} = \frac{1}{648} \).

Key Concepts

Common Ratio

In the context of an infinite geometric series, the common ratio is the factor by which we multiply each term to get the next term. It is denoted by \( r \). To find it, we divide any term in the sequence by the previous term. This step is essential as it helps determine whether the series is convergent.

In our series, we had terms like \( \frac{1}{3^6} \) and \( \frac{1}{3^8} \). By dividing the second term by the first term, you get the common ratio \( r = \frac{1}{9} \). This ratio tells us about the shrinking size of each subsequent term in the series.

A common ratio with an absolute value less than 1 indicates convergence, while an absolute value equal to or greater than 1 indicates divergence.

Geometric Sequence

A geometric sequence is a list of numbers where each term after the first is found by multiplying the previous term by a non-zero constant known as the common ratio. In simple terms, it's a number pattern created through consistent multiplication.

In our exercise, the sequence starts with the initial term \( \frac{1}{3^6} \), followed by \( \frac{1}{3^8} \), and so on. Here, each term is obtained by multiplying the previous one by \( \frac{1}{9} \). The terms steadily become smaller, illustrating how the pattern unfolds in a precise, predictable manner.

Series Convergence

Series convergence is a key concept in understanding infinite geometric series. We say a series converges if the sum of its terms approaches a definite value as more terms are added.

For a geometric series to converge, the common ratio must have an absolute value less than 1 (i.e., \( |r| < 1 \)). In our example, the common ratio \( \frac{1}{9} \) clearly satisfies this condition, indicating that the series will converge. Through convergence, the series allows us to calculate a finite sum, demonstrating the remarkable property of infinite series that yield tangible results.

Sum Formula

The sum formula for an infinite geometric series offers a way to find the sum of all terms when the series converges. The formula is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio.

In our particular series, with the first term \( a = \frac{1}{3^6} \) and the common ratio \( r = \frac{1}{9} \), we substitute these values into the sum formula:

• Calculate the denominator: \( 1 - r = 1 - \frac{1}{9} = \frac{8}{9} \).
• Use the formula: \( S = \frac{\frac{1}{3^6}}{\frac{8}{9}} = \frac{1}{3^6} \times \frac{9}{8} \).
• Simplify further: \( S = \frac{9}{8 \times 729} = \frac{9}{5832} = \frac{1}{648} \).

This computation shows that even for an infinite set of terms, the overall sum can settle at a precise value.