Problem 41 - Algebra 2

Question

Find the sum of each infinite geometric series, if it exists. $$ \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots $$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{1}{6} \).

Step 1: Identify the First Term

The first term of the geometric series is represented by \( a \). Here, the first term \( a = \frac{1}{8} \).

Step 2: Determine the Common Ratio

The common ratio \( r \) can be found by dividing the second term by the first term of the series. So, \( r = \frac{1}{32} \div \frac{1}{8} = \frac{1}{4} \).

Step 3: Verify the Condition for Convergence

For the sum of an infinite geometric series to exist, the absolute value of the common ratio \( |r| \) must be less than 1. Here, \( |r| = |\frac{1}{4}| = \frac{1}{4} < 1 \), so the series converges, and a sum exists.

Step 4: Apply the Formula for Sum of Infinite Series

The sum \( S \) of an infinite geometric series is given by the formula \( S = \frac{a}{1-r} \). Substitute \( a = \frac{1}{8} \) and \( r = \frac{1}{4} \) into the formula: \( S = \frac{\frac{1}{8}}{1-\frac{1}{4}} = \frac{\frac{1}{8}}{\frac{3}{4}} \).

Step 5: Simplify the Result

Calculate \( \frac{\frac{1}{8}}{\frac{3}{4}} \) by flipping the divisor and multiplying: \( \frac{1}{8} \times \frac{4}{3} = \frac{1 \times 4}{8 \times 3} = \frac{4}{24} = \frac{1}{6} \). Thus, the sum of the series is \( \frac{1}{6} \).

Key Concepts

Common Ratio

The common ratio is a crucial concept in understanding geometric series, including infinite ones. It is the factor by which we multiply each term to get the next term in the series. To find the common ratio in a series, divide any term except the first by the term immediately before it.
In our provided example of the series \( \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots \), we obtain the common ratio \( r \) by dividing the second term \( \frac{1}{32} \) by the first term \( \frac{1}{8} \). Thus, we have:

• \( r = \frac{1}{32} \div \frac{1}{8} = \frac{1}{4} \).

This value, \( \frac{1}{4} \), is our series' common ratio. Knowing the common ratio helps us understand how quickly the terms in a series decrease or increase and is essential when determining if a series converges.

Convergence of Series

When working with infinite geometric series, one of the primary concerns is whether the series converges, meaning it approaches a finite sum. For this to happen, the common ratio \( |r| \) must be less than 1. This tells us that the terms are getting smaller and smaller as we proceed further in the series.
In our example, the common ratio \( r \) is \( \frac{1}{4} \). Calculate the absolute value to check for convergence:

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

Since this condition is satisfied, the series converges, meaning it will have a finite sum. This criterion helps us quickly determine which types of infinite geometric series can be summed.

Sum of Series

Once we establish that an infinite geometric series converges, calculating the sum is straightforward using a specific formula. The formula states:
\[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. This formula only applies when the series converges, that is, when \( |r| < 1 \).
For the series \( \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots \), we have already identified \( a = \frac{1}{8} \) and \( r = \frac{1}{4} \). Plug these into the formula:

• \( S = \frac{\frac{1}{8}}{1 - \frac{1}{4}} = \frac{\frac{1}{8}}{\frac{3}{4}} \)

To get the numerical result, carry out the division: flip the denominator (the number under 1) and multiply:

• \( S = \frac{1}{8} \times \frac{4}{3} = \frac{4}{24} = \frac{1}{6} \)

Therefore, the sum of the series is \( \frac{1}{6} \), showcasing how infinite repeated additions can converge into a finite number.