Problem 56 - Algebra 2

Question

What is the sum of the infinite geometric series \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\ldots ?\) F. \(\frac{2}{3}\) G. 1 H. 1\(\frac{1}{3}\) J. 1\(\frac{2}{3}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the series

The given series is \( \frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots \), which is a geometric series.

Step 2: Find the first term

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

Step 3: Determine the common ratio

The common ratio \( r \) is found by dividing the second term by the first term: \( \frac{1}{6} \div \frac{1}{3} = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2} \). So, the common ratio \( r = \frac{1}{2} \).

Step 4: Use the infinite geometric series formula

The formula for the sum of an infinite geometric series \( S \) is \( S = \frac{a}{1-r} \), where \( |r| < 1 \).

Step 5: Calculate the sum

Plug the values of \( a = \frac{1}{3} \) and \( r = \frac{1}{2} \) into the formula: \[ S = \frac{\frac{1}{3}}{1-\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3} \].

Key Concepts

geometric series formula

An infinite geometric series is a sum of infinitely many terms that form a geometric sequence. Each term in this sequence is found by multiplying the previous term by a constant known as the common ratio. This series can converge to a certain value when the absolute value of the common ratio is less than 1.

To find the sum of an infinite geometric series, we use a specific formula:

• \( S = \frac{a}{1-r} \)

Here, \( S \) represents the sum of the series, \( a \) is the first term, and \( r \) is the common ratio.

This formula only applies when \( |r| < 1 \). This is important because it ensures that the values of the terms get smaller and approach zero as you go further in the sequence, allowing the series to have a finite sum.

common ratio

The common ratio is a key characteristic of a geometric series. It is the factor by which each term of the series is multiplied to get the next term.

To find the common ratio \( r \), simply divide any term by the previous term.

• For example, in the series \( \frac{1}{3}, \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \ldots \), calculate the common ratio by dividing the second term \( \frac{1}{6} \) by the first term \( \frac{1}{3} \), which gives \( r = \frac{1}{2} \).

The common ratio must satisfy the condition \( |r| < 1 \) for the formula for the sum of an infinite series to apply, ensuring the series converges and has a finite sum.

Understanding the common ratio helps clarify how quickly the terms in the series shrink, setting the pace for how fast the series approaches its limit.

sum of series

The sum of an infinite geometric series is the total value when all terms in the series are added together. This is possible only if the common ratio \( r \) is such that \( |r| < 1 \).

Using the formula \( S = \frac{a}{1-r} \) where \( a \) is the first term and \( r \) is the common ratio, helps in calculating this sum.

• For instance, for the series \( \frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \ldots \), the sum is calculated by substituting \( a = \frac{1}{3} \) and \( r = \frac{1}{2} \) into the formula, resulting in a sum of \( \frac{2}{3} \).

The sum is not just an abstract number but represents the combined value of infinitely many terms that, due to the nature of geometric series, is finite in well-defined conditions. The ability to find this sum gives insight into how large the infinite sequence approach over its domain.