Problem 51 - A Graphical Approach to College Algebra

Question

Find each sum that converges. $$16+2+\frac{1}{4}+\frac{1}{32}+\cdots$$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{128}{7} \).

Step 1: Identify the Sequence

Observe that the given sequence is 16, 2, \( \frac{1}{4} \), \( \frac{1}{32} \),... This is a geometric series where each term after the first is obtained by multiplying the previous term by a constant ratio. This ratio is known as the common ratio \( r \).

Step 2: Determine the Common Ratio

To find the common ratio \( r \), divide the second term by the first term: \( r = \frac{2}{16} = \frac{1}{8} \). Verify with the next terms: \( \frac{1}{4} \div 2 = \frac{1}{8} \) and \( \frac{1}{32} \div \frac{1}{4} = \frac{1}{8} \). So, \( r = \frac{1}{8} \).

Step 3: Recognize the Geometric Series Formula

For a geometric series \( a + ar + ar^2 + ar^3 + \cdots \) where \(|r| < 1\), the sum \( S \) converges to \( \frac{a}{1 - r} \), where \( a \) is the first term of the sequence.

Step 4: Use the Sum Formula for Convergence

Substitute the first term \( a = 16 \) and the common ratio \( r = \frac{1}{8} \) into the sum formula: \[ S = \frac{16}{1 - \frac{1}{8}} \]

Step 5: Simplify the Expression

Calculate the denominator: \( 1 - \frac{1}{8} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8} \). Substitute back into the sum formula: \[ S = \frac{16}{\frac{7}{8}} \]

Step 6: Final Calculation

Simplify the fraction: \[ S = 16 \times \frac{8}{7} = \frac{128}{7} \]. Thus, the sum that converges for the series is \( \frac{128}{7} \).

Key Concepts

Sequence Convergence

Understanding sequence convergence is crucial when dealing with infinite series. In simple terms, a sequence converges if it approaches a specific number as the number of terms increases.
For a geometric series like the one in the exercise, 16, 2, \( \frac{1}{4} \), \( \frac{1}{32} \),..., the terms get progressively smaller due to the effect of the common ratio.
In such cases, if the absolute value of the common ratio \( r \) is less than 1, the series converges.

• The absolute value of \( r \) tells us whether the terms are getting smaller.
• The sum of the series approaches a certain finite number, which we can calculate using the sum formula for convergence.

Because \( |r| = \frac{1}{8} \) is less than 1, we can confidently say the series converges.

Common Ratio

The common ratio \( r \) is a key factor in determining the behavior of a geometric series.
It's the consistent factor by which each term is multiplied to get the next term in the sequence. To find the common ratio:

• Take any term in the series and divide it by the previous term.
• For our series: \( r = \frac{2}{16} = \frac{1}{8} \).
• Verification: \( \frac{1}{4} \div 2 = \frac{1}{8} \) and \( \frac{1}{32} \div \frac{1}{4} = \frac{1}{8} \).

This ratio tells us how quickly the series will converge. The smaller the common ratio, the faster the terms reduce in size, ensuring convergence.

Sum Formula for Geometric Series

The sum formula for a geometric series is a powerful tool used to find the sum of an infinite series that converges. This formula only applies when the absolute value of the common ratio \( |r| \) is less than 1, ensuring the series converges to a finite sum. The formula is:

• \[ S = \frac{a}{1 - r} \]
• Where \( a \) is the first term and \( r \) is the common ratio.

For the series in the exercise with \( a = 16 \) and \( r = \frac{1}{8} \), the sum is:

• \[ S = \frac{16}{1 - \frac{1}{8}} \]
• Calculate the denominator: \( 1 - \frac{1}{8} = \frac{7}{8} \).
• The sum converges to: \[ S = \frac{16}{\frac{7}{8}} = 16 \times \frac{8}{7} = \frac{128}{7} \].

This result shows the finite sum to which the infinite series converges.