Problem 54 - Algebra and Trigonometry

Question

Find the sum of the infinite geometric series. $$ \frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2 \sqrt{2}}+\frac{1}{4}+\cdots $$

Step-by-Step Solution

Answer

The sum of the infinite geometric series is 2.

Step 1: Identify the series terms

The series provided is: \( \frac{1}{\sqrt{2}} + \frac{1}{2} + \frac{1}{2\sqrt{2}} + \frac{1}{4} + \cdots \). Observe the denominators or numerators to start noticing the pattern and confirm that indeed it is a geometric series.

Step 2: Determine the first term (a)

The first term of the series, often represented as \(a\), is the first element of the series: \(a = \frac{1}{\sqrt{2}}\).

Step 3: Identify the common ratio (r)

To find the common ratio \(r\), divide the second term by the first term. Thus, \( r = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \).

Step 4: Check that \(|r| < 1\)

For the sum of an infinite geometric series to exist, the absolute value of the common ratio, \(|r|\), must be less than 1. Here, \(|\frac{\sqrt{2}}{2}| < 1\) is satisfied.

Step 5: Use the formula for the sum of an infinite geometric series

The sum \(S\) of an infinite geometric series is given by the formula: \[ S = \frac{a}{1-r} \] Substitute \(a = \frac{1}{\sqrt{2}}\) and \(r = \frac{\sqrt{2}}{2}\) into the formula.

Step 6: Calculate the sum of the series

Substitute the values we determined into the formula: \[ S = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{\sqrt{2}}{2}} \] Simplify the denominator: \[ 1 - \frac{\sqrt{2}}{2} = \frac{2}{2} - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \] Thus, \[ S = \frac{\frac{1}{\sqrt{2}}}{\frac{2 - \sqrt{2}}{2}} = \frac{1}{\sqrt{2}} \times \frac{2}{2 - \sqrt{2}} \] \[ S = \frac{2}{\sqrt{2}(2 - \sqrt{2})} \]

Step 7: Simplify the result

To simplify, we multiply numerator and denominator by \(\sqrt{2}+\sqrt{2}\), leveraging the difference of squares: \[ S = \frac{2(\sqrt{2}+\sqrt{2})}{(\sqrt{2})(2 - \sqrt{2})(\sqrt{2}+\sqrt{2})} \] You can simplify the new denominator: \[ (\sqrt{2})(2^2 - (\sqrt{2})^2) = (\sqrt{2})(4-2) =(\sqrt{2})(2) = 2\sqrt{2} \] Thus, \[ S = \frac{2(\sqrt{2}+\sqrt{2})}{2\sqrt{2}} \] Simplifying further, \[ S = \frac{2 \times 2\sqrt{2}}{2\sqrt{2}} = 2 \].Therefore, the sum of the infinite geometric series is 2.

Key Concepts

Sum of Geometric Series

One of the fascinating aspects of a geometric series is its ability to converge to a finite sum even if it contains an infinite number of terms. This property is particularly interesting in an infinite geometric series where the sum, denoted by \( S \), is determined using the formula: \[ S = \frac{a}{1-r} \] Here, \( a \) represents the first term, and \( r \) is the common ratio of the series. It's essential for the convergence of the series that the absolute value of the common ratio \( |r| \) is less than 1. If this condition is met, we can successfully calculate the sum, otherwise, the series does not converge. In our specific problem, we applied this formula after confirming that \( |r| < 1 \). The calculations led us to a sum of 2 for the infinitely ongoing series. This powerful tool allows us to summarize endless patterns into comprehensible finite sums, bridging the infinite and the finite in everyday mathematics.

Common Ratio

The common ratio, a crucial element in understanding geometric series, is the consistent factor multiplied to each term to produce the next. It is denoted by \( r \) and defined by dividing any term in the series by the preceding term. For example, in our series: - The first few terms are \( \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2\sqrt{2}}, \frac{1}{4}, \ldots \)- Calculate \( r \) by taking the second term and dividing it by the first: \[ r = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{2} \]An essential criterion for the sum of an infinite geometric series to exist is that \(|r| \) must be less than 1. For this particular sequence, \(|\frac{\sqrt{2}}{2}| < 1\) ensures the series transitions towards a finite sum, allowing us to apply the infinite sum formula effectively. Recognizing and calculating the common ratio helps us predict the nature of the series, whether it converges or not.

First Term in Series

The first term of a geometric series, often symbolized by \( a \), sets the stage for the entire sequence. It's not only the starting point but also a crucial component in calculating the series sum. In an infinite series, the first term offers the initial value which the rest of the terms, multiplied by a sequence of common ratios, build upon. For our infinite series, the first term was straightforward to identify as \( \frac{1}{\sqrt{2}} \). Having this initial term was key because:

• It determined the starting value for subsequent calculations.
• It formed the basis for applying the sum of the infinite series formula \( S = \frac{a}{1-r} \).

Always ensure you've accurately identified the first term, as any errors here can cascade into larger mistake throughout the solution. Understanding its meaning and how it ties the series together is fundamental in depicting the series’ overall behavior.