Problem 61 - Algebra and Trigonometry

Question

Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. $$ \frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2 \sqrt{2}}+\frac{1}{4}+\cdots $$

Step-by-Step Solution

Answer

The series is convergent, and its sum is approximately 2.

Step 1: Identify the First Term and Common Ratio

The given series is \( \frac{1}{\sqrt{2}} + \frac{1}{2} + \frac{1}{2\sqrt{2}} + \frac{1}{4} + \cdots \). The first term \( a \) is \( \frac{1}{\sqrt{2}} \). To find the common ratio \( r \), divide the second term by the first term: \( r = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{2} \cdot \sqrt{2} = \frac{\sqrt{2}}{2} \approx 0.707 \).

Step 2: Determine Convergence or Divergence

An infinite geometric series converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( |r| = \left| \frac{\sqrt{2}}{2} \right| = 0.707 \), which is less than 1. Therefore, the series is convergent.

Step 3: Calculate the Sum of the Convergent Series

The sum \( S \) of a convergent 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 \( a = \frac{1}{\sqrt{2}} \) and \( r = \frac{\sqrt{2}}{2} \): \[ S = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{\sqrt{2}}{2}} \] Simplifying, the sum \( S \) is approximately \( 2 \).

Key Concepts

Convergent Series

A convergent series is a sequence of numbers where the sum approaches a specific limit as more terms are added. In the context of an infinite geometric series, this means that even though the series is endless, the total sum of all its terms will settle at a certain value.
For a geometric series to be convergent, an important condition must be met: the absolute value of the common ratio must be less than one, i.e., \(|r| < 1\). This ensures that each subsequent term in the series becomes progressively smaller, leading the overall sum to stabilize.
In the provided exercise, after identifying the common ratio \( r = \frac{\sqrt{2}}{2} \), it was found that \(|r| \approx 0.707\), which is indeed less than one. Therefore, we can conclude that the series is convergent.

Common Ratio

The common ratio is a crucial element of a geometric progression. It is the constant factor by which each term of the series is multiplied to get the next term. The formula to find the common ratio \( r \) in a series is to divide any term by its preceding term.
In the given problem, the common ratio was determined by dividing the second term \(\frac{1}{2}\) by the first term \(\frac{1}{\sqrt{2}}\), resulting in \( r = \frac{\sqrt{2}}{2}\). Knowing this ratio allows us to ascertain how each term relates to the previous one and is essential for determining the series' behavior, including whether it converges or diverges.

Sum of Series

An infinite geometric series has a sum if it is convergent. There is a specific formula used to find this sum, which is: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term of the series and \( r \) is the common ratio.
Applying this formula to the series from the exercise, with \( a = \frac{1}{\sqrt{2}} \) and \( r = \frac{\sqrt{2}}{2} \), we can calculate the sum. After simplification, it was determined that the sum \( S \) of the series approaches approximately 2. This formula is powerful because it provides a manageable expression for calculating the sum of an infinitely long series, provided it converges.

Geometric Progression

In mathematics, a geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This type of sequence can either be finite or infinite.
Key aspects of a geometric progression include:

• The first term, which initiates the sequence.
• The common ratio, which determines how the sequence evolves.

The series presented in the exercise is an example of an infinite geometric progression, beginning at \( \frac{1}{\sqrt{2}} \) and continuing indefinitely. As clarified by its common ratio, \( \frac{\sqrt{2}}{2} \), the terms decreased progressively, confirming the series' nature and its eventual convergence.