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Chapter 11: Problem 35

In Exercises \(33-38,\) find the series' interval of convergence and, within this interval, the sum of the series as a function of \(x .\) $$ \sum_{n=0}^{\infty}\left(\frac{\sqrt{x}}{2}-1\right)^{n} $$

Short Answer

Expert verified
The interval of convergence is \(0 < x < 16\), and the sum is \(\frac{2}{4 - \sqrt{x}}\).

Step by step solution

01

Identify the Type of Series and Determine Convergence Criteria

The given series is \(\sum_{n=0}^{\infty}\left(\frac{\sqrt{x}}{2}-1\right)^{n}\). This is a geometric series, which converges if the absolute value of the common ratio \(r\) is less than 1, i.e., \(|r| < 1\). In this series, the common ratio \(r\) is \(\frac{\sqrt{x}}{2} - 1\).
02

Solve the Inequality for Convergence

To find the interval of convergence, solve the inequality \(\left|\frac{\sqrt{x}}{2} - 1\right| < 1\).1. \(\frac{\sqrt{x}}{2} - 1 < 1\) implies \(\frac{\sqrt{x}}{2} < 2\) or \(\sqrt{x} < 4\).2. \(\frac{\sqrt{x}}{2} - 1 > -1\) implies \(\frac{\sqrt{x}}{2} > 0\) or \(\sqrt{x} > 0\).Therefore, \(0 < \sqrt{x} < 4\), which means \(0 < x < 16\).
03

Determine the Sum of the Series

For a convergent geometric series with first term \(a\) and common ratio \(r\), the sum is given by \(\frac{a}{1 - r}\). Here, the first term \(a = 1\) and the common ratio \(r = \frac{\sqrt{x}}{2} - 1\).Thus, the sum \(S\) is\[S = \frac{1}{1 - \left(\frac{\sqrt{x}}{2} - 1\right)} = \frac{1}{2 - \frac{\sqrt{x}}{2}} = \frac{2}{4 - \sqrt{x}}.\]
04

Confirm the Sum and Interval of Convergence

Within the open interval \(0 < x < 16\), the series \(\sum_{n=0}^{\infty}\left(\frac{\sqrt{x}}{2}-1\right)^{n}\) converges to \(\frac{2}{4 - \sqrt{x}}\). Check the endpoints to see if they might be included:- At \(x = 16\), \(\sqrt{16} = 4\), so \(\left(\frac{4}{2} - 1\right) = 1\), which does not satisfy \(|r| < 1\).- At \(x = 0\), \(\sqrt{0} = 0\), so \(\left(\frac{0}{2} - 1\right) = -1\), which does not satisfy \(|r| < 1\).Thus, the series converges only strictly within \(0 < x < 16\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. These series play a crucial role in mathematics due to their simple yet profound properties.
In our given exercise, the series is \(\sum_{n=0}^{\infty}\left(\frac{\sqrt{x}}{2}-1\right)^{n}\). Here, we clearly spot a geometric series with the common ratio of \(r = \frac{\sqrt{x}}{2} - 1\). This common ratio determines if a geometric series converges or diverges. The series will converge only if the absolute value of this common ratio \(|r|\) is less than 1.
Convergence Criteria
Understanding the convergence criteria for a geometric series is essential. Convergence refers to the behavior of a series where its sum approaches a fixed number rather than growing infinitely. For a geometric series to converge, the absolute value of the common ratio \( r \) must be strictly less than 1:
\[ |r| < 1 \]
In our specific series, the common ratio is \(r = \frac{\sqrt{x}}{2} - 1\). To determine where our series converges, we solve the inequality:
\[ \left| \frac{\sqrt{x}}{2} - 1 \right| < 1 \]
This inequality helps us find the allowable values of \( x \) that ensure convergence.
Sum of Series
Once we establish that a geometric series converges, we can find its sum using a simple formula. The sum \( S \) of a geometric series is given by:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term of the series, and \( r \) is the common ratio. In our exercise, the first term \( a \) is 1, as the series starts at \( n=0 \), and the corresponding value is 1. The common ratio, as defined, is \( \frac{\sqrt{x}}{2} - 1 \).
Therefore, the sum of our series for the interval \(0 < x < 16\) is calculated as:
\[ S = \frac{1}{1 - \left(\frac{\sqrt{x}}{2} - 1\right)} = \frac{2}{4 - \sqrt{x}} \]
This tells us how much the series "adds up" to when it converges.
Inequality for Convergence
The process of solving convergence inequalities is crucial to finding the interval within which a geometric series converges. The inequality for convergence is derived from the condition \(|r| < 1\), where \( r \) here is \( \frac{\sqrt{x}}{2} - 1 \). To solve this inequality:
  • First, consider \( \frac{\sqrt{x}}{2} - 1 < 1 \), simplifying to \( \frac{\sqrt{x}}{2} < 2 \), hence \( \sqrt{x} < 4 \).
  • Next, examine \( \frac{\sqrt{x}}{2} - 1 > -1 \), which simplifies to \( \frac{\sqrt{x}}{2} > 0 \) or \( \sqrt{x} > 0 \).
Through this, we find the series converges for values satisfying \(0 < \sqrt{x} < 4\), meaning \(0 < x < 16\). Checking endpoints, neither \(x = 0\) nor \(x = 16\) meet the \(|r| < 1\) condition; hence, they are excluded, ensuring convergence lies strictly within \(0 < x < 16\).

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