### Problem Statement**6. Show that the sequence $ \left( \frac{1}{2} \right)^n $ is Cauchy by using the Cauchy definition.**### Detailed ExplanationIn this problem, you are asked to prove that the sequence $ \left( \frac{1}{2} \right)^n $ is a Cauchy sequence. To do this, you need to use the Cauchy definition for a sequence.### Cauchy Sequence DefinitionA sequence $ (a_n) $ is said to be a Cauchy sequence if for every $ \epsilon > 0 $, there exists an integer $ N $ such that for all $ m, n \geq N $, the inequality $ |a_n - a_m| < \epsilon $ holds.### Steps to Prove1. **Identify the Sequence**: The sequence given is $ a_n = \left( \frac{1}{2} \right)^n $.2. **Write the Cauchy Condition**: For every $ \epsilon > 0 $, find an integer $ N $ such that for all $ m, n \geq N $, $ | \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m | < \epsilon $.### Solution Outline1. Start by considering the absolute difference $ | a_n - a_m | = \left| \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m \right| $.2. Without loss of generality, assume $ n \leq m $. Then rewrite the difference as $ \left| \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m \right| = \left( \frac{1}{2} \right)^n \left| 1 - \left( \frac{1}{2} \right)^{m-n} \right| $.3. Notice that $ \left( \frac{1}{2} \right)^{m-n} $ is a positive number between 0 and 1, thus \( \left| 1 - \left( \frac{1}{

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Description: ### Problem Statement**6. Show that the sequence \( \left( \frac{1}{2} \right)^n $ is Cauchy by using the Cauchy definition.**### Detailed ExplanationIn this problem, you are asked to prove that the sequence $ \left( \frac{1}{2} \right)^n $ is a

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Answered: 6. Show that the sequence (5)" is…

Math

Advanced Math

6. Show that the sequence (5)" is Cauchy by using the Cauchy definition.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Problem 1RQ

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Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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### Problem Statement
**6. Show that the sequence $ \left( \frac{1}{2} \right)^n $ is Cauchy by using the Cauchy definition.**

### Detailed Explanation
In this problem, you are asked to prove that the sequence $ \left( \frac{1}{2} \right)^n $ is a Cauchy sequence. To do this, you need to use the Cauchy definition for a sequence.

### Cauchy Sequence Definition
A sequence $ (a_n) $ is said to be a Cauchy sequence if for every $ \epsilon > 0 $, there exists an integer $ N $ such that for all $ m, n \geq N $, the inequality $ |a_n - a_m| < \epsilon $ holds.

### Steps to Prove
1. **Identify the Sequence**: The sequence given is $ a_n = \left( \frac{1}{2} \right)^n $.
2. **Write the Cauchy Condition**: For every $ \epsilon > 0 $, find an integer $ N $ such that for all $ m, n \geq N $, $ | \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m | < \epsilon $.

### Solution Outline
1. Start by considering the absolute difference $ | a_n - a_m | = \left| \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m \right| $.
2. Without loss of generality, assume $ n \leq m $. Then rewrite the difference as $ \left| \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^m \right| = \left( \frac{1}{2} \right)^n \left| 1 - \left( \frac{1}{2} \right)^{m-n} \right| $.
3. Notice that $ \left( \frac{1}{2} \right)^{m-n} $ is a positive number between 0 and 1, thus \( \left| 1 - \left( \frac{1}{

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