### 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

Section: Chapter Questions

Problem 1RQ

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