C **Title: Convergence of Sequences****Introduction:**In this exercise, we will determine whether each given sequence is convergent and provide a justification for our conclusion.**Sequence A:**\[(x_n) \text{ where } x_n = \frac{\sin(n\pi/6)}{n}\]**Sequence B:**\[\left( \frac{c^{n+2} - d^{n+2}}{c^n + d^n} \right) \text{ where } 0 < c < d\]**Sequence C:**\[(x_n) \text{ where } x_n = (-1)^n \left(\frac{n-1}{n}\right)\]**Explanation:**You will explore each sequence, analyze their behavior as $ n $ approaches infinity, and determine whether they converge to a specific limit. Consider any known theorems or properties of limits that can aid in the justification of your conclusions.

Answered: Determine whether each of the following…

Determine whether each of the following Sequences is convergent. Justify!! @@ (x₁) where x := n 16. с 142 d^+2 n c't dn ← Sin (nπ/6) where окска (X₁) where x ₁² = (-1)^ ()

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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Question

**Title: Convergence of Sequences**

**Introduction:**
In this exercise, we will determine whether each given sequence is convergent and provide a justification for our conclusion.

**Sequence A:**
\[
(x_n) \text{ where } x_n = \frac{\sin(n\pi/6)}{n}
\]

**Sequence B:**
\[
\left( \frac{c^{n+2} - d^{n+2}}{c^n + d^n} \right) \text{ where } 0 < c < d
\]

**Sequence C:**
\[
(x_n) \text{ where } x_n = (-1)^n \left(\frac{n-1}{n}\right)
\]

**Explanation:**
You will explore each sequence, analyze their behavior as $ n $ approaches infinity, and determine whether they converge to a specific limit. Consider any known theorems or properties of limits that can aid in the justification of your conclusions.

Expert Solution

Step 1: Question Description

Determine whether each of the following sequences is convergent. Justify.

C. $open parentheses x subscript n close parentheses$ where $x subscript n equals open parentheses negative 1 close parentheses to the power of n open parentheses fraction numerator n minus 1 over denominator n end fraction close parentheses$ .

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