suppose Cn is a segn of real numbers such that lim Ich exists and is non-zero. If the ROC h→0 of the power series & Chah is equal to then ROC of the power serlis Σ n² ch is = ? chach n=1 n=1

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Title:** Problem Solving in Sequence and Series

**Content:**

Suppose \( C_n \) is a sequence of real numbers such that 
\[ \lim_{n \to \infty} |C_n|^{1/n} \]
exists and is non-zero. If the Radius of Convergence (ROC) of the power series 
\[ \sum_{n=1}^{\infty} C_n x^n \]
is equal to \( r \), then what is the ROC of the power series 
\[ \sum_{n=1}^{\infty} n^2 C_n x^n \]?

**Explanation:**

- **Sequence \( C_n \)**: Given a sequence of real numbers \( C_n \).
- **Limit Condition**: The limit \( \lim_{n \to \infty} |C_n|^{1/n} \) exists and is non-zero.
- **Power Series 1**: \( \sum_{n=1}^{\infty} C_n x^n \) has a radius of convergence \( r \).
- **Power Series 2**: We are asked to determine the radius of convergence for the series \( \sum_{n=1}^{\infty} n^2 C_n x^n \).

In a power series, the radius of convergence (ROC) is determined by analyzing the limit of the nth term as \( n \) approaches infinity. For the series \( \sum_{n=1}^{\infty} C_n x^n \), the ROC is given by:
\[ r = \frac{1}{\limsup_{n \to \infty} |C_n|^{1/n}}. \]

Considering the augmented series \( \sum_{n=1}^{\infty} n^2 C_n x^n \), the presence of \( n^2 \) changes the ROC:
\[ \frac{1}{\limsup_{n \to \infty} |n^2 C_n|^{1/n}}. \]

Analyzing further, we have \( |n^2 C_n|^{1/n} = (n^2)^{1/n} |C_n|^{1/n} \), where \( (n^2)^{1/n} \rightarrow 1 \) as \( n \rightarrow \infty \). Thus, the ROC remains affected primarily by \(
Transcribed Image Text:**Title:** Problem Solving in Sequence and Series **Content:** Suppose \( C_n \) is a sequence of real numbers such that \[ \lim_{n \to \infty} |C_n|^{1/n} \] exists and is non-zero. If the Radius of Convergence (ROC) of the power series \[ \sum_{n=1}^{\infty} C_n x^n \] is equal to \( r \), then what is the ROC of the power series \[ \sum_{n=1}^{\infty} n^2 C_n x^n \]? **Explanation:** - **Sequence \( C_n \)**: Given a sequence of real numbers \( C_n \). - **Limit Condition**: The limit \( \lim_{n \to \infty} |C_n|^{1/n} \) exists and is non-zero. - **Power Series 1**: \( \sum_{n=1}^{\infty} C_n x^n \) has a radius of convergence \( r \). - **Power Series 2**: We are asked to determine the radius of convergence for the series \( \sum_{n=1}^{\infty} n^2 C_n x^n \). In a power series, the radius of convergence (ROC) is determined by analyzing the limit of the nth term as \( n \) approaches infinity. For the series \( \sum_{n=1}^{\infty} C_n x^n \), the ROC is given by: \[ r = \frac{1}{\limsup_{n \to \infty} |C_n|^{1/n}}. \] Considering the augmented series \( \sum_{n=1}^{\infty} n^2 C_n x^n \), the presence of \( n^2 \) changes the ROC: \[ \frac{1}{\limsup_{n \to \infty} |n^2 C_n|^{1/n}}. \] Analyzing further, we have \( |n^2 C_n|^{1/n} = (n^2)^{1/n} |C_n|^{1/n} \), where \( (n^2)^{1/n} \rightarrow 1 \) as \( n \rightarrow \infty \). Thus, the ROC remains affected primarily by \(
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