Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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 \(](https://content.bartleby.com/qna-images/question/588b9222-edf2-44c3-a5bc-71e88a46ec33/46d403b0-9737-4c78-8cd4-b47570c5e117/ann06da_processed.jpeg)
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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