First state the power series center, then use either the root or ratio test to find R and I. Please provide step by step solution using simple steps. ### Infinite Series RepresentationThe given mathematical expression describes an infinite series. The series is represented as follows:\[\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n+1}}{2^n n}\]#### Explanation- **Summation Symbol (∑):** This symbol indicates that we are dealing with a series, which is a sum of a sequence of terms. - **Limits of Summation (n=1 to ∞):** The variable $ n $ starts at 1 and approaches infinity (∞), meaning we are summing an infinite number of terms starting from $ n = 1 $.- **Term of the Series ($\frac{(-1)^n x^{2n+1}}{2^n n} $):** - $ (-1)^n $ oscillates between -1 and 1 for different values of $ n $. - $ x^{2n+1} $ raises $ x $ to the power of $ 2n+1 $, which means the exponent is always an odd number. - $ 2^n $ raises 2 to the power of $ n $. - $ n $ is a simple multiplier for each term.Therefore, each term in the series is a fraction where: - The numerator is $ (-1)^n $ times $ x^{2n+1} $, - The denominator is $ 2^n $ times $ n $.This series may be used in contexts such as Taylor expansions or Fourier series in mathematics and engineering.

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Answered: 8. x2n+1 E(-1)" 2nn n=1

Math

Advanced Math

8. x2n+1 E(-1)" 2nn n=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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First state the power series center, then use either the root or ratio test to find R and I.

Please provide step by step solution using simple steps.

### Infinite Series Representation

The given mathematical expression describes an infinite series. The series is represented as follows:

\[
\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n+1}}{2^n n}
\]

#### Explanation

- **Summation Symbol (∑):** This symbol indicates that we are dealing with a series, which is a sum of a sequence of terms.

- **Limits of Summation (n=1 to ∞):** The variable $ n $ starts at 1 and approaches infinity (∞), meaning we are summing an infinite number of terms starting from $ n = 1 $.

- **Term of the Series ($\frac{(-1)^n x^{2n+1}}{2^n n} $):**
- $ (-1)^n $ oscillates between -1 and 1 for different values of $ n $.
- $ x^{2n+1} $ raises $ x $ to the power of $ 2n+1 $, which means the exponent is always an odd number.
- $ 2^n $ raises 2 to the power of $ n $.
- $ n $ is a simple multiplier for each term.

Therefore, each term in the series is a fraction where:
- The numerator is $ (-1)^n $ times $ x^{2n+1} $,
- The denominator is $ 2^n $ times $ n $.

This series may be used in contexts such as Taylor expansions or Fourier series in mathematics and engineering.

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