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Find the radius of convergence and interval of convergence of the series. $$\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}$$

Short Answer

Expert verified
The radius of convergence is infinite and the interval of convergence is \((-\infty, \infty)\).

Step by step solution

01

Determine the type of series

The given series is \( \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !} \). This is a power series centered at zero (0) with terms of the form \( a_n = (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \).
02

Apply the Ratio Test for Convergence

The ratio test is a common method to find the radius of convergence. We take the limit of the absolute value of the ratio of consecutive terms:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{x^{2n+3}}{(2n+3)!}}{(-1)^{n} \frac{x^{2n+1}}{(2n+1)!}} \right| = \lim_{n \to \infty} \left| \frac{x^{2n+3}}{(2n+3)!} \times \frac{(2n+1)!}{x^{2n+1}} \right| \]which simplifies to:\[ \lim_{n \to \infty} \left| \frac{x^{2}}{(2n+2)(2n+3)} \right| \]
03

Simplify the Ratio Test formula

Simplifying further, we see that the factors \((2n+2)(2n+3)\) grow much faster than \(x^2\) as \(n\) approaches infinity. Thus:\[ \lim_{n \to \infty} \left| \frac{x^2}{(2n+2)(2n+3)} \right| = 0 \] for all \(x\), meaning the series converges for any \(x\).
04

Evaluate the Radius and Interval of Convergence

Since the ratio converges to 0 for all \(x\), the radius of convergence \(R\) is infinite. Therefore, the interval of convergence is \(-\infty < x < \infty\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interval of Convergence
When working with power series, it's crucial to determine where they converge. The interval of convergence is essentially all the values of the variable, typically denoted as \( x \), where the series converges. This interval might be finite or infinite, depending on the behavior of the terms in the series.
In this particular exercise, because the Ratio Test showed convergence for all \( x \), the interval of convergence is \(-\infty < x < \infty \). This means the series will converge regardless of the \( x \) value you choose, which results in the entire real line being the interval of convergence.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of a series. It involves comparing the terms of the series as \( n \) approaches infinity. Here's how it works:
  • Consider the series terms \( a_n \). Find the ratio of the \( n+1 \) term to the \( n \) term, \( \left| \frac{a_{n+1}}{a_n} \right| \).
  • Take the limit of this ratio as \( n \) approaches infinity.
  • If the limit is less than 1, the series converges absolutely. If greater than 1, it diverges. If exactly 1, the test is inconclusive.
In this exercise, applying the Ratio Test revealed that the limit was 0, less than 1, indicating convergence for any value of \( x \).
Thus, it played a crucial role in determining the radius and interval of convergence.
Power Series
Power series are a fundamental concept in calculus, composed of terms involving powers of a variable\( x \). It's written as \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where \( a \) is the center of the series and \( c_n \) are coefficients. In this scenario, the given series is \( \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \), which is a type of power series centered at zero.
Understanding power series is essential because they allow us to represent functions that are otherwise complex with infinite polynomials. Power series are used to approximate complicated functions and solve equations that are not otherwise easily handled.
They help in exploring properties of functions, especially near a point \( x = a \), the center of the series.
Infinite Convergence
Infinite convergence refers to situations where a series converges for every value of \( x \). This means there isn't any specific restriction on the values for which the series compresses to a finite sum.
The series \( \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !} \) in the exercise converges for every \( x \) because of the diminishing effect of the factorial in the denominator of each term. The factorial grows quickly and outpaces the growth of \( x^{2n+1} \), leading to a series that adds up to a finite value for any \( x \).
Infinite convergence often leads to significant implications in mathematical analysis, especially when working with power series. It indicates the series represents a certain function over the entire real line.

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