91Ó°ÊÓ

A geometric series is a series of the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term of the series, and \( r \) is the common ratio. Each term in the series is obtained by multiplying the previous term by the common ratio \( r \). This makes the series predictable and easy to manipulate. The series involves sums like \( a, ar, ar^2, ar^3, \ldots \).

In this exercise, the given series \( \sum_{n=0}^{\infty} \left(\frac{x^2-1}{2}\right)^n \) can be identified as a geometric series with the first term \( a = 1 \) and common ratio \( r = \frac{x^2-1}{2} \). This recognition is the key to solving the problem as the properties of geometric series can be directly applied. Understanding the structure of a geometric series allows you to quickly deduce the conditions for convergence and compute the sum when it converges.

An inequality is a mathematical statement that relates two expressions and shows that one is larger or smaller than the other. Importantly, inequalities are used in determining the interval of convergence for a series.

For geometric series, the convergence criterion demands that the absolute value of the common ratio \( r \) must be less than 1. That is, \( |r| < 1 \). In the exercise, we apply this inequality to the common ratio \( r = \frac{x^2-1}{2} \), resulting in the inequality \( \left|\frac{x^2-1}{2}\right| < 1 \).

To solve this inequality, we simplify step-by-step:

• \( -1 < \frac{x^2-1}{2} < 1 \)
• Multiply each part by 2: \( -2 < x^2 - 1 < 2 \)
• Add 1 to each part: \( -1 < x^2 < 3 \)

Recognizing that \( x^2 \) must be non-negative refines this to \( 0 \leq x^2 < 3 \), leading to the interval \( -\sqrt{3} < x < \sqrt{3} \). This is the range of \( x \) where the series converges.

The sum of a series, especially for a geometric series, can be found using a specific formula. For a geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is given by \( S = \frac{a}{1-r} \), when \( |r| < 1 \). This is due to the fact that as \( n \) approaches infinity, the terms become negligibly small as long as \( |r| < 1 \), ensuring convergence.

In the provided exercise, since \( a = 1 \) and \( r = \frac{x^2-1}{2} \), the sum within its interval of convergence is:\[ S(x) = \frac{1}{1 - \frac{x^2 - 1}{2}}\] By simplifying the denominator: \[ 1 - \frac{x^2 - 1}{2} = \frac{3 - x^2}{2}\]This expression for the sum can be further simplified to get:\[ S(x) = \frac{2}{3 - x^2}\] This result provides the sum of the series as a function of \( x \) within the derived interval of convergence.

A function of \( x \) expresses how a certain mathematical entity changes with respect to \( x \). In this context, the function \( S(x) = \frac{2}{3 - x^2} \) represents the sum of the series as a variable function dependent on \( x \). Such a representation is critical when analyzing how the behavior of the series changes across its interval of convergence.

This sum is only valid within the interval of convergence, which is \( -\sqrt{3} < x < \sqrt{3} \). Outside this interval, the series does not converge, and thus \( S(x) \) does not exist. The function \( S(x) \) hence helps in predicting and understanding the behavior of the series within valid boundaries.

Exploring the function further, as \( x \) approaches the boundary \( \sqrt{3} \), the denominator approaches zero, causing the value of \( S(x) \) to increase dramatically, illustrating the critical nature of interval boundaries.