Problem 48 - Thomas Calculus

Question

In Exercises \(41-48\) , use Theorem 20 to find the series' interval of convergence and, within this interval, the sum of the series as a function of \(x .\) $$ \sum_{n=0}^{\infty}\left(\frac{x^{2}-1}{2}\right)^{n} $$

Step-by-Step Solution

Answer

The interval of convergence is \( -\sqrt{3} < x < \sqrt{3} \), and the sum is \(\frac{2}{3 - x^2}\).

Step 1: Identify the Type of Series

The series \( \sum_{n=0}^{\infty} \left( \frac{x^{2}-1}{2} \right)^{n} \) is a geometric series where each term is of the form \( ar^n \) with \( a = 1 \) and \( r = \frac{x^2 - 1}{2} \).

Step 2: Determine the Interval of Convergence

For a geometric series \( \sum_{n=0}^{\infty} ar^n \) to converge, \( |r| < 1 \) must hold. Here, the common ratio \( r = \frac{x^{2}-1}{2} \), so the condition becomes \( \left| \frac{x^{2} - 1}{2} \right| < 1 \). Solving \( |x^2 - 1| < 2 \), we get \(-2 < x^2 - 1 < 2\). This simplifies to \(-1 < x^2 < 3\). Since \(x^2\) must be non-negative, the interval of convergence is \( -\sqrt{3} < x < \sqrt{3} \).

Step 3: Find the Sum of the Series

Within the interval of convergence, the sum of a geometric series \( \sum_{n=0}^{\infty} ar^n \) is \( \frac{a}{1-r} \). Here, \( a = 1 \) and \( r = \frac{x^2 - 1}{2} \), so the sum is given by: \[ \frac{1}{1 - \frac{x^2 - 1}{2}} = \frac{1}{\frac{3 - x^2}{2}} = \frac{2}{3 - x^2} \].

Key Concepts

Interval of Convergence

Understanding the interval of convergence in a series is crucial, as it tells us where a series will behave as expected. For a geometric series, such as \( \sum_{n=0}^{\infty} \left( \frac{x^2 - 1}{2} \right)^n \), this concept focuses on the common ratio \( r \). We start by determining \( r \) and ensuring it satisfies the condition \( |r| < 1 \). Here, our ratio is \( r = \frac{x^2 - 1}{2} \).To find when the series converges:

• Set \( \left| \frac{x^2 - 1}{2} \right| < 1 \).
• Solve the inequality \( -2 < x^2 - 1 < 2 \).
• This reduces to \( -1 < x^2 < 3 \).

Non-negative property of \( x^2 \) implies \( x^2 \geq 0 \). Therefore, the interval of convergence is \(-\sqrt{3} < x < \sqrt{3}\). This interval lets us use the series reliably for calculating results.

Sum of Series

Once we've determined where a series converges, the next step is finding its sum. The sum within the interval of convergence for a geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by the formula \( \frac{a}{1-r} \). For our specific series, \( a = 1 \) and \( r = \frac{x^2 - 1}{2} \).Let’s calculate:

• Substitute into the formula to get \( \frac{1}{1 - \frac{x^2 - 1}{2}} \).
• Simplify this expression to find \( \frac{1}{\frac{3 - x^2}{2}} \).
• Further simplification gives the sum as \( \frac{2}{3 - x^2} \).

This result is the sum of the series as a function of \( x \) within the interval \(-\sqrt{3} < x < \sqrt{3}\), allowing any valid \( x \) to produce an accurate summation value.

Series Convergence

The convergence of a series is a fundamental concept in understanding the behavior of sequences and series. In simpler terms, when a series converges, the infinite addition approaches a particular value, meaning it doesn't simply grow indefinitely. For a geometric series, this convergence depends on the absolute value of the common ratio \( r \).Key points to remember about series convergence:

• A geometric series converges when \( |r| < 1 \).
• Otherwise, it diverges, indicating the series sum isn't finite.
• This gives a reliable framework to decide when our mathematical model (the series) will produce meaningful outputs.

In this example, knowing \( \frac{x^2 - 1}{2} \) must be less than one in absolute value enables us to ensure convergence and appropriately calculate sums when \(-\sqrt{3} < x < \sqrt{3}\). This principle not only applies to this series but also to any geometric series, making it an essential part of mathematical understanding.