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 series converges for \(-\sqrt{3} < x < \sqrt{3}\), and the sum is \(\frac{2}{3-x^{2}}\).

Step 1: Identify the Series Type

First, observe that the series \( \sum_{n=0}^{\infty}\left(\frac{x^{2}-1}{2}\right)^{n} \) is a geometric series. A geometric series has the form \( \sum_{n=0}^{\infty} ar^n \) with common ratio \( r \). In this case, \( r = \frac{x^{2}-1}{2} \).

Step 2: Determine the Interval of Convergence

A geometric series converges when the absolute value of the common ratio is less than 1, i.e., \( |r| < 1 \).Therefore, solve the inequality:\[\left| \frac{x^{2}-1}{2} \right| < 1\]This can be rewritten as:\[|x^{2} - 1| < 2\]which leads to:\[-2 < x^{2} - 1 < 2\]Add 1 to all sides to get:\[-1 < x^{2} < 3\]

Step 3: Solve for x

For \( -1 < x^{2} < 3 \), the inequality implies: From the left, \( x^{2} > -1 \) is always true for all real \( x \). Thus, we only consider \( x^{2} < 3 \):\[-\sqrt{3} < x < \sqrt{3}\]Therefore, the interval of convergence for the series is \((-\sqrt{3}, \sqrt{3})\).

Step 4: Find the Sum of the Series

For a geometric series \( \sum_{n=0}^{\infty} ar^n \), the sum is \( \frac{a}{1-r} \) when \( |r| < 1 \). Here \( a = 1 \) and \( r = \frac{x^{2}-1}{2} \). The sum of the series is:\[S(x) = \frac{1}{1 - \frac{x^{2}-1}{2}}\]Simplify the denominator:\[1 - \frac{x^{2}-1}{2} = \frac{2 - (x^{2} - 1)}{2} = \frac{3 - x^{2}}{2}\]Thus, the sum as a function of \( x \) is:\[S(x) = \frac{2}{3-x^{2}}\]

Key Concepts

Geometric Series

A geometric series is a particular type of infinite series where each term after the first is found by multiplying the previous term by a constant known as the "common ratio". If you remember, the definition of a geometric series follows the pattern:

• A starting term, usually denoted by the symbol \( a \).
• A common ratio \( r \).

Thus, the series can be written as: \( a, ar, ar^2, ar^3, \ldots \). That's the same as summing terms \( \sum_{n=0}^{\infty} ar^n \). In our problem, look at the series \(\sum_{n=0}^{\infty} \left(\frac{x^{2}-1}{2}\right)^n\). Here, \( a = 1 \) and the common ratio is \( r = \frac{x^{2}-1}{2} \). Recognizing a geometric series is crucial in determining how it behaves under certain conditions.

Sum of Series

Once we've identified a geometric series, the next step is finding its sum when it converges. The formula to find the sum of an infinite geometric series is given by \( S = \frac{a}{1 - r} \) provided that the common ratio \( |r| < 1 \). In our series, \( a = 1 \) and \( r = \frac{x^{2}-1}{2} \). Applying the formula, we substitute the values:

• Start with \( S(x) = \frac{1}{1 - \frac{x^{2}-1}{2}} \).
• Simplify: \( S(x) = \frac{2}{3 - x^{2}} \).

This function \( S(x) \) now gives the sum of the series within its interval of convergence for various "x" values.

Inequality Solving

To determine where a geometric series converges, we need to solve an inequality to find the range of values for \( x \). The series converges when \( |r| < 1 \). In our exercise, the inequality to solve is: \[ \left| \frac{x^{2} - 1}{2} \right| < 1 \]. This simplifies to \( |x^{2} - 1| < 2 \).

• Break it down to: \(-2 < x^{2} - 1 < 2\).
• Add 1 to each part: \(-1 < x^{2} < 3\).

Since \( -1 < x^2 \) is always true for real \( x \), focus on \( x^2 < 3 \) which gives \( -\sqrt{3} < x < \sqrt{3} \). This process determines the interval \((-\sqrt{3}, \sqrt{3})\), where the series converges.

Convergence Criteria

The main criteria for determining if an infinite series like a geometric series converges is based on the absolute value of its common ratio. For convergence:

• Geometric series converge if \( |r| < 1 \).

This rule helps understand why the inequality \( |\frac{x^{2} - 1}{2}| < 1 \) was solved in our exercise. A series converges in its interval of convergence, meaning inside this range, the series sums to a finite number. If these criteria are not met, the series may diverge, returning infinite sums.