Problem 67 - Calculus Early Transcendentals

Question

Find the function represented by the following series and find the interval of convergence of the series. $$\sum_{k=0}^{\infty}\left(\frac{x^{2}-1}{3}\right)^{k}$$

Step-by-Step Solution

Answer

Answer: The function represented by the given geometric series is $S(x) = \frac{3}{4-x^{2}}$. Its interval of convergence is $\sqrt{-2}< x < 2$.

Step 1: Identify the series as a geometric series

The given series is: $$\sum_{k=0}^{\infty}\left(\frac{x^{2}-1}{3}\right)^{k}$$ This is a geometric series, where the common ratio is: $$r=\frac{x^{2}-1}{3}$$

Step 2: Find the sum of the infinite geometric series

For a geometric series with first term a and common ratio r, the sum of the series, S, converges if |r| < 1, and is given by the formula: $$S = \frac{a}{1-r}$$ In this case, the first term, a, when k=0, is: $$a = \left(\frac{x^{2}-1}{3}\right)^{0} = 1$$ Using the sum formula and plugging in a and r, we get the function: $$S(x) = \frac{1}{1-\frac{x^{2}-1}{3}}$$

Step 3: Simplify the function

Now, we simplify the function: $$S(x) = \frac{1}{1-\frac{x^{2}-1}{3}} = \frac{1}{\frac{3-(x^{2}-1)}{3}} = \frac{1}{\frac{4-x^{2}}{3}} = \frac{3}{4-x^{2}}$$

Step 4: Find the interval of convergence using the Ratio Test

To find the interval of convergence, we apply the Ratio Test. For the given series, we take the limit as k approaches infinity: $$\lim_{k\to\infty}\left|\frac{(\frac{x^{2}-1}{3})^{k+1}}{(\frac{x^{2}-1}{3})^{k}}\right| = \lim_{k\to\infty}\left|\frac{x^{2}-1}{3}\right|$$ For the series to converge, this limit must be less than 1: $$\left|\frac{x^{2}-1}{3}\right| < 1$$ Solving for x, we find the interval of convergence: $$-1 < \frac{x^{2}-1}{3} < 1$$ $$-3 < x^{2}-1 < 3$$ $$-2 < x^{2} < 4$$ $$\sqrt{-2} < x < \sqrt{4}$$ Thus, the interval of convergence is: $$\sqrt{-2}< x < 2$$ To summarize, the function represented by the given geometric series is: $$S(x) = \frac{3}{4-x^{2}}$$ And its interval of convergence is: $$\sqrt{-2}< x < 2$$

Key Concepts

Understanding the Interval of Convergence

The **interval of convergence** of a series is a crucial concept in determining where a series converges on the real number line. For the geometric series \( \sum_{k=0}^{\infty}\left(\frac{x^{2}-1}{3}\right)^{k} \), we are interested in finding the values of \( x \) that make the series converge.
The convergence depends on the absolute value of the common ratio \( r = \frac{x^{2}-1}{3} \). The geometric series converges when the absolute value of the ratio is strictly less than one:
\[ \left| \frac{x^{2}-1}{3} \right| < 1 \]
Solving this inequality involves:

• Manipulating the inequality: \(-1 < \frac{x^{2}-1}{3} < 1\)
• Clearing the fraction by multiplying by 3: \(-3 < x^2 - 1 < 3\)
• Resulting in: \(-2 < x^2 < 4\)

This simplifies further, leading to: \[ -\sqrt{2} < x < 2 \]
However, because \(-\sqrt{2}\) is not real, we interpret this as \(x^2\) needing to be in (0,4), giving the interval \(x \in (-2, 2)\). This interval means the series converges only when \(x\) lies strictly between these bounds.

Sum of the Series

Finding the **sum of the series** involves determining the expression that the infinite series represents. For a geometric series where the first term \(a\) and the common ratio \(r\) are known, the sum \(S\) of the series is given by the formula:
\[ S = \frac{a}{1-r} \]
In our series, when \(k=0\), the first term \(a\) is \(1\), since
\( a = \left(\frac{x^2-1}{3}\right)^0 = 1 \).
The common ratio \(r\) is \(\frac{x^{2}-1}{3}\). Plug these into the sum formula:
\[ S(x) = \frac{1}{1-\frac{x^2-1}{3}} \]
We simplify this: 

• First handle the fraction \(\frac{x^2-1}{3}\) by substituting it into the denominator.
• From there: \( 1 - \frac{x^2-1}{3} = \frac{3-(x^2-1)}{3} \)
• This gives us: \( S(x) = \frac{1}{\frac{4-x^2}{3}} \)
• Finally, simplifying to: \( S(x) = \frac{3}{4-x^2} \)

This function represents the sum of the infinite series within its interval of convergence.

Utilizing the Ratio Test

The **ratio test** is a tool to determine the convergence of a series, particularly useful for series with a varying coefficient. Applying the ratio test involves examining the limit of the ratio of consecutive terms. For our geometric series, this involves:
Taking the limit as \(k\) approaches infinity:
\[ \lim_{k \to \infty} \left| \frac{\left(\frac{x^2-1}{3}\right)^{k+1}}{\left(\frac{x^2-1}{3}\right)^k} \right| = \lim_{k \to \infty} \left| \frac{x^2-1}{3} \right| \]
This simplifies to the absolute value of the ratio itself, \(|r|<1\).
We solve \(|\frac{x^2-1}{3}| < 1\) and interpret it to find the valid \(x\) interval:

• Calculate the constraint: \(-1 < \frac{x^2-1}{3} < 1\)
• Clear fractions: \(-3 < x^2 - 1 < 3\)
• Simplify the inequality to find: \(-2 < x^2 < 4\)

This ultimately provides a valid interval for convergence: \(x \in (-2, 2)\).
Thus, the ratio test confirms where the original series converges effectively.