Problem 9 - Essential Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Radius of convergence is 2; interval of convergence is \((-2, 2)\).

Step 1: Recognize the Series

The given series is \( \sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2} x^{n}}{2^{n}} \). Our task is to find the radius and interval of convergence for this power series.

Step 2: Use the Ratio Test to Find the Radius of Convergence

The ratio test is used to find the radius of convergence for a power series. We analyze the absolute value of the ratio of successive terms in the series:\[\lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{(n+1)^{2} x^{n+1}}{2^{n+1}}}{(-1)^{n} \frac{n^{2} x^{n}}{2^{n}}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^{2} x}{2n^{2}} \right| = \left| x \right| \lim_{n \to \infty} \frac{(n+1)^{2}}{2n^{2}} \].

Step 3: Simplify the Limit for the Ratio Test

Simplifying the limit expression, we have:\[\lim_{n \to \infty} \frac{(n+1)^{2}}{2n^{2}} = \lim_{n \to \infty} \frac{n^{2} + 2n + 1}{2n^{2}} = \lim_{n \to \infty} \left(\frac{1}{2} + \frac{1}{n} + \frac{1}{2n^{2}}\right) = \frac{1}{2}\]Thus, the expression from Step 2 simplifies to \( \left| x \right| \cdot \frac{1}{2} \).

Step 4: Find Condition for Convergence from the Ratio Test

For the series to converge, the following condition must be satisfied:\[ \left| x \right| \cdot \frac{1}{2} < 1 \implies \left| x \right| < 2\]This gives the radius of convergence, \( R = 2 \).

Step 5: Determine the Interval of Convergence

To find the interval of convergence, test the endpoints \(x = -2\) and \(x = 2\):- **At \(x = -2\):** The series becomes \( \sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2} (-2)^{n}}{2^{n}} = \sum_{n=1}^{\infty} (-1)^{2n} n^{2} = \sum_{n=1}^{\infty} n^{2} \), which diverges.- **At \(x = 2\):** The series becomes \( \sum_{n=1}^{\infty} (-1)^{n} n^{2} \), an alternating series which diverges as \( n^{2} \) does not tend to 0.Thus, the interval of convergence is \((-2, 2)\).

Key Concepts

Radius of Convergence

When we talk about the radius of convergence of a power series, we refer to the distance from the center of the series (often around zero) within which the series converges. For the given series \( \sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2} x^{n}}{2^{n}} \), we determined the radius of convergence using the Ratio Test.

During this process, after simplifying the ratio of consecutive terms, we found that the limit simplified to \( \frac{1}{2} \). This allowed us to express the convergence condition as \( |x| \cdot \frac{1}{2} < 1 \). Solving this inequality yields \( |x| < 2 \), indicating that the power series converges for \( x \) values within 2 units of the center, leading us to conclude that the radius of convergence \( R = 2 \).

The radius of convergence is crucial, as it forms part of the fundamental framework required to check where exactly the series maintains its convergence behavior, forming a solid basis to further explore the interval of convergence.

Interval of Convergence

Part of understanding a series is knowing not just where it converges, but precisely the set of \( x \) values over which this happens. This set of values is called the "interval of convergence."

With a radius of convergence \( R = 2 \) for our series \( \sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2} x^{n}}{2^{n}} \), the interval of potential convergence spans from \( -2 \) to \( 2 \). However, we must check the behavior at these endpoints individually:

• At \( x = -2 \), substituting in the series results in a divergence, as the terms become larger due to constant powers of \(-2\).
• Similarly, at \( x = 2 \), the series remains divergent because the terms \( n^2 \) do not approach zero, rendering the series non-convergent despite being alternating.

Hence, the interval of convergence, excluding the endpoints where we found divergence, is strictly \((-2, 2)\). This complete insight into the interval empowers students to understand the precise range within which the series is valid and converges.

Ratio Test

The ratio test is a method particularly useful for determining the radius and interval of convergence for power series. It involves taking the absolute value of the ratio of successive terms as \( n \) approaches infinity.

For the series \( \sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2} x^{n}}{2^{n}} \), we began by calculating \[ \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{(n+1)^{2} x^{n+1}}{2^{n+1}}}{(-1)^{n} \frac{n^{2} x^{n}}{2^{n}}} \right| \].

This simplified to \( \left|x\right| \lim_{n \to \infty} \frac{(n+1)^2}{2n^2} \), which further reduced to \( \left|x\right| \cdot \frac{1}{2} \). The condition for convergence is when this result is less than 1, simplifying to \( \left|x\right| < 2 \).

Using the ratio test helps delineate the set of \( x \) values over which a series converges, making it invaluable in series analysis. Through it, we determined both the radius \( R = 2 \) and began exploring the interval for our specific series, displaying how crucial the ratio test is when working with power series.