Problem 10 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The radius of convergence is \( \frac{1}{2} \) and the interval is \( (-\frac{1}{2}, \frac{1}{2}) \).

Step 1: Apply the Ratio Test

The ratio test is used to find the radius of convergence of a series. Consider the given series \( \sum_{n=1}^{\infty} 2^n n^2 x^n \). Define \( a_n = 2^n n^2 x^n \). We need to find \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Start by finding \( a_{n+1} = 2^{n+1} (n+1)^2 x^{n+1} \).

Step 2: Simplify the Terms

Now, calculate the ratio \( \frac{a_{n+1}}{a_n} = \frac{2^{n+1} (n+1)^2 x^{n+1}}{2^n n^2 x^n} = 2 \cdot \frac{(n+1)^2}{n^2} \cdot x \).

Step 3: Calculate the Limit

Find the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| 2 \cdot \frac{(n+1)^2}{n^2} \cdot x \right| = \lim_{n \to \infty} |2x| \cdot \left( 1 + \frac{2}{n} + \frac{1}{n^2} \right) \approx 2|x| \]. As \( n \to \infty \), \( \frac{2}{n} \) and \( \frac{1}{n^2} \) approach 0. Thus, the limit is \( 2|x| \).

Step 4: Apply the Convergence Condition

For convergence via the ratio test, \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), which implies \( 2|x| < 1 \). Solving for \( |x| \), we get \( |x| < \frac{1}{2} \). This means the radius of convergence \( R = \frac{1}{2} \).

Step 5: Determine the Interval of Convergence

The interval of convergence can initially be written as \( -\frac{1}{2} < x < \frac{1}{2} \). We need to check convergence at the endpoints \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).

Step 6: Test Endpoints x = -1/2

For \( x = -\frac{1}{2} \), the series becomes \( \sum_{n=1}^{\infty} 2^n n^2 (-\frac{1}{2})^n = \sum_{n=1}^{\infty} (-1)^n n^2 \), which does not converge (since terms do not approach zero).

Step 7: Test Endpoints x = 1/2

For \( x = \frac{1}{2} \), the series becomes \( \sum_{n=1}^{\infty} 2^n n^2 \frac{1}{2}^n = \sum_{n=1}^{\infty} n^2 \), which diverges (the terms \( n^2 \) themselves diverge as \( n \to \infty \)). Thus, there is no convergence at \( x = \frac{1}{2} \).

Step 8: State the Interval of Convergence

Since the series does not converge at \( x = \pm \frac{1}{2} \), the interval of convergence is \( (-\frac{1}{2}, \frac{1}{2}) \).

Key Concepts

Radius of Convergence

In the study of power series, the radius of convergence is a crucial concept. It helps determine the range of values for which the series will converge. For a given power series, the radius of convergence \( R \) is the distance from the center of the series to the boundary of convergence.
In our original exercise, for the series \( \sum_{n=1}^{\infty} 2^n n^2 x^n \), the radius was found using the ratio test.
When the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 2|x| \) is calculated, it gives us the condition for convergence, which is \( 2|x| < 1 \).

• Solving \( |x| < \frac{1}{2} \), we conclude the radius to be \( \frac{1}{2} \).

The radius \( R \) tells us the series converges for all \( x \) within \( \frac{1}{2} \) units from the center.Understanding and calculating the radius of convergence is essential when working with power series since it dictates where the series is valid.

Interval of Convergence

Once the radius of convergence is determined, the interval of convergence comes next.
This interval specifies the set of \( x \) values for which the whole series converges, not just within the radius. In the case of our exercise, after determining the radius as \( \frac{1}{2} \),
we establish the interval based on the inequality \(-\frac{1}{2} < x < \frac{1}{2} \).

• We need to check the endpoints separately as they require special consideration and aren't always included in the interval.

Understanding the interval of convergence is key, as it fully defines where the series is useful and well-behaved.

Ratio Test

The ratio test is a powerful tool in analysis, especially when determining convergence of series.
It focuses on comparing the sizes of consecutive terms in a series to determine convergence.
For the given series \( \sum_{n=1}^{\infty} 2^n n^2 x^n \), we used the ratio test by computing:
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} |2x| \cdot \left( 1 + \frac{2}{n} + \frac{1}{n^2} \right) \approx 2|x| \] This step simplifies to say, if \( 2|x| < 1 \), then the series converges. This is a simple yet effective method to determine the basic convergence behavior
Keeping in mind its assumptions is crucial for accurate results.
Using the ratio test efficiently helps in understanding many series without computing large amounts of terms.

Endpoint Testing

After finding the interval of convergence, endpoint testing ensures completeness in convergence analysis.
Endpoints may or may not be part of the interval, so they require individual checks.
For a series with endpoints related to \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \):

• At \( x = -\frac{1}{2} \), the series becomes \( \sum_{n=1}^{\infty} (-1)^n n^2 \), which does not converge, as the terms don't get closer to zero.
• At \( x = \frac{1}{2} \), the series simplifies to \( \sum_{n=1}^{\infty} n^2 \), which diverges too, since the terms grow indefinitely.

This testing confirms the endpoints are not included, leading to the interval of convergence being strictly \( (-\frac{1}{2}, \frac{1}{2}) \). Endpoint testing demystifies whether the boundaries hold convergence or not, ensuring the interval is perfectly identified.