Problem 26 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The radius of convergence is 1 and the interval of convergence is \((-1, 1)\).

Step 1: Identify the type of series

The series given is \( \sum_{n = 2}^{\infty} \frac{x^{2n}}{n(\ln n)^2} \). This is a power series where the general term is \( a_n = \frac{x^{2n}}{n(\ln n)^2} \). We will use the Ratio Test to find the radius of convergence.

Step 2: Apply the Ratio Test

To apply the Ratio Test, calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_{n+1} = \frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2} \) and \( a_n = \frac{x^{2n}}{n(\ln n)^2} \).Calculate the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2} \cdot \frac{n (\ln n)^2}{x^{2n}} \right| = |x^2| \cdot \frac{n(\ln n)^2}{(n+1)(\ln(n+1))^2} \].

Step 3: Simplify the Limit Expression

Now, simplify the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = |x^2| \cdot \frac{n(\ln n)^2}{(n+1)(\ln(n+1))^2} \].As \( n \to \infty \), it becomes: \[ |x^2| \cdot \frac{1}{1} = |x^2| \].

Step 4: Find the Radius of Convergence

For the series to converge by the Ratio Test, the limit must be less than 1. Hence, we set:\[ |x^2| < 1 \].This simplifies to \( |x| < 1 \). Thus, the radius of convergence \( R \) is \( 1 \).

Step 5: Determine the Interval of Convergence

The radius of convergence is \( R = 1 \), which suggests the interval is \( -1 < x < 1 \). However, we must check the endpoints \( x = -1 \) and \( x = 1 \) for convergence separately.

Step 6: Check endpoint x = 1

Plug \( x = 1 \) into the series:\( \sum_{n=2}^{\infty} \frac{1^{2n}}{n(\ln n)^2} = \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} \).This series is known to diverge by the integral test or the comparison test, as \( \int_{2}^{\infty}\frac{1}{n(\ln n)^2} dn \) diverges.

Step 7: Check endpoint x = -1

Plug \( x = -1 \) into the series:\( \sum_{n=2}^{\infty} \frac{(-1)^{2n}}{n(\ln n)^2} = \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} \).Similar to \( x = 1 \), this series also diverges.

Step 8: Conclusion of Interval of Convergence

Since the series diverges at both endpoints \( x = 1 \) and \( x = -1 \), the interval of convergence is not inclusive of the endpoints. The final interval of convergence is \( (-1, 1) \).

Key Concepts

Radius of Convergence

In the context of a power series, the radius of convergence is a crucial concept. It defines the range of values for which the series converges. In simpler terms, imagine it as a radius extending from a central point on a number line where the series behaves nicely and adds up to a finite value. For any power series centered at zero, the radius tells you how far out you can go from zero before things get messy and the series stops summing to a finite result.

To find this radius, we often use specific methods such as the Ratio Test. In our example, the series \( \sum_{n=2}^{\infty} \frac{x^{2n}}{n(\ln n)^2} \) has the general term \( a_n = \frac{x^{2n}}{n(\ln n)^2} \). By applying the Ratio Test, we simplify the ratio of successive terms and find the limit. When we simplify it, we find that the condition for convergence is \( |x| < 1 \). Therefore, the radius of convergence \( R \) is \( 1 \).

• Radius gives the range of convergence from the center, here it is 1.
• X values within this radius lead to convergence.
• Outside this radius, the series diverges.

Interval of Convergence

The interval of convergence takes the radius a step further. It not only tells you about the area around the center where the series converges but also whether the endpoints of this interval should be included or not. This is crucial because even if the radius is clear, the behavior at the edge (endpoint) might vary.

For our specific series, with a radius of 1 found previously, the preliminary interval of convergence extends from \(-1\) to \(1\). But we need to check the endpoints separately. By substituting \( x = 1 \), the series becomes \( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} \), which diverges. Similarly, substituting \( x = -1 \), the series also diverges. Thus, neither endpoint should be part of the convergence interval.

• Interval requires checking whether endpoints result in convergence or divergence.
• At \(x = 1\), the series diverges, so this endpoint is excluded.
• At \(x = -1\), the series diverges, so this endpoint is also excluded.
• The final interval of convergence is open: \((-1, 1)\).

Ratio Test

The Ratio Test is a powerful tool used to find the radius of convergence for a power series. It's simple yet effective, especially when dealing with factorials or exponential expressions. The test involves taking the limit of the absolute value of the ratio of consecutive terms.Here's how it works: given a series \( \sum a_n \), calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Depending on this limit, you can determine the behavior of your series:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \), the series diverges.
• If \( L = 1 \), the test is inconclusive and other methods are needed.

The beauty of the Ratio Test in our example is its ability to distill the complex expression into something simple and manageable. We examined the series \( \frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2} \) versus \( \frac{x^{2n}}{n(\ln n)^2} \) and discovered the limit term \(|x^2|\). For convergence, we found that \(|x^2| < 1\), ultimately leading to \(|x| < 1\), a clear and concise result for the radius of convergence.