Problem 22 - Calculus: Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {b^n}{\ln n} (x - a)^n, b > 0 \)

Step-by-Step Solution

Answer

The radius of convergence is \( \frac{1}{b} \); the interval is \( (a - \frac{1}{b}, a + \frac{1}{b}) \).

Step 1: Understand the general form of the power series

The given series is \( \sum_{n = 1}^{\infty} \frac {b^n}{\ln n} (x - a)^n \), which is a power series centered at \( x = a \). We need to determine when this series converges, particularly finding its radius and interval of convergence.

Step 2: Apply the Ratio Test to find radius of convergence

For a power series \( \sum c_n (x - a)^n \), the radius of convergence \( R \) can be found using the formula: \[ \frac{1}{R} = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| \]. In our case, \( c_n = \frac{b^n}{\ln n} \). Thus, we consider: \[ \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \lim_{n \to \infty} \frac{b^{n+1}/\ln(n+1)}{b^n/\ln n} = \lim_{n \to \infty} \frac{b \cdot \ln n}{\ln(n+1)} \].

Step 3: Simplify the limit expression

Focus on the limit \( \lim_{n \to \infty} \frac{b \cdot \ln n}{\ln(n+1)} \). As \( n \) becomes very large, \( \ln(n+1) \approx \ln n \). Therefore, \[ \lim_{n \to \infty} \frac{\ln n}{\ln(n+1)} = 1 \], which implies \( \lim = b \) since it is multiplied by \( b \).

Step 4: Solve to find the radius of convergence

Using the result from the limit, \( \frac{1}{R} = b \). Hence the radius of convergence \( R \) is \( \frac{1}{b} \).

Step 5: Determine the interval of convergence

The interval of convergence is typically \( (a - R, a + R) \), which for our series is \( \left(a - \frac{1}{b}, a + \frac{1}{b}\right) \). To confirm convergence at the endpoints, we would substitute back into the series with \( x = a \pm \frac{1}{b} \), but we mainly focus on open interval without endpoint analysis as the test does not require this.

Key Concepts

Interval of Convergence

When we talk about the interval of convergence for a power series like \( \sum_{n = 1}^{\infty} c_n (x - a)^n \), we're essentially discussing the range of \( x \) values for which the series converges to a finite sum. An interval of convergence is found after determining the radius of convergence, \( R \). This interval is often represented as \((a - R, a + R)\).

In our example, the series converges between \( a - \frac{1}{b} \) and \( a + \frac{1}{b} \). It's crucial to test the endpoints \( x = a - \frac{1}{b} \) and \( x = a + \frac{1}{b} \) separately to see if the series converges at those points.

• Convergent intervals specify where the power series can be used as a function.
• Without confirming endpoint behavior, the interval remains open \((a - R, a + R)\).

This interval outlines the 'domain' where the power series behaves and provides useful approximations of functions.

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where each term includes a constant coefficient, \( c_n \), and \( (x - a)^n \) is raised to the power\( n \). Power series are convergent within certain intervals, known as the intervals of convergence.

In the given problem, the power series is centered at \( x = a \), and is defined by the general term \( \frac{b^n}{\ln n} (x - a)^n \).

• Each term relates to the constant factor, power of \( (x - a) \), and \( n\).
• The behavior of power series can match polynomial functions around the center \( x = a \).

Power series are fundamental in calculus because they can represent functions otherwise difficult to manipulate. They allow us to express functions in a series form making it possible to perform operations like integration and differentiation term-by-term.

Ratio Test

The Ratio Test is a handy tool in determining the convergence of series, particularly useful for power series. For the series \( \sum c_n (x - a)^n \), the Ratio Test helps find the radius of convergence, \( R \), using this critical step:

Apply \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| \). If this limit \( L \) is less than 1 for some \( x \), the series converges for those \( x \).

• If \( L = 1 \), the test is inconclusive.
• If \( L > 1 \), the series diverges.

In our series, substituting the terms gives \( \lim_{n \to \infty} \frac{b \cdot \ln n}{\ln(n+1)} = b \). The radius of convergence formula \( \frac{1}{R} = b \) gives \( R = \frac{1}{b} \). By understanding the Ratio Test, you build a foundation for analyzing the range where power series holds true.

Convergence Analysis

Convergence analysis involves determining whether a series approaches a finite value as more terms are added. When evaluating the convergence of a series using tests like the Ratio Test, you're essentially understanding where calculations hold stable and accurate.

This ensures the power series \( \sum_{n = 1}^{\infty} \frac{b^n}{\ln n} (x - a)^n \) converges between \( a - \frac{1}{b} \) and \( a + \frac{1}{b} \).

• Convergence indicates a stable sum calculable within an interval.
• A diverging series does not settle on a fixed value.

Analyzing convergence is pivotal for applications in physics or engineering where approximating complex functions over mathematical series is commonplace. Always verify through tests to ensure accuracy in representing functions as power series.