Problem 14 - Thomas Calculus

Question

In Exercises \(1-36,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely (c) conditionally? $$ \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} 3^{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely for \(-2 \leq x \leq 4\) and conditionally nowhere.

Step 1: Identify Series Form

The given series is \( \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} 3^{n}} \). This is a power series of the form \( \sum_{n=1}^{\infty} a_n (x-c)^n \) where \( a_n = \frac{1}{n^3 3^n} \) and \( c = 1 \).

Step 2: Determine Radius of Convergence

To find the radius of convergence \( R \), use the ratio test. Consider \( a_n = \frac{1}{n^3 3^n} \). Compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1}{(n+1)^3 3^{n+1}} \cdot \frac{n^3 3^n}{1} = \frac{1}{3} \lim_{n \to \infty} \frac{n^3}{(n+1)^3} = \frac{1}{3} \).

Step 3: Calculate the Radius

From Step 2, the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) is \( \frac{1}{3} \), the radius is therefore \( R = 3 \).

Step 4: Find the Interval of Convergence

Since the radius \( R = 3 \), the interval of convergence is centered at \( x = 1 \). Therefore, the series converges for \( |x-1| < 3 \) or \( -2 < x < 4 \).

Step 5: Test Endpoints for Absolute Convergence

Check \( x = -2 \): The series becomes \( \sum_{n=1}^{\infty} \frac{(-3)^n}{n^3 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \), which converges absolutely by the p-series test with \( p = 3 > 1 \). Check \( x = 4 \): The series becomes \( \sum_{n=1}^{\infty} \frac{3^n}{n^3 3^n} = \sum_{n=1}^{\infty} \frac{1}{n^3} \), which also converges absolutely.

Step 6: Conclusion on Absolute and Conditional Convergence

Since both \( x = -2 \) and \( x = 4 \) points converge absolutely, the series converges absolutely throughout the interval \( -2 \leq x \leq 4 \). Thus, there is no interval where the series converges conditionally.

Key Concepts

Power Series

A power series is a type of infinite series that can help us represent functions in a form that is easier to work with for analysis. It usually looks like this:

• \( \sum_{n=0}^{\infty} a_n (x-c)^n \)

Here, each term in the series involves raising a variable, say \( x \), to a power \( n \), then multiplying it by a coefficient \( a_n \). The variable \( c \) is the center of the series. For example, the given series is \( \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} 3^{n}} \), where the series is centered at \( x = 1 \).
Power series make it possible to approximate functions, analyze their behavior around specific points, and determine their convergence properties.

Interval of Convergence

The interval of convergence of a power series is the range of values for which the series converges. This range of \( x \) values is especially important because it defines where the series accurately represents a desired function. To find this, we use the radius of convergence \( R \):

• If \( R \) is positive, the series converges when \( |x-c| < R \).

For our example, the series is centered at \( c = 1 \) and the radius of convergence is \( R = 3 \). Therefore, the interval is:

• \( -2 < x < 4 \)

We also test the endpoints at \( x = -2 \) and \( x = 4 \) separately to determine the nature of convergence at those points.

Absolute Convergence

A series converges absolutely if the series of absolute values also converges. Practically, this means that regardless of the original series' terms’ signs, their sum tends towards a specific number. It's a stronger form of convergence because:

• If a series converges absolutely, rearranging the series' terms won't affect the sum.

For the power series in our exercise, testing at endpoints \( x = -2 \) and \( x = 4 \) shows absolute convergence in the intervals since:

• The corresponding p-series with \( p = 3 \) converge.

This verifies absolute convergence over the entire interval \( -2 \leq x \leq 4 \). Therefore, there are no portions of this interval with conditional convergence.

Conditional Convergence

Conditional convergence occurs when a series converges, but does not converge absolutely. This is more complex because:

• The series 'delicately balances' the sums of positive and negative terms to converge.
• Rearranging terms in a conditionally convergent series can change its sum.

In our exercise, however, both endpoints \( x = -2 \) and \( x = 4 \) show absolute convergence. As a result, the series does not exhibit conditional convergence within its interval of convergence.