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 on \([-2, 4]\) with no conditional convergence.

Step 1: Analyze the Form of the Series

The series is given by \( \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} 3^{n}} \). Observe that this is a power series centered at \( x = 1 \) with the general term \( a_n = \frac{(x-1)^n}{n^3 \cdot 3^n} \).

Step 2: Apply the Ratio Test for Convergence

To find the radius of convergence, use the Ratio Test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Compute this ratio:\[\lim_{n \to \infty} \left| \frac{(x-1)^{n+1}}{(n+1)^3 \cdot 3^{n+1}} \cdot \frac{n^3 \cdot 3^n}{(x-1)^n} \right| = \lim_{n \to \infty} \left| \frac{x-1}{3} \cdot \frac{n^3}{(n+1)^3} \right|.\]due to \( \lim_{n \to \infty} \frac{n^3}{(n+1)^3} = 1 \), it simplifies to:\[ \left| \frac{x-1}{3} \right|. \]The series converges when \( \left| \frac{x-1}{3} \right| < 1 \), leading to \( -3 < x-1 < 3 \), or \( -2 < x < 4 \). So, the radius of convergence is \( R = 3 \), and the interval is initially \( (-2, 4) \).

Step 3: Test Endpoints for Conditional or Absolute Convergence

Check the endpoints \( x = -2 \) and \( x = 4 \) separately to see if the series converges there.- At \( x = -2 \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-3)^n}{n^3 \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \), which converges by the Alternating Series Test (since \( \frac{1}{n^3} \) decreases and approaches 0).- At \( x = 4 \), the series becomes \( \sum_{n=1}^{\infty} \frac{3^n}{n^3 \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n^3} \), which converges by the p-series test, as \( p = 3 > 1 \). So \( x = -2 \) and \( x = 4 \) are included in the interval of convergence.

Step 4: Determine Absolute and Conditional Convergence

- The series converges absolutely for all \( x \) such that \( -2 \leq x \leq 4 \) because the function inside is positive at both endpoints after simplification.- It does not converge conditionally at any \( x \) in its interval because both endpoints already satisfy absolute convergence conditions with the p-series and alternating series tests.

Key Concepts

Power Series

A power series is essentially an infinite series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are the coefficients and \( a \) is the center of the series. In our problem, the power series is centered at \( x = 1 \), with each term represented by \( \frac{(x-1)^n}{n^3 \cdot 3^n} \).
This setup gives us a function of \( x \) described by a sum of powers of \( x-1 \), making it a series expansion around \( x=1 \).
Power series are significant because they allow us to represent a wide variety of functions using polynomials, which are easier to handle and solve.
The radius of convergence determines where this series maintains validity and represents a real number within which the series converges to a function. When dealing with functions, understanding the radius and point of convergence helps anticipate behavior and possible divergences in calculus and analysis.

Ratio Test

The Ratio Test is a common technique used to find the convergence of a series, especially useful with power series. This test helps determine the radius of convergence by comparing the limit of the ratio of successive terms of the series. If we consider two successive terms \( a_n \) and \( a_{n+1} \), the Ratio Test involves computing the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
The series converges absolutely if this limit is less than 1, diverges if greater than 1, and is inconclusive if equal to 1.

• If the limit is less than 1, then the power series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit is equal to 1, the test is inconclusive, and further analysis is required.

For the series \( \sum_{n=1}^{\infty} \frac{(x-1)^n}{n^3 \cdot 3^n} \), the ratio test leads us to conclude \( \left| \frac{x-1}{3} \right| < 1 \), which results in a radius of convergence \( R = 3 \) with the initial interval \( (-2, 4) \).
Understanding and applying the Ratio Test is critical as it lays down the groundwork for determining where the series is reliable or well-behaved.

Absolute Convergence

Absolute convergence refers to a series \( \sum a_n \) that converges absolutely if \( \sum |a_n| \) converges. This means that even if you consider the absolute values of the terms of the series, it still sums to a finite number.
This concept is important because absolute convergence implies convergence itself, but the reverse is not necessarily true. In this problem, for all \( x \) within \(-2 \leq x \leq 4\), the series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) converges, proving absolute convergence at these endpoints. This is due to \( \frac{1}{n^3} \) being a classic \( p \)-series with \( p = 3 > 1 \).
When a series converges absolutely, it indicates strong convergence properties, making the function represented by the series more robust over its interval. For example, both endpoints in this series satisfy conditions for absolute convergence, meaning that the series converges regardless of the sign of its terms.

Conditional Convergence

While absolute convergence means an entire series converges when considering the absolute values of its terms, conditional convergence is slightly different. A series converges conditionally if it converges when taken normally but not when considering the absolute values of its terms.
This typically occurs in alternating series where terms switch in sign and converge due to the alternating pattern, not because their absolute values converge. However, in the series given by \( \sum_{n=1}^{\infty} \frac{(x-1)^n}{n^3 \cdot 3^n} \), the series does not converge conditionally at any \( x \) value within \( -2 \leq x \leq 4 \). This occurs because the series at the endpoints already exhibits absolute convergence.
Thus, for this specific exercise, there is no value of \( x \) where the series converges conditionally but not absolutely. Understanding conditional convergence is vital in identifying and analyzing series behavior through alternating series and recognizing the utility beyond absolute convergence.