Problem 20 - 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? $$ \sum_{n=1}^{\infty} \sqrt[n]{n}(2 x+5)^{n} $$

Step-by-Step Solution

Answer

Radius of convergence: 1; Interval: \((-3, -2)\); No absolute or conditional convergence.

Step 1: Identify the Given Power Series

The series given is \( \sum_{n=1}^{\infty} \sqrt[n]{n}(2x+5)^{n} \). This is a power series in the form \( \sum a_n (x - c)^n \) where \( a_n = \sqrt[n]{n} \) and \( c = -\frac{5}{2} \).

Step 2: Use the Root Test to Find Radius of Convergence

Apply the root test, which gives the radius of convergence for this type of series. For the root test, consider:\[ \lim_{n \to \infty} \sqrt[n]{|a_n (2x+5)^n|} = \lim_{n \to \infty} \sqrt[n]{|\sqrt[n]{n} (2x+5)^n|} = \lim_{n \to \infty} n^{1/n} |2x+5| \]We know \( n^{1/n} \to 1 \) as \( n \to \infty \). Therefore, the limit simplifies to \( |2x+5| \). For convergence, we need:\[ |2x+5| < 1 \]

Step 3: Solve for the Interval of Convergence

Solve the inequality \( |2x+5| < 1 \) to find the interval of convergence:1. \( -1 < 2x + 5 < 1 \)2. Solve for \( x \): - Subtract 5 from all parts: \( -6 < 2x < -4 \) - Divide all parts by 2: \( -3 < x < -2 \)Thus, the interval of convergence is \((-3, -2)\).

Step 4: Check the Endpoints for Absolute and Conditional Convergence

Now, test the endpoints \( x = -3 \) and \( x = -2 \) for convergence.1. For \( x = -3 \): - The series becomes \( \sum \sqrt[n]{n}(-1)^n \). The absolute value grows as \( n^{1/n} \). Thus, it doesn't converge absolutely or conditionally.2. For \( x = -2 \): - The series becomes \( \sum \sqrt[n]{n}(1)^n \). Similarly, \( n^{1/n} \) prevents it from converging absolutely or conditionally.Hence, there is no conditional or absolute convergence on the endpoints.

Key Concepts

Interval of Convergence

When examining a power series, one essential aspect is identifying where the series converges. This is known as the interval of convergence. For the series \( \sum_{n=1}^{\infty} \sqrt[n]{n}(2x+5)^{n} \), the interval of convergence tells us the range of \(x\) values at which the series converges. To determine this interval, we use the root test, which involves taking the \(n\)-th root of the absolute terms of the series and finding the limit as \(n\) approaches infinity. For the given series, we have\[ |2x+5| < 1 \]By solving this inequality, we find that \(-3 < x < -2\), which represents the interval within which the series converges. So, whenever \(x\) lies between \(-3\) and \(-2\) (not including the endpoints), the series will converge. This defines our interval of convergence, giving us a clear picture of where our series behaves nicely.

Absolute Convergence

Absolute convergence of a series occurs when the series of the absolute values of its terms converges. Looking at the endpoints of our interval of convergence, we will test for absolute convergence by replacing \(x\) with the endpoint values. First, consider \(x = -3\), the series becomes \( \sum_{n=1}^{\infty} \sqrt[n]{n}(-1)^n \). Evaluating the behavior of this series, the absolute terms \(\sqrt[n]{n}\) still diverge as \(n\) grows larger. Thus, it fails to converge absolutely at \(x = -3\). Similarly, for \(x = -2\), the series reduces to \( \sum_{n=1}^{\infty} \sqrt[n]{n}(1)^n \). Again, the term \( \sqrt[n]{n} \) diverges, preventing absolute convergence at this endpoint. In this particular case, neither endpoint values lead to convergence of the absolute series, confirming that absolute convergence is not present at the boundaries of our interval.

Conditional Convergence

Conditional convergence is a slightly less restrictive form of convergence. A series is conditionally convergent if the series itself converges, but the series of its absolute values does not. To explore this in the context of our series, we again test the endpoints of the interval of convergence. For \(x = -3\), as observed earlier, the series \( \sum_{n=1}^{\infty} \sqrt[n]{n}(-1)^n \) doesn't converge, neither absolutely nor conditionally. Similarly, for \(x = -2\), the series becomes \( \sum_{n=1}^{\infty} \sqrt[n]{n}(1)^n \). Once more, it doesn't meet the criteria for convergence. Thus, at neither endpoint do we find conditional convergence. In situations like this, where neither kind of convergence is occurring at endpoints, it highlights the specific range where the series maintains regular convergence.