Problem 7 - 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? $$ \sum_{n=0}^{\infty} \frac{n x^{n}}{n+2} $$

Step-by-Step Solution

Answer

Radius of convergence is 1; the series converges absolutely on \(-1 < x < 1\). No conditional convergence.

Step 1: Identify the Series Type

The given series is \[ \sum_{n=0}^{\infty} \frac{n x^n}{n+2} \]It looks like a power series, but the factor \( \frac{n}{n+2} \) requires special attention, as it might affect convergence.

Step 2: Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test to the series.Evaluate the limit:\[ L = \lim_{n \to \infty} \left| \frac{(n+1)x^{n+1}}{(n+3)} \frac{n+2}{nx^n} \right| = \lim_{n \to \infty} \left| x \right| \frac{n+1}{n} \frac{n+2}{n+3} \]Simplify:\[ L = \lim_{n \to \infty} |x| \frac{n+1}{n} \cdot \frac{n+2}{n+3} = |x| \]The Ratio Test tells us the series converges when \( |x| < 1 \). Thus, the radius of convergence \( R \) is 1.

Step 3: Determine the Interval of Convergence

Given the radius of convergence \( R = 1 \), the interval of convergence is initially \(-1 < x < 1\). We must check the endpoints of the interval separately.

Step 4: Check Convergence at x = 1

When \( x = 1 \), the series becomes:\[ \sum_{n=0}^{\infty} \frac{n}{n+2} \]This simplifies to a divergent series because the terms do not approach zero as \( n \to \infty \). The series diverges at \( x = 1 \).

Step 5: Check Convergence at x = -1

For \( x = -1 \), the series is:\[ \sum_{n=0}^{\infty} \frac{n (-1)^n}{n+2} \]This is an alternating series. We check the Alternating Series Test:The terms \( a_n = \frac{n}{n+2} \) are not decreasing to zero, hence it does not satisfy the conditions for convergence.Therefore, the series diverges at \( x = -1 \).

Step 6: Conclude Interval of Convergence

Since the series diverges at both \( x = 1 \) and \( x = -1 \), the series converges only within the open interval \(-1 < x < 1\).

Step 7: Find Absolute Convergence

The series converges absolutely wherever it converges within the radius of convergence. For \(-1 < x < 1\), the series is also absolutely convergent.

Step 8: Determine Conditional Convergence

The series does not converge at the endpoints. It converges within the interval \(-1 < x < 1\) and is absolute within this interval. Therefore, there is no conditional convergence.

Key Concepts

Interval of Convergence

When dealing with power series like \( \sum_{n=0}^{\infty} \frac{n x^n}{n+2} \), finding the interval of convergence is crucial. The interval of convergence tells us the values of \( x \) for which the series converges. We usually start by finding the radius of convergence using tests such as the Ratio Test, which helps identify the "reach" of the series' convergence.
In this exercise, the radius of convergence is found to be 1, determined through the Ratio Test calculation. This means the series converges absolutely for \( |x| < 1 \).

• The interval of convergence is initially determined as \(-1 < x < 1\). This is because, within this range, the series behaves in a convergent manner.
• We then individually check the endpoints \( x = -1 \) and \( x = 1 \) by substituting them into the series to see if they converge.

After testing the endpoints and verifying that the series diverges for both \( x = -1 \) and \( x = 1 \), the conclusion is that the series converges only for \( -1 < x < 1 \). This interval excludes the endpoints, confirming it’s a strict open interval of convergence.

Absolute Convergence

Absolute convergence is a stronger form of convergence that occurs if a series converges when all terms are replaced by their absolute values. It is an important notion because when a series is absolutely convergent, it ensures not only the regular convergence of the series but also allows us to rearrange its terms without altering the sum.
For the series in question, \( \sum_{n=0}^{\infty} \frac{n x^n}{n+2} \), it converges absolutely within the interval \(-1 < x < 1\). Here's why:

• Since the radius of convergence is 1, the series will be absolutely convergent wherever it is within this radius, according to the Ratio Test results.
• If we consider the absolute values of the terms, the series still converges because \( |x| < 1 \) ensures that the individual terms of the series become increasingly small and tend towards zero.

This absolute nature of convergence remains effective for all \( x \) values within the interval \(-1 < x < 1\), meaning wherever the series converges, it, in fact, converges absolutely as well.

Conditional Convergence

Conditional convergence is a type of convergence where a series converges, but it does not converge absolutely. However, this series \( \sum_{n=0}^{\infty} \frac{n x^n}{n+2} \) does not demonstrate conditional convergence within its interval.
Here’s a deeper look into why conditional convergence doesn't apply here:

• The series converges for \( -1 < x < 1 \), as already determined by the radius of convergence and the evaluation of endpoints.
• Given that the series also converges absolutely within this interval, this automatically excludes the possibility of conditional convergence.

Since the series fully acquires absolute convergence between \( -1 < x < 1 \) and fails to demonstrate convergence at the endpoints, no conditional convergence occurs, concluding that for \( x = -1 \) and \( x = 1 \), the series is divergent.

Ratio Test

The Ratio Test is a fundamental tool for determining the radius of convergence of a power series. It involves comparing the magnitudes of successive terms and is particularly handy in calculating the convergence behavior of the series like \( \sum_{n=0}^{\infty} \frac{n x^n}{n+2} \).
Here's how the Ratio Test is typically executed:

• Firstly, calculate the ratio of the \( (n+1)^{th} \) term to the \( n^{th} \) term of the series and consider its absolute value.
• The limit \( L \) of these ratios as \( n \to \infty \) determines the nature of convergence:

\[ L = \lim_{n \to \infty} \left| \frac{(n+1)x^{n+1}}{(n+3)} \frac{n+2}{nx^n} \right| = |x| \]
This limit results in \( L = |x| \), leading directly to the conclusion that the series converges when \( |x| < 1 \). The result provides a clear indication of the convergence range of the series.