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

The series converges absolutely for \(-1 < x < 1\) and conditionally at \(x = -1\).

Step 1: Identify the Series Structure

The given series is \( \sum_{n=0}^{\infty} \frac{n x^{n}}{n+2} \). We notice that this is a power series of the form \( \sum a_n x^n \), where \( a_n = \frac{n}{n+2} \). To find the radius and interval of convergence, we can use the Ratio Test.

Step 2: Apply the Ratio Test

For the Ratio Test, we compute \( \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \). Thus, \[ \lim_{n \to \infty} \left| \frac{(n+1)x^{n+1}/(n+3)}{n x^n/(n+2)} \right| = \lim_{n \to \infty} \left| \frac{(n+1)x}{(n+3)} \cdot \frac{(n+2)}{n} \right|. \] Simplifying gives \(|x| \cdot 1 = |x|\). Now, consider \( \lim_{n \to \infty} \frac{(n+1)(n+2)}{n(n+3)} = 1\). The radius of convergence is thus \( R = 1 \).

Step 3: Determine the Interval of Convergence

With \( R = 1 \), the series converges for \(-1 < x < 1\). We need to check endpoints \(x = -1\) and \(x = 1\).

Step 4: Check Convergence at Endpoints

For \(x = 1\): The series becomes \(\sum_{n=0}^{\infty} \frac{n}{n+2}\) which diverges (as \(n/(n+2)\) does not tend to zero). For \(x = -1\): The series becomes \(\sum_{n=0}^{\infty} \frac{n(-1)^{n}}{n+2}\) which also does not converge absolutely, but further testing for conditional convergence can be done.

Step 5: Determine Absolute and Conditional Convergence

(b) The series converges absolutely for \( -1 < x < 1 \) because the Ratio Test implies convergence in this interval without conditions. (c) Although the series does not converge absolutely at \(x = -1\) and \(x = 1\), we must check for conditional convergence. It turns out, using the Alternating Series Test, that \(x = -1\) indeed converges conditionally because \(\sum \frac{n}{n+2} \cdot (-1)^n\) satisfies the alternating series criteria.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when dealing with power series. It indicates the distance from the center of the series within which the series converges. To determine this radius, we often use the Ratio Test or the Root Test. For a power series \(\sum a_n x^n\), the Radius of Convergence \(R\) can be found using:

• Ratio Test: \( R = \lim_{{n \to \infty}} \left| \frac{a_n}{a_{n+1}} \right| \)
• Root Test: \( R = 1/\limsup_{{n \to \infty}} \sqrt[n]{|a_n|} \)

For the given series \(\sum_{{n=0}}^{\infty} \frac{n x^n}{n+2}\), applying the Ratio Test shows that the terms simplify to \(|x|\) as \(n\) grows larger. Consequently, we find that \(R = 1\), so the series will converge as long as \(x\) is within 1 unit from the origin.

Interval of Convergence

Once the radius of convergence is known, the next step is finding the interval of convergence. It is the range of \(x\) values for which the series converges. Typically expressed as \((c-R, c+R)\), where \(c\) is the center of the series, and \(R\) is the radius of convergence.

For our series, centered at 0, and with \(R = 1\), this interval is \(-1 < x < 1\). However, we cannot forget about the endpoints, \(-1\) and \(1\). Checking these endpoints separately is crucial:

• At \(x = 1\), the series \( \sum \frac{n}{n+2} \) diverges.
• At \(x = -1\), the series transforms into an alternating form \( \sum \frac{n(-1)^n}{n+2} \).

Thus, the interval is open at both ends without inclusion of \(1\), but further testing can find conditional convergence at \(-1\).

Conditional Convergence

Conditional convergence occurs when a series converges but does not converge absolutely. This can happen if the alternating series test confirms convergence after absolute convergence has been found lacking.

For the given series at \(x = -1\), absolute convergence is not present, but since terms \((-1)^n\) can alternate, we apply the Alternating Series Test:

• The series \(\sum \frac{(-1)^n n}{n+2}\)
• Terms \(b_n = \frac{n}{n+2}\) decrease in magnitude and approach zero.

These satisfy the conditions for the Alternating Series Test, allowing us to conclude that at \(x = -1\), the series converges conditionally. This highlights the importance of testing for both absolute and conditional convergence when dealing with boundary values.