Problem 27 - 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? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely for \(-4 < x < 0\) and conditionally at \(x = -4\). The radius of convergence is 2.

Step 1: Identifying the series and applying the Ratio Test

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x+2)^{n}}{n 2^{n}} \). To find the radius of convergence, we'll use the Ratio Test. According to the Ratio Test, the series \( \sum a_n \) with terms \( a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Here, \( a_n = \frac{(-1)^{n+1} (x+2)^n}{n 2^n} \).

Step 2: Calculate the Ratio \( \left| \frac{a_{n+1}}{a_n} \right| \)

First, determine \( a_{n+1} = \frac{(-1)^{n+2} (x+2)^{n+1}}{(n+1) 2^{n+1}} \).Now, compute the ratio: \[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{n+2}(x+2)^{n+1}}{(n+1) 2^{n+1}} \cdot \frac{n 2^n}{(-1)^{n+1}(x+2)^n} \right|\]Simplifying, we get \[\left| \frac{(x+2)(n)}{2(n+1)} \right| = \left| \frac{x+2}{2} \right| \cdot \frac{n}{n+1} \].

Step 3: Evaluate the limit as \( n \to \infty \)

As \( n \to \infty \), \( \frac{n}{n+1} \to 1 \). Therefore, \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x+2}{2} \right|\].Thus, for convergence, \[\left| \frac{x+2}{2} \right| < 1\]Solving gives the interval \(-2 < x+2 < 2\), or equivalently \(-4 < x < 0\).

Step 4: Identify the radius and interval of convergence

The interval of convergence is \(-4 < x < 0\). The radius of convergence is half the length of this interval, so the radius is \( R = 2 \).

Step 5: Determine absolute convergence

For absolute convergence, use the Ratio Test again on the series \( \sum \frac{(x+2)^n}{n2^n} \) without the alternating factor \((-1)^{n+1}\). The series converges absolutely in the same interval \(-4 < x < 0\) because absolute values don't affect the ratio \(\left| \frac{x+2}{2} \right| < 1\).

Step 6: Check endpoints for conditional convergence

Evaluate the convergence at the endpoints \(x = -4\) and \(x = 0\).- When \(x = -4\), the series becomes \(\sum \frac{(-1)^{n+1}(-2)^n}{n2^n} = \sum \frac{(-1)^n}{n}\), which converges by the Alternating Series Test.- When \(x = 0\), the series becomes \(\sum \frac{(-1)^{n+1}0^n}{n2^n} = 0\), trivially convergent.Thus, both endpoints must be checked; the series converges conditionally at \(x = -4\), but absolutely at \(x = 0\).

Key Concepts

Ratio Test

The Ratio Test is a method used to determine the convergence of a series. It's especially helpful for series involving powers of a variable. You use this test to find the radius of convergence of a power series, which tells us where the series converges absolutely.

Here's how the Ratio Test works. For a given series with terms \( a_n \), consider the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L \) is infinity, the series diverges.
• If \( L = 1 \), the test is inconclusive.

In the context of the given problem, the Ratio Test helps determine the values of \( x \) for which the series converges by analyzing the expression \( \left| \frac{x+2}{2} \right| < 1 \). Solving this inequality gives us the interval \(-4 < x < 0\), creating a boundary for convergence, with a computed radius of convergence \( R = 2 \).

Absolute Convergence

Absolute convergence of a series means that when you take the absolute value of all terms, the series still converges. This is stronger than regular convergence.

For absolute convergence, particularly in power series like the one given, we use the same steps as when applying the Ratio Test but we ignore any alternating terms like \((-1)^{n+1}\). This simplifies the series and ensures that only positive terms influence the ratio. In the exercise, it reveals that the series converges absolutely over \(-4 < x < 0\) as any potential negative terms from the alternating sign disappear. This indicates that without the negative signs flipping the series, convergence still holds over this interval.

Conditional Convergence

A series is said to converge conditionally if it converges but does not do so absolutely. This often involves alternating series where the convergence is heavily reliant on the change of signs.

In the exercise provided, after determining the absolute convergence interval, the next step was to evaluate the endpoints \( x = -4 \) and \( x = 0 \). With conditional convergence, the property of alternating signs can play a crucial role in finite sums. At \( x = -4 \), the given series \( \sum \frac{(-1)^n}{n} \), known as the Alternating Harmonic Series, converges but not absolutely. This means the convergence is conditional since removing the alternating sign would result in the divergent harmonic series.

Alternating Series Test

The Alternating Series Test is a convergence test used specifically for series where the terms change signs in a regular pattern. This method is useful when you encounter terms like \((-1)^{n+1}\) in your series.

The test states that if you have a series \( \sum (-1)^n b_n \), and if it meets the following criteria:

• \( b_n > 0 \) for all \( n \)
• \( b_{n+1} \leq b_n \) (terms are decreasing)
• \( \lim_{n \to \infty} b_n = 0 \)

Then, the series converges.

In the exercise, using the Alternating Series Test at \( x = -4 \) confirmed the series' convergence because \( b_n = \frac{1}{n} \) is positive, decreasing, and approaches zero.