Problem 62 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the interval of convergence of the series. Explain your reasoning fully. \(\sum_{k=1}^{\infty}(-1)^{k} \frac{(x+1)^{k}}{3 k}\)

Step-by-Step Solution

Answer

The interval of convergence for the given series is (-2, 0).

Step 1: Define the series

The given series for this problem is \(\sum_{k=1}^{\infty}(-1)^{k} \frac{(x+1)^{k}}{3 k}\)

Step 2: Apply the Ratio Test

The Ratio Test is a converged test for infinite series and states that if the absolute value of the ratio of the (k+1) term to the kth term is less than 1, the series converges. The ratio is \[\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(-1)^{k+1} \frac{(x+1)^{k+1}}{3 (k+1)}}{(-1)^{k} \frac{(x+1)^{k}}{3 k}}\right| \]Simplify this to \(\left|\frac{k(x+1)}{k+1}\right|\)

Step 3: Determining the Interval of Convergence

The series converges if the ratio is less than 1, so set \(\left|\frac{k(x+1)}{k+1}\right| < 1\) and solve for x. The inequalities -1 < x+1 < 1 yield -2 < x < 0. This is the interval of convergence.

Key Concepts

Ratio Test

The Ratio Test is an essential tool when examining whether an infinite series converges. It's particularly useful here because it provides a straightforward way to find out if our series will converge based on the behavior of the terms as they progress. To apply the Ratio Test, we examine the limit of the absolute value of the ratio of consecutive terms in our series. For our series \[ \sum_{k=1}^{\infty} (-1)^{k} \frac{(x+1)^{k}}{3k} \]we observe each term and its subsequent term. The critical part is the limit calculation of \[ \left| \frac{a_{k+1}}{a_k} \right|. \]If this value is less than 1, the series converges. If it's more than 1, the series diverges. Lastly, if it equals 1, the test is inconclusive.This test is named so because it examines the ratio between terms, making it ideal for evaluating series like ours, where terms grow or shrink steadily.

Infinite Series

Infinite Series are essentially sums of infinite sequences. These sequences can be represented in various forms, such as arithmetic sequences or geometric sequences. In our case, the series is neither of these classic types since it involves powers of \((x+1)\) and an additional factor in the denominator.Understanding infinite series means recognizing how the terms might behave over time:

• Convergent series, where terms approach a limit as they progress.
• Divergent series, where terms grow without bound.

The terms themselves, like \((-1)^{k} \frac{(x+1)^{k}}{3k},\)add complexity because the power of \((-1)^{k}\)causes an alternating sign. Such alternating series can also have unique convergence properties.

Convergence Test

A Convergence Test is a method used to determine whether an infinite series converges. There are several tests available, but using the appropriate one is key. In our exercise, the Ratio Test is selected, but there are other tests such as:

• Integral Test
• Comparison Test
• Alternating Series Test
• Root Test

Each test has its own criteria and is suited for certain types of series. We use the Ratio Test here to determine the behavior of the interval convergence for our specific series.Applying a convergence test involves setting up inequalities to ascertain where the terms stabilize or if they oscillate. Solving these inequalities informs us of the interval on the x-axis where our series remains within bounds. For instance, simplifying our ratio inequality helps to isolate \(x\)to \(-2 < x < 0\)indicating the series converges in this region.