Problem 35 - Calculus: One and Several Variables

Question

Find the interval of convergence. $$\sum(-1)^{k}\left(\frac{2}{3}\right)^{k}(x+1)^{k}$$

Step-by-Step Solution

Answer

The interval of convergence for the power series \(\sum_{k=0}^{\infty} (-1)^k \left(\frac{2}{3}\right)^k (x + 1)^k\) is \((-2.5, 0.5)\).

Step 1: Identify the series to be analyzed

We are given a series of the form: \[ \sum_{k=0}^{\infty} (-1)^k \left(\frac{2}{3}\right)^k (x + 1)^k \] We will now apply the Ratio Test to analyze its convergence.

Step 2: Apply the Ratio Test

The Ratio Test states that the limit of the ratio of consecutive terms can be used to determine the convergence of a series. The test is given by: \[ L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \] If \( L < 1 \), the series converges. If \( L > 1 \), the series diverges. If \( L = 1 \), the test is inconclusive.

Step 3: Calculate the ratio of consecutive terms

To apply the Ratio Test, we need to find the ratio of consecutive terms. We will calculate \(\frac{a_{k+1}}{a_k}\) using the given series. \[ \frac{a_{k+1}}{a_k} = \frac{(-1)^{k+1} \left(\frac{2}{3}\right)^{k+1} (x + 1)^{k+1}}{(-1)^k \left(\frac{2}{3}\right)^k (x + 1)^k} \] Simplify by canceling out terms: \[ \frac{a_{k+1}}{a_k} = \frac{(-1)^{k+1}}{(-1)^k} \cdot \frac{\left(\frac{2}{3}\right)^{k+1}}{\left(\frac{2}{3}\right)^k} \cdot \frac{(x + 1)^{k+1}}{(x + 1)^k} = -\left(\frac{2}{3}\right) (x + 1) \]

Step 4: Apply the Ratio Test

Now, apply the Ratio Test by taking the limit of the ratio of consecutive terms as k→∞: \[ L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left|- \left(\frac{2}{3}\right) (x + 1)\right| \] Since the limit is not dependent on k, the calculation simplifies to: \[ L = \left|\frac{2}{3}(x + 1)\right| \]

Step 5: Determine the interval of convergence

For the series to converge, we need the limit L to be less than 1. Therefore, \[ \left|\frac{2}{3}(x + 1)\right| < 1 \] To find the interval of convergence, solve the inequality for x: \[ \frac{2}{3}(x + 1) < 1 \quad \text{and} \quad -\frac{2}{3}(x + 1) < 1 \\ 2(x + 1) < 3 \quad \text{and} \quad -(2)(x + 1) < 3 \] Divide by 2: \[ x + 1 < \frac{3}{2} \quad \text{and} \quad -(x + 1) < \frac{3}{2} \] Now, solve each inequality for x: \[ x < \frac{1}{2} \quad \text{and} \quad x > -\frac{5}{2} \] The interval of convergence for the series is \((-2.5, 0.5)\): \[ -2.5 < x < 0.5 \] In conclusion, the interval of convergence for the given power series is \((-2.5, 0.5)\).

Key Concepts

Power Series

In mathematics, a power series is a series in the form of \(\sum_{k=0}^{\infty} a_k (x - c)^k\), where \(a_k\) are the coefficients of the series, \(x\) is the variable, and \(c\) is the center of the series. The power series can represent a function within an interval around the center, known as the interval of convergence.

The power series in the given exercise is \(\sum (-1)^{k}\left(\frac{2}{3}\right)^{k}(x+1)^{k}\), which can be seen as centered around \(c = -1\), with \(a_k = (-1)^{k}\left(\frac{2}{3}\right)^{k}\). Understanding power series is crucial because they allow for functions to be expressed in a form that makes them easier to analyze, differentiate, and integrate. Furthermore, they are indispensable in fields such as physics and engineering, where they are used for approximation and solving differential equations.

Ratio Test

The Ratio Test is a method to determine the convergence of a series by examining the ratio of consecutive terms. Specifically, the test involves calculating the limit \(L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|\) where \(a_k\) is the \(k\)-th term of the series.

If \(L < 1\), the series converges absolutely; if \(L > 1\), it diverges; and if \(L = 1\), the test is inconclusive. In our exercise, after simplification, we found that \(L = \left|\frac{2}{3}(x + 1)\right|\). The Ratio Test is particularly effective for series that resemble geometric series or when terms have factorial expressions. It's also useful when terms include exponents that depend on the variable \(k\) because it provides a clear criteria for convergence or divergence.

Series Convergence

Series convergence is the idea that as you add more and more terms of a series, the sum approaches a finite value. In the case of the power series, we want to find the values of \(x\) for which the series converges to a sum.

To find the interval of convergence for the series in the exercise, we used the Ratio Test. The inequalities derived, \(x < \frac{1}{2}\) and \(x > -\frac{5}{2}\), established the boundaries of our interval \( (-2.5, 0.5) \). This interval is where the series approaches a finite value as \(k\) increases without bounds. Series convergence is not just a theoretical interest; in practice, it allows for the computation of sums of infinitely many terms, which is essential in various physics, engineering, and economics applications to represent complex phenomena with simpler expressions.