Problem 23 - Calculus: One and Several Variables

Question

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

Step-by-Step Solution

Answer

The interval of convergence for the given series is \(-\frac{3}{2} \leq x \leq \frac{3}{2}\).

Step 1: Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test on the given series: \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \] with \(a_k = (-1)^k \frac{2^k}{3^{k+1}} x^k\).

Step 2: Calculate the Ratio of Consecutive Terms

Divide the term \(a_{k+1}\) by \(a_k\): \[ \left| \frac{(-1)^{k+1} \frac{2^{k+1}}{3^{k+2}} x^{k+1}}{(-1)^k \frac{2^k}{3^{k+1}} x^k} \right| \]

Step 3: Simplify the Expression

Simplify the expression to get: \[ \lim_{k \to \infty} \left|\frac{2}{3}x\right| \]

Step 4: Apply the Ratio Test Condition for Convergence

For the series to converge, the result of the Ratio Test must be less than 1, so: \[ \left|\frac{2}{3}x\right| < 1 \]

Step 5: Find the Interval of Convergence

Solve the inequality and express the interval as a range for x: \[ -1 < \frac{2}{3}x < 1 \] Divide by \(\frac{2}{3}\) on each side: \[ -\frac{3}{2} < x < \frac{3}{2} \] So we now have the potential interval of convergence: \(-\frac{3}{2} < x < \frac{3}{2}\).

Step 6: Test the Endpoints of the Interval

Consider the endpoints \(x = -\frac{3}{2}\) and \(x = \frac{3}{2}\). Substitute these values into the original series and check for convergence/divergence: 1) \(x = -\frac{3}{2}\): \[ \sum (-1)^k \frac{2^k}{3^{k+1}} \left(-\frac{3}{2}\right)^k \] This is an alternating series. Since the terms are decreasing and the limit of their absolute value is zero, the series converges at \(x = -\frac{3}{2}\). 2) \(x = \frac{3}{2}\): \[ \sum (-1)^k \frac{2^k}{3^{k+1}} \left(\frac{3}{2}\right)^k \] This series converges due to the alternating sign and decreasing terms, similar to the previous case.

Step 7: State the Interval of Convergence

Since the series converges at both endpoints, the interval of convergence is: \[ -\frac{3}{2} \leq x \leq \frac{3}{2} \]

Key Concepts

Ratio Test

The Ratio Test is a powerful method for determining the convergence of infinite series. Its main strength lies in its ability to identify the radius and, consequently, the interval of convergence for power series. For a given series \(\sum a_k\), the Ratio Test considers the limit:

• Find the limit \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \]
• If the limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In our case, we used the Ratio Test on \(\sum(-1)^{k} \frac{2^{k}}{3^{k+1}} x^{k}\). By simplifying the ratio of consecutive terms, we found that the series converges if \[ \left| \frac{2}{3}x \right| < 1 \]. This leads us to the radius of convergence.

Alternating Series

An alternating series is a series whose terms alternate in sign. This occurs because each consecutive term is multiplied by \(-1\), making the sequence flip back and forth between positive and negative values. This type of series can converge even if its non-alternating counterpart does not.

• The series \(\sum(-1)^{k} \frac{2^{k}}{3^{k+1}} x^{k}\) is an example of an alternating series.
• We used the Alternating Series Test in this series, which states that if the terms \(|a_k|\) are decreasing and tend to zero, then the series converges.

The presence of \((-1)^k\) ensures that our series swaps signs regularly, contributing to convergence at specific points within the interval.

Radius of Convergence

The radius of convergence is a central concept when working with power series. It defines the region around the center of the series where the series converges. Once you have the radius, finding the interval becomes a straightforward task. To determine the radius of convergence:

• Apply the Ratio Test and solve \(\lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1\), which in our case simplified to \(\left| \frac{2}{3}x \right| < 1\)
• From this, the radius is computed by solving \(\left| x \right| < \frac{3}{2}\).

Thus, the radius of convergence in our exercise is \frac{3}{2}\. This is the distance from the center, usually at zero, to the endpoints of convergence.

Series Convergence

Series convergence is about whether a sequence's infinite sum reaches a finite limit. Recognizing convergence in a series is a key skill in calculus and analysis. For convergence:

• The terms of the series must approach zero as \(k\) increases.
• Tests like the Ratio Test or Alternating Series Test can confirm convergence.

In today's problem, we confirmed convergence for our series in the interval \(-\frac{3}{2} \leq x \leq \frac{3}{2}\). Both endpoints, \(x = -\frac{3}{2}\) and \(x = \frac{3}{2}\), were tested by substituting into the original series. The alternating nature and the fact that terms diminish to zero confirmed that the series converges everywhere within this range. By understanding and applying these concepts, one can easily determine convergence intervals for complex series.