Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=0}^{\infty} \frac{(2 x)^{n}}{3^{n}}$$

Short answer

The interval of convergence is \ ( -\frac{3}{2}, \frac{3}{2} \).

Step-by-step solution

- Identify the General Form

Write down the general form of the given power series. The given series is: \[\frac{(2x)^n}{3^n} = \bigg(\frac{2x}{3}\bigg)^n\]

- Apply the Ratio Test

The Ratio Test is often used to determine the interval of convergence. Compute the limit: \[ \lim_{{n \to \infty}} \frac{a_{n+1}}{a_n} \] where \ a_n = \bigg(\frac{2x}{3}\bigg)^n \.

- Simplify the Ratio

Substitute \ a_n = \bigg(\frac{2x}{3}\bigg)^n \ and \ a_{n+1} = \bigg(\frac{2x}{3}\bigg)^{n+1} \ into the Ratio Test limit: \[ \lim_{{n \to \infty}} \frac{\bigg(\frac{2x}{3}\bigg)^{n+1}}{\bigg(\frac{2x}{3}\bigg)^n} = \lim_{{n \to \infty}} \bigg(\frac{2x}{3}\bigg) = \bigg|\frac{2x}{3}\bigg| \]

- Set Up the Convergence Condition

According to the Ratio Test, the series converges if the limit of the ratio is less than 1: \[ \bigg|\frac{2x}{3}\bigg| < 1 \]

- Solve the Inequality

Solve the inequality \[ \bigg|\frac{2x}{3}\bigg| < 1 \] This leads to: \[ \bigg| 2x \bigg| < 3 \] or \[ -3 < 2x < 3 \] Then divide by 2: \[ -\frac{3}{2} < x < \frac{3}{2} \]

- Check the Endpoints

Check the convergence at the endpoints \ x = -\frac{3}{2} \ and \ x = \frac{3}{2} \ by substituting these values back into the original series: \For \ x = -\frac{3}{2} \:\[ \sum_{n=0}^{\infty} \bigg(\frac{2(-\frac{3}{2})}{3}\bigg)^n = \sum_{n=0}^{\infty} (-1)^n \] which is a divergent series (alternating series that does not tend towards 0).For \ x = \frac{3}{2} \:\[ \sum_{n=0}^{\infty} \bigg(\frac{2(\frac{3}{2})}{3}\bigg)^n = \sum_{n=0}^{\infty} 1^n \] which is also a divergent series.

- State the Interval of Convergence

Since the series does not converge at \ x = -\frac{3}{2} \ or \ x = \frac{3}{2} \, the interval of convergence is: \ \( -\frac{3}{2}, \frac{3}{2} \)

Key concepts

Power Series

A power series is an infinite series of the form $$\sum_{n=0}^{\infty} a_n (x - c)^n,$$ where

• \(a_n\) represents the coefficients,
• \(c\) is the center or the point where the series is expanded,
• and \(x\) is the variable.

Power series allow functions to be expressed as an infinite sum of terms involving those functions. This gives mathematicians a powerful tool to explore functions behavior over different intervals.
For the given exercise, the series is $$\sum_{n=0}^{\infty} \frac{(2x)^n}{3^n},$$ which is already centered at \(c=0\) and can be rewritten as $$\sum_{n=0}^{\infty} \left(\frac{2x}{3}\right)^n.$$ Understanding power series helps us to break down complex functions into simpler polynomials sum, analyzed within a specific interval of convergence.

Ratio Test

The Ratio Test is a method used to determine the convergence of a series. Given a series $$\sum_{n=0}^{\infty} a_n,$$ we examine the limit: $$\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|.$$ The test dictates:

• If the limit is less than 1, the series converges absolutely.
  
• If the limit is greater than 1, the series diverges.
  
• If the limit equals 1, the test is inconclusive, and further examination is needed.

For our power series $$\left(\frac{2x}{3}\right)^n,$$ we set $$a_n = \left(\frac{2x}{3}\right)^n.$$ Using the ratio test, we find the limit as $$\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{{n \to \infty}} \left| \frac{\left(\frac{2x}{3}\right)^{n+1}}{\left(\frac{2x}{3}\right)^n} \right| = \left| \frac{2x}{3} \right|.$$ Setting this less than 1, $$\left| \frac{2x}{3} \right| < 1.$$ With further simplification, we find the interval for \(x\) is $$-\frac{3}{2} < x < \frac{3}{2}.$$ This helps us understand where the series converges.

Convergence

Convergence in the context of power series implies that as \(n\) approaches infinity, the series sums to a finite value within a specific interval.
For the interval \(-\frac{3}{2} < x < \frac{3}{2}\), we check the endpoints to make sure if they belong to the convergence interval:
For \(x = -\frac{3}{2}\), substituting in the series gives $$\sum_{n=0}^{\infty} \left(\frac{2(-\frac{3}{2})}{3}\right)^n = \sum_{n=0}^{\infty} (-1)^n,$$ which is a divergent series.
For \(x = \frac{3}{2}\), substituting in the series gives $$\sum_{n=0}^{\infty} \left(\frac{2(\frac{3}{2})}{3}\right)^n = \sum_{n=0}^{\infty} 1^n,$$ which also diverges.
Thus, the series converges only within the open interval $$-\frac{3}{2} < x < \frac{3}{2}.$$ Knowing convergence is crucial for understanding where the power series behaves correctly and where it does not.