Question

In Exercises \(101-104,\) find the values of \(x\) for which the series converges. $$ \sum_{n=0}^{\infty} 2\left(\frac{x}{3}\right)^{n} $$

Short answer

The series converges for all \(x\) in the interval \(-3 < x < 3\).

Step-by-step solution

- Identify the series

The series provided is \(\sum_{n=0}^{\infty} 2\left(\frac{x}{3}\right)^{n}\). As is evident, this is a geometric series with a common ratio of \(\frac{x}{3}\) and a first term of \(2\). The general term is \(2\left(\frac{x}{3}\right)^n\).

- Apply the geometric series test

A geometric series \(\sum ar^n\) will converge if \(|r| < 1\). The general test for this is: if \(|r| >= 1\), then the series will diverge. In our case, \(r = \frac{x}{3}\). To apply the geometric series test, we set \(|\frac{x}{3}| < 1\).

- Solve the inequality

We solve the inequality \(|\frac{x}{3}| < 1\) by breaking it into two inequalities depending on the sign of \(x\). The solution to the inequalities is \(x < 3\) and \(x > -3\). Thus, the series converges for all \(x\) in the interval \(-3 < x < 3\).