1. **State the problem:** Determine if the infinite series $$\sum_{n=1}^\infty \frac{2^n - 1}{3^n}$$ converges. 2. **Rewrite the series:** Split the sum into two separate series: $$\sum_{n=1}^\infty \frac{2^n}{3^n} - \sum_{n=1}^\infty \frac{1}{3^n}$$ 3. **Recognize geometric series:** Each series is geometric with common ratios: - For the first: $$r_1 = \frac{2}{3}$$ - For the second: $$r_2 = \frac{1}{3}$$ 4. **Recall convergence rule for geometric series:** A geometric series $$\sum_{n=1}^\infty ar^{n-1}$$ converges if $$|r| < 1$$. 5. **Check convergence:** Both $$|r_1| = \frac{2}{3} < 1$$ and $$|r_2| = \frac{1}{3} < 1$$, so both series converge. 6. **Sum of geometric series:** The sum from $$n=1$$ to infinity for $$ar^{n-1}$$ is $$\frac{a}{1-r}$$. 7. **Calculate each sum:** - First series: $$\sum_{n=1}^\infty \frac{2^n}{3^n} = \sum_{n=1}^\infty \left(\frac{2}{3}\right)^n = \frac{\frac{2}{3}}{1 - \frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$$ - Second series: $$\sum_{n=1}^\infty \frac{1}{3^n} = \sum_{n=1}^\infty \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$ 8. **Find the total sum:** $$\sum_{n=1}^\infty \frac{2^n - 1}{3^n} = 2 - \frac{1}{2} = \frac{3}{2}$$ **Final answer:** The series converges and its sum is $$\frac{3}{2}$$.