$\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n = 1 + \frac{1}{3} + \frac{1}{9} + \cdots \infty$
Step-by-Step Solution
Key Concept: Apply the infinite geometric series formula $S = \frac{a}{1-r}$ when $|r| < 1$.
This is an infinite geometric series with first term $a = 1$ and common ratio $r = \frac{1}{3}$. Since $|r| < 1$, the series converges. Using the formula for the sum of an infinite geometric series: $S = \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$.
Correct Answer: 1