1. The problem asks to find the sum of the infinite geometric series: $\frac{3}{4} + \frac{3}{8} + \frac{3}{16} + \ldots$ 2. The formula for the sum $S$ of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r| < 1$) is: $$S = \frac{a}{1 - r}$$ 3. Identify the first term $a$ and the common ratio $r$: - $a = \frac{3}{4}$ - To find $r$, divide the second term by the first term: $$r = \frac{\frac{3}{8}}{\frac{3}{4}} = \frac{3}{8} \times \frac{4}{3} = \frac{1}{2}$$ 4. Since $|r| = \frac{1}{2} < 1$, the sum converges. 5. Substitute $a$ and $r$ into the sum formula: $$S = \frac{\frac{3}{4}}{1 - \frac{1}{2}} = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{4} \times 2 = \frac{3}{2}$$ 6. Therefore, the sum of the infinite geometric series is $\boxed{\frac{3}{2}}$.