1. The problem asks to express the repeating decimal $3.999\ldots$ as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. 2. Let $x = 3.999\ldots$. 3. Multiply both sides by 10 to shift the decimal point: $$10x = 39.999\ldots$$ 4. Subtract the original $x$ from this equation: $$10x - x = 39.999\ldots - 3.999\ldots$$ 5. Simplify the left and right sides: $$9x = 36$$ 6. Divide both sides by 9 to solve for $x$: $$x = \frac{\cancel{9}x}{\cancel{9}} = \frac{36}{9}$$ 7. Simplify the fraction: $$x = 4$$ 8. Therefore, the value of $3.999\ldots$ as a fraction is $\frac{4}{1}$. This shows that $3.999\ldots$ is exactly equal to 4.