2016 AMC 12B Problem 14

With ratio $r$ and second term $1,$ the first term is $\frac{1}{r},$ so $S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}$

$S$ is smallest when $r-r^2$ is largest; maximize the downward parabola $r-r^2$

Let $r$ be the common ratio. Since the second term is $1,$ the first term is $\dfrac1r,$ so $S=\dfrac{\frac{1}{r}}{1-r}=\dfrac{1}{r-r^2}.$ Convergence requires $|r|\lt1,$ and $S\gt0$ then forces $0\lt r\lt1.$ Therefore $S$ is smallest when $r-r^2$ is largest. The parabola $r-r^2$ peaks at $r=\tfrac12,$ where it equals $\tfrac14,$ so the smallest value of $S$ is $\dfrac{1}{\frac{1}{4}}=4.$

Thus, the correct answer is E.