smc

The sum of the geometric sequence is $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar=1\Rightarrow a=\frac{1}{r}.$ This means, $$S=\frac{a}{1-r}=\frac{1/r}{1-r}=\frac{1}{r(1-r)}.$$ In order to minimize $S,$ we maximize the denominator. By AM-GM, $$\frac{(r)+(1-r)}{2}\ge \sqrt{r(1-r)}\Rightarrow \frac{1}{4}\ge r(1-r).$$ Equality occurs at $r=1-r\Rightarrow r=\frac{1}{2}.$ This gives the minimum value of $S$ as $\frac{\frac{1}{1/2}}{1-\frac{1}{2}}=\boxed{4}.$