To find the sum of an infinite geometric series, I need to know the first term and the common ratio. Let the first term be $a$ and the common ratio be $r$. Then the second term is $ar$, and the sum is $\frac{a}{1-r}$. I know that $ar = 1$, so I can substitute that into the formula for the sum and get $\frac{1}{r(1-r)}$. To minimize this expression, I want to make the denominator as large as possible. The denominator is a quadratic function of $r$, and it has zeros at $r = 0$ and $r = 1$. The vertex of the parabola is at $r = \frac{1}{2}$, and that is where the denominator reaches its maximum value of $\frac{1}{4}$. Therefore, the smallest possible value of $S$ is when $r = \frac{1}{2}$, and it is $S = \frac{1}{\frac{1}{4}} = 4$. 
The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S?$