2016 AMC 10B Problem 16

1. Call the first term $a$ and the ratio $r$. For the sum to make sense, $r$ must be between $-1$ and $1$.
2. The second term is $a \cdot r = 1$, so the first term is $a = \frac1r$.
3. The sum of an infinite geometric series is $\frac{\text{first term}}{1 - r}$, so
$$S = \frac{1/r}{1 - r} = \frac{1}{r(1 - r)}.$$
4. $S$ is positive, so $r(1 - r)$ must be positive. With $r$ between $-1$ and $1$, that means $r$ is between $0$ and $1$.
5. $S$ is smallest when the bottom, $r(1 - r)$, is largest. Try a few values:

It looks largest at $r = \frac12$. To be sure, rewrite it:
$$r(1 - r) = \tfrac14 - \left(r - \tfrac12\right)^2.$$
A square is never negative, so this is never more than $\frac14$, and it equals $\frac14$ exactly at $r = \frac12$.
6. So $S = \frac{1}{r(1 - r)}$ is at least $\frac{1}{1/4} = 4$.
7. And $S = 4$ really happens: with $r = \frac12$ the series is $2 + 1 + \frac12 + \frac14 + \cdots = 4$, and its second term is $1$.

The answer is $\boxed{\textbf{(E)}\ 4}$.

Fixing the second term links the first term and the ratio, so $S$ depends on $r$ alone. Making a fraction small means making its bottom big, and a product like $r(1 - r)$, whose two parts add to $1$, is biggest when the two parts are equal.

Use the fact that for two positive numbers with a fixed sum, the product is largest when they are equal. Here $r$ and $1 - r$ always add to $1$, so their product is at most $\frac12 \cdot \frac12 = \frac14$. Then $S = \frac{1}{r(1 - r)} \ge 4$, with $S = 4$ at $r = \frac12$.