2016 AMC 10B Problem 16

Let the first term be $a$ and the ratio $r$, with $|r| < 1$ so the series converges. The second term is $ar = 1$, so $a = \frac{1}{r}$ and
$$S = \frac{a}{1 - r} = \frac{1}{r(1 - r)}.$$

For $S$ to be positive we need $r(1-r) > 0$, i.e. $0 < r < 1$. On that interval
$$r(1 - r) = \tfrac14 - \left(r - \tfrac12\right)^2 \le \tfrac14,$$
with equality at $r = \tfrac12$. A smaller denominator makes $S$ larger, so
$$S = \frac{1}{r(1-r)} \ge \frac{1}{1/4} = 4,$$
and $S = 4$ is attained by the series $2 + 1 + \tfrac12 + \tfrac14 + \cdots$.

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

Fixing the second term ties $a$ and $r$ together, leaving $S$ as a function of $r$ alone. Minimizing $\frac{1}{r(1-r)}$ is the same as maximizing the quadratic $r(1-r)$, whose vertex sits midway between its roots $0$ and $1$. Whenever a problem fixes a product and asks for an extreme of a sum-like quantity, expect a completed square or AM-GM to finish it.

By AM-GM, $\sqrt{r(1-r)} \le \frac{r + (1-r)}{2} = \frac12$, so $r(1-r) \le \frac14$ and $S \ge 4$, with equality when $r = 1 - r$.

• Assuming the first term is $1$ (so $S = \frac{1}{1-r}$ and "smallest" looks like $1$ or $2$) instead of the second term.
• Letting $r$ approach $0$ or $1$ to "shrink" $S$; both make the denominator $r(1-r)$ vanish and $S$ blow up, so the extreme is in the middle, not at the ends.