2016 AMC 12B Problems/Problem 14

Problem

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 $S?$

$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

The second term in a geometric series is $a_2 = a \cdot r$ , where $r$ is the common ratio for the series and $a$ is the first term of the series. So we know that $a\cdot r = 1$ and we wish to find the minimum value of the infinite sum of the series. We know that: $S_\infty = \frac{a}{1-r}$ and substituting in $a=\frac{1}{r}$ , we get that $S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}$ . From here, you can either use calculus or AM-GM.

Calculus: Let $f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}$ , then $f'(x) = -(x-x^2)^{-2}\cdot (1-2x)$ . Since $f(0)$ and $f(1)$ are undefined $x \neq 0,1$ . This means that we only need to find where the derivative equals $0$ , meaning $1-2x = 0 \Rightarrow x =\frac{1}{2}$ . So $r = \frac{1}{2}$ , meaning that $S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4$

AM-GM For 2 positive real numbers $a$ and $b$ , $\frac{a+b}{2} \geq \sqrt{ab}$ . Let $a = \frac{1}{r}$ and $b = \frac{1}{1-r}$ . Then: $\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}$ . This implies that $\frac{S_\infty}{2} \geq \sqrt{S_\infty}$ .

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2016 AMC 12B Problems/Problem 14

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