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.

$\textbf{Calculus}$

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 $S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4$

$\textbf{AM-GM}$

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 $\frac{S_\infty}{2} \geq \sqrt{S_\infty}$ . or $S_\infty^2 \geq 4 \cdot S_\infty$ . Rearranging : $(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4$ . Thus, the smallest value is $S_\infty = 4$ .

Solution 2

A geometric sequence always looks like

$a,ar,ar^2,ar^3,\dots$

and they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.

$\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}$

$\textbf{Solving in terms of \textit{a} then graphing}$

$S=\frac{a^2}{a-1}$

We seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{r} then graphing}$

$S=\frac{1}{-r^2+r}$

We seek the smallest positive value of $S$ . We proceed by graphing in the $rS$ plane and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{a} then calculus-ing}$

$S=\frac{a^2}{a-1}$

We seek the smallest positive value of $S$ . $\frac{(a-2)a}{(a-1)^2}=S'$ and $\frac{(a-2)a}{(a-1)^2}=0$ at $a=0$ and $a=2$ . $\frac{2}{(a-1)^3}=S''$ and $\frac{2}{(0-1)^3}$ is negative (implying a maximum occurs at $a=0$ ) and $\frac{2}{(2-1)^3}$ is positive (implying a minimum occurs at $a=2$ ). At $a=2$ , $S=\boxed{\textbf{(E)}\ 4}.$ This is the only positive minimum $S$ has, so it must be the answer.

$\textbf{Solving in terms of \textit{r} then being a clever cow}$

$S=\frac{1}{-r^2+r}$

We seek the smallest positive value of $S$ . We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at $x=\frac{1}{2}$ and $\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}$ .

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

Contents

Problem

Solution

Solution 2

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