Second Term Is All You Need

The sum of infinite geometric progression with first term $a$ and common ratio $r$ is given by:

$\begin{aligned} S & = \frac a{1-r} & \small \color{#3D99F6}{\text{Multiply up and down of RHS by }r.} \\ & = \frac {ar}{r(1-r)} & \small \color{#3D99F6}{\text{Note that the second term is }ar=1} \\ & = \frac {\color{#3D99F6}{1}}{r(1-r)} \\ & = \color{#3D99F6}{\frac 1r + \frac 1{1-r}} & \small \color{#3D99F6}{\text{By AM-HM inequality (see Note)}} \\ & \ge \frac 4{r+1-r} \\ & \ge \boxed{4} \end{aligned}$

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$\color{#3D99F6}{\text{Note:}}$

We note that for $S>0$ , $\implies a >0$ and $r > 0$ and for the infinite GP to converge $r < 1$ . Therefore, $\dfrac 1r, \dfrac 1{1-r} > 0$ and AM-HM inequality applies.

Relevant wiki: Geometric Progression Problem Solving

The sum $S$ of an infinite geometric series is $S=\dfrac{a_1}{1-r}$ , where $a_1$ is the first term and $r$ is the common ratio between consecutive terms.

Since it is a geometric series, a term multiplied by the common ratio yields the consecutive term. So, $a_1\cdot r=a_2$ . It is given that $a_2=1$ ; thus $a_1=\dfrac{1}{r}$ .

Substituting that into the infinite geometric series sum formula, we have

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

We are asked to minimize $S$ . So, in order to minimize $\dfrac{1}{-r^2+r}$ we must maximize the denominator. The denominator, $-r^2+r$ , is a quadratic expression in the form $Ax^2+Bx+C$ , so its maximum/minimum (in this case maximum since $A<0$ ) occurs when $r=-\dfrac{B}{2A}=0.5$ .

Plugging $r=0.5$ back in the equation for $S$ , we get $S=\dfrac{1}{-0.5^2+0.5}=4$ . Therefore, the smallest possible value of $S$ is $\boxed{4}$ .