In an infinite geometric sequence the fourth term is 18, what is the smallest possible sum of all the terms?

Question

In an infinite geometric sequence the fourth term is 18, what is the smallest possible sum of all the terms?

My steps

If $s=\frac{a}{1-r}$ we can do the following:

$$\frac{a}{1-r}*\frac{r^3}{r^3}=\frac{ar^3}{r^3-r^4}$$

and then we do substitution to get the following:

$$\frac{18}{r^3(1-r)}$$

so then I thought that in order to make the sum the smallest possible number we ought to make the denominator super small, so then I know that the "zeroes" of $r^3-r^4$ are $0$ & $1$ so then the minimum must be at when x=0.5 but then according to a "math genius" he said that I was wrong.

How do I actually go about solving this problem?

Answer 1

Observe that $|r| < 1$, and look at $f(r) = r^3 - r^4\implies f'(r) = 3r^2 - 4r^3 = r^2(3-4r) = 0 \iff r = 0,\frac{3}{4}=0.75$, and $f''(r) = 6r-12r^2\implies f''(0) = 0, f''(\frac{3}{4}) = -\dfrac{9}{4} < 0\implies r = \frac{3}{4}$ yields a maximum by the $2^{\text{nd}}$ derivative test $\implies S_{\text{min}} = \dfrac{18}{0.75^3 - 0.75^4}= ...$

Answer 2

The minimum of $r^3-r^4$ on the interval $(0,1)$ is not at $1/2$. To find it, you can use calculus (solve $f'(x)=0$ where $f(x)=x^3-x^4$). Also, are you allowing $r$ to be negative? If you are, then for $r$ negative and near zero, the sum is dominated by the first term which is negative, and has large absolute value.