In the textbook I'm working from, the formula for the sum of a geometric progression is given as
$S^n=\frac{a(1-r^n)}{1-r}$
It then adds that the formula may also be written as
$S^n=\frac{a(r^n-1)}{r-1}$
...if $r>1$
I can't work out why we would require $r>1$ to simply multiply the formula by $\frac{-1}{-1}$. Obviously we need $r\neq 1$ but that's true in both cases, not just the alternate form. Could someone shed some light on this, or else just confirm this is a mistake?