Answer
Proved.
Work Step by Step
If $r_1$ is the common ratio with $|r_1|<1$ and $a_1$ is the first term of geometric series, then
$\sum^{\infty}_{n=1}a_1r_1^n=\frac{a_1}{1-r_1}$ if $|r_1|<1$
For $\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}+...$
$a_1=\frac{1}{r}$ and $r_1=\frac{1}{r}$
Therefore,
$\sum^{\infty}_{n=1}(\frac{1}{r})(\frac{1}{r})^n=\frac{\frac{1}{r}}{1-\frac{1}{r}}$ if $|\frac{1}{r}|<1$
$\Rightarrow \frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}+...=\frac{1}{r-1}$ if $|r|>0$ since $\frac{a}{b}<1\Rightarrow \frac{b}{a}>1 $
