**Proof** If $|r|<1$ then $\sum_{n=0}^{\infty} r^{n} = \frac{1}{1-r}$ Indeed, the partial sums are $\sum_{n=0}^{k} r^{n} = \frac{1-r^{k+1}}{1-r}. $Provided that $r \neq 1$. We can show by induction that $(1-r)(1+r+r^{2}+\dots+r^{k}) = 1-r^{k+1}$ We can then take limits as $k \to \infty$ to show that the partial sums converge.