Partial Sums for Geometric Series

For $r\ne 1$

$S_n=\frac{a}{1-r}(1-r^{n+1})$ Alternatively, ${\sum }_{i=0}^n{a}^i=\frac{1-a^{n+1}}{1-a}$

For $r=1$

$S_n=(n+1)(a)$

For $n\to \infty$

$S_{\infty }=\frac{a}{1-r}$