Geometric series
$$ \sum_{i = 0}^{n-1} a_{i} = a(\frac{1-r^n}{1-r}) $$
For |r| < 1
$$ \sum_{i = 0}^{\infty} a_{i} = (\frac{a}{1-r}) $$
$$ \sum_{i = 0}^{n-1} a_{i} = a(\frac{1-r^n}{1-r}) $$
For |r| < 1
$$ \sum_{i = 0}^{\infty} a_{i} = (\frac{a}{1-r}) $$