# Question #83975

Dec 5, 2014

Geometric Series

$a + a r + a {r}^{2} + \cdots = \frac{a}{1 - r} \text{ }$ if $\text{ } - 1 < r < 1$

Since

$b \left(1 + b + {b}^{2} + \cdots\right) = b + {b}^{2} + {b}^{3} + \cdots$

is a geometric series with $\left\{\begin{matrix}a = b \\ r = b\end{matrix}\right.$, its sum $S$ can be found by

$S = \frac{b}{1 - b} \text{ }$ if $\text{ } - 1 < b < 1$.

For other values of $b$, the series diverges.

I hope that this was helpful.