# The Formula

The geometric series $$\sum\limits_{n=0}^{\infty}ar^{n}$$ converges when $$|r|\le 1$$ and sums to $$\frac{a}{1-r}$$

# Examples

$$\sum\limits_{n=0}^{\infty}3(\frac{3}{2})^{n}$$
Since $$r \ge 1$$ , this series diverges .
$$\sum\limits_{n=0}^{\infty}5(-1.05)^{n}$$
Since $$r \ge 1$$ , this series also diverges .
$$\sum\limits_{n=0}^{\infty}\frac{7}{3}(\frac{1}{3})^{n}$$
$$\frac{\frac{7}{3}}{1-\frac{1}{3}} = \boxed{ \frac{7}{2} }$$ Since $$r \lt 1$$ , this series converges .
$$\sum\limits_{n=1}^{\infty}\frac{1}{e^{n}}$$
Remember $$\sum\limits_{n=1}^{\infty}\frac{1}{e^{n}} \text{can be rewritten} \sum\limits_{n=1}^{\infty}1(\frac{1}{e})^{n}$$
$$r \le 1$$, converges, $$\frac{\frac{1}{e}}{1-\frac{1}{e}}= \boxed{ \frac{1}{e-1} }$$