Geometric Series

A very special type of series is called a geometric series. It is the sum of some constant times another constant raised to varying exponents.

$$\displaystyle \sum^{\infty}_{n=0}k(r)^n$$

For example, $\sum^{\infty}_{n=0}2(1/3)^n$ is a geometric series with $k=2$ and $r=1/3$. So is the following series:

$$\displaystyle \sum^{\infty}_{n=0}\displaystyle \frac{1}{e^{3n}}= \displaystyle \sum^{\infty}_{n=0}\left(\displaystyle \frac{1}{e^3}\right)^n$$

where $k=1$ and $r=1/e^3$.

Geometric series converge if and only if $r<1$. If the indices start at $n=0$ we have proven that the series adds up to

$$\displaystyle \sum^{\infty}_{n=0}k(r)^n=\displaystyle \frac{k}{1-r}$$

If the indices start at $n=k$ for some $k>0$ then we must subtract $a_0+a_1+a_2+\dots+a_{k-1}$ from $k/(1-r)$. For example,

$$\sum^{\infty}_{n=2}2(1/3)^n = \displaystyle \frac{2}{1-1/3}-2(1/3)^0-2(1/3)$$

$$= 3-2(1)-2/3$$

$$= 1/3$$