geometric seriesGeometricSeries2002-01-03 17:47:492006-10-25 00:28:36Definitioninfinite geometric seriesinfinite series\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}A \emph{geometric series} is a series of the form
\begin{align*}
\sum_{i=1}^n ar^{i-1}
\end{align*}
(with $a$ and $r$ real or complex numbers). The partial sums of a geometric series are given by
\begin{align}
s_n=\sum_{i=1}^n ar^{i-1} = \frac{a(1 -r^n)}{1-r}.
\end{align}
An \emph{infinite geometric series} is a geometric series, as above, with $n \rightarrow \infty$. It is denoted by
\begin{align*}
\sum_{i=1}^\infty ar^{i-1}
\end{align*}
If $|r|\ge 1$, the infinite geometric series diverges. Otherwise it converges to
\begin{align}
\sum_{i=1}^\infty ar^{i-1} = \frac{a}{1-r}
\end{align}
Taking the limit of $s_n$ as $n \rightarrow \infty$, we see that $s_n$ diverges if $|r| \ge 1$. However, if $|r| < 1$, $s_n$ approaches (2).
One way to prove (1) is to take
\begin{align*}
s_n = a + ar + ar^2 + \cdots + ar^{n-1}
\end{align*}
and multiply by $r$, to get
\begin{align*}
r s_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^{n}
\end{align*}
subtracting the two removes most of the terms:
\begin{align*}
s_n - rs_n = a - ar^n
\end{align*}
factoring and dividing gives us
\begin{align*}
s_n = \frac{a(1 -r^n)}{1-r}
\end{align*}
$\square$