Geometric Series Geometric Series

Introduction

A geometric series is of the form:

\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + \ldots
where
a
and
r
are real numbers.

The following are geometric series:
\begin{align*} &\sum_{n=0}^{\infty} 2(0.5)^n &=&\, 2 + 1 + 0.5 + 0.25 + \ldots \\ &\sum_{n=0}^{\infty} 3(2)^n &=&\, 3 + 6 + 12 + 24 + \ldots \\ &\sum_{n=0}^{\infty} 4(-3)^n &=&\, 4 - 12 + 36 - 108 + \ldots \end{align*}

Convergence and Divergence

The series converges if:

|r| < 1
And diverges if:
|r| \geq 1

Sum of a Convergent Geometric Series

The sum of a convergent geometric series is:

\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}
Proof:
\begin{align*} S &= \sum_{n=0}^{N} ar^n \\ &= a + ar + ar^2 + ar^3 + \ldots + ar^N \\ rS &= ar + ar^2 + ar^3 + ar^4 + \ldots + ar^{N+1} \\ S - rS &= a + \left( ar^2 + ar^3 + \ldots + ar^N \right) - \left( ar^2 + ar^3 + \ldots + ar^{N+1} \right) \\ &= a - ar^{N+1} \\ S(1 - r) &= a(1-r^{N+1}) \\ S &= \frac{a(1-r^{N+1})}{1 - r} \end{align*}
\begin{align*} \text{Note that if }\, |r| &< 1 \\ \text{} \\ \lim_{N \to \infty} r^{N+1} &= 0 \\ \text{} \\ S &= \frac{a}{1 - r} \end{align*}

Examples



Example 1: Basic Convergent Series
Find the sum of the geometric series

\sum_{n=0}^{\infty} 3 \left(\frac{1}{2}\right)^n
.
\begin{align*} a &= 3 \\ r &= \frac{1}{2} \\ S &= \frac{a}{1 - r} \\ &= \frac{3}{1 - \frac{1}{2}} \\ &= \frac{3}{\frac{1}{2}} = 6 \end{align*}
The sum of the geometric series is 6.

Example 2: Divergent Geometric Series
Does the geometric series
\sum_{n=0}^{\infty} 2(3)^n
converge or diverge?
\begin{align*} a &= 2 \\ r &= 3 \\ |r| &= 3 \geq 1 \end{align*}
Since
|r| \geq 1
, the geometric series diverges. Therefore, it does not have a finite sum.

Example 3: Repeating Decimal as Geometric Series
Express the repeating decimal
0.\overline{37}
as a fraction using a geometric series.
\begin{align*} 0.\overline{37} &= 0.373737\ldots = \frac{37}{99} \\\\ &= \frac{37}{100} + \frac{37}{10000} + \frac{37}{1000000} + \ldots \\\\ &= \frac{37}{100} + \frac{37}{100}\left(\frac{1}{100}\right) + \frac{37}{100}\left(\frac{1}{100}\right)^2 + \ldots \\\\ &= \sum_{n=0}^{\infty} \frac{37}{100}\left(\frac{1}{100}\right)^n \end{align*}
Therefore,
\begin{align*} S &= \frac{a}{1 - r} \\\\ &= \frac{\frac{37}{100}}{1 - \frac{1}{100}} \\\\ &= \frac{\frac{37}{100}}{\frac{99}{100}} \\\\ &= \frac{37}{99} \end{align*}