3.2 Geometric Series

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We find the sum of a geometric series.

Definition of Infinte Series

Infinite Series We define $\sum _{n=1}^\infty a_n = \lim _{N \to \infty } \sum _{n=1}^N a_n.$ If this limit is a finite number, then we say that the infinite series converges. If this limit is not finite or if it does not exist, then we say that the infinite series diverges.

The sum, $\displaystyle {\sum _{n=1}^N a_n}$ , is called the $N^{th}$ partial sum of the infinite series $\displaystyle {\sum _{n=1}^\infty a_n}$ . Thus, in words, we define an infinite series to be the limit of its partial sums. This is similar to an improper integral which is also defined in terms of a limit.

Geometric Series

Geometric Series A geometric series is an infinite series of the form $\sum _{n=0}^\infty ar^n = a + ar + ar^2 + ar^3 + \cdots .$ The parameter $r$ is called the common ratio.

example 1 Consider the series: $\sum _{n=0}^\infty \frac {3}{2^n}$ Write out the first four terms of the series. If the series is geometric, find the common ratio.

The first four terms are $3 + \frac 32 + \frac 34 + \frac 38 + \cdots .$

The series is geometric and the common ratio is $\frac 12$ .

(problem 1a) Write the first four terms of the infinite series and answer the questions below: $\sum _{n=0}^\infty \frac {2}{3^n}$

True or false: this is a geometric series TrueFalse

The first term is $a = \answer {2}$
The common ratio is $r = \answer {1/3}$

(problem 1b) Write the first four terms of the infinite series and answer the questions below: $\sum _{n=2}^\infty \frac {3}{2^n}$

True or false: this is a geometric series TrueFalse

The first term is $a = \answer {3/4}$
The common ratio is $r = \answer {1/2}$

example 2 Consider the series: $\sum _{n=1}^\infty \frac {3}{n^2}$ Write out the first four terms of the series. If the series is geometric, find the common ratio.

The first four terms are $3 + \frac 34 + \frac 39 + \frac {3}{16} + \cdots .$ The ratio of the second term to the first is $\frac 14$ but the ratio of the third term to the second is $\frac 49$ . The series is not geometric because it fails to have a common ratio.

(problem 2a) Write the first four terms of the infinite series and answer the questions below: $\sum _{n=1}^\infty \frac {1}{n}$

True or false: this is a geometric series TrueFalse

(problem 2b) Write the first four terms of the infinite series and answer the questions below: $\sum _{n=0}^\infty \sqrt n$

True or false: this is a geometric series TrueFalse

Convergence

Convergence of Geometric Series The geometric series $\sum _{n=0}^\infty ar^n$ converges if and only if the common ratio $r$ satisfies the condition $-1 < r < 1$ . This condition can also be written as $|r| < 1$ .

example 3 Does the geometric series $\displaystyle {\sum _{n=0}^\infty \frac {3}{2^n}}$ converge or diverge?
The common ratio of this geometric series is $r = \frac 12$ . Since $-1 < \frac 12 < 1$ , the series converges.

(problem 3a) Does the geometric series $\sum _{n=2}^\infty \frac {3}{2^n}$ converge or diverge?
The common ratio is $r = \answer {1/2}$
The series

converges diverges

(problem 3b) Does the geometric series $\sum _{n=1}^\infty (-1)^n\frac {2}{3^n}$ converge or diverge?
The common ratio is $r = \answer {-1/3}$
The series

converges diverges

(problem 3c) Does the geometric series $\sum _{n=0}^\infty (-2)^n$ converge or diverge?
The common ratio is $r = \answer {-2}$
The series

converges diverges

example 4 Does the geometric series $\displaystyle {\sum _{n=0}^\infty \frac {3^n}{2^{n+1}}}$ converge or diverge?
The common ratio of this geometric series is $r = \frac 32$ . Since $\frac 32$ is not between $-1$ and $1$ , the series does not converge, i.e., it diverges.

(problem 4a) Does the geometric series $\sum _{n=0}^\infty \frac {2^{n+1}}{3^n}$ converge or diverge?
The common ratio is $r = \answer {2/3}$
The series

converges diverges

(problem 4b) Does the geometric series $\sum _{n=0}^\infty \frac {2^{2n}}{3^n}$ converge or diverge?
The common ratio is $r = \answer {4/3}$
The series

converges diverges

example 5 Does the geometric series $\displaystyle {\sum _{n=0}^\infty \left (-\frac 23\right )^n }$ converge or diverge?
The common ratio of this geometric series is $r = -\frac 23$ . Since $-1 < -\frac 23 < 1$ , the series converges.

(problem 5) Does the geometric series $\sum _{n=0}^\infty \left (-\frac 56 \right )^n$ converge or diverge?
The common ratio is $r = \answer {-5/6}$
The series

converges diverges

Sum of a Geometric Series

Consider the geometric series $\sum _{n=0}^\infty ar^n$ where $|r| < 1$ (so that the series converges). The $N^{th}$ partial sum of this series is given by $S_N = \sum _{n=0}^N ar^n = a + ar + ar^2 + \cdots + ar^{N-1} + ar^N.$ Multiply both sides by $r$ : $rS_N = ar + ar^2 + + ar^3 + \cdots + ar^N + ar^{N+1}.$ Now subtract $rS_N$ from $S_N$ :

$\begin{align*} S_N - rS_N &= \left(a + ar + \cdots + ar^{N-1} + ar^N\right) - \left(ar + ar^2 + \cdots + ar^N + ar^{N+1} \right)\\ &= a - ar^{N+1}. \end{align*}$ We can factor out $S_N$ on the left side and then divide by $1-r$ to obtain $S_N = \frac {a - ar^{N+1}}{1-r}.$ We can now compute the sum of the geometric series by taking the limit as $N \to \infty$ : $S = \sum _{n=0}^\infty ar^n = \lim _{N \to \infty } \sum _{n=0}^N ar^n = \lim _{N \to \infty } S_N = \lim _{N \to \infty } \frac {a - ar^{N+1}}{1-r} = \frac {a}{1-r}.$ We present this formula in the theorem below.

Sum of a Geometric Series If $-1 < r < 1$ , then the sum of the geometric series $\sum _{n=0}^\infty ar^n = a + ar + ar^2 + \cdots$ is given by $S = \frac {a}{1-r}.$ That is, if $-1 < r < 1$ , then $\sum _{n=0}^\infty ar^n = \frac {a}{1-r}.$

example 6 Find the sum of the geometric series $\sum _{n=0}^\infty \frac {3}{2^n}.$ The first term is $a = 3$ , and the common ratio is $r = \frac 12$ . The sum of the series is $\sum _{n=0}^\infty \frac {3}{2^n} = \frac {3}{1-\frac 12} = \frac {3}{\left (\frac 12\right )} = 6.$

(problem 6a) Find the sum of the geometric series $\sum _{n=0}^\infty \frac {5}{3^n}.$ The first term is $a = \answer {5}$
The common ratio is $r = \answer {1/3}$
The sum is $S= \answer {15/2}$

(problem 6b) Find the sum of the geometric series $\sum _{n=1}^\infty \frac {4}{2^n}.$ The first term is $a = \answer {2}$
The common ratio is $r = \answer {1/2}$
The sum is $S= \answer {4}$

(problem 6c) Find the sum of the geometric series $\sum _{n=0}^\infty \frac {3^n}{2^{2n}}.$ The first term is $a = \answer {1}$
The common ratio is $r = \answer {3/4}$
The sum is $S= \answer {4}$

example 7 Find the sum of the geometric series $\sum _{n=0}^\infty \left (-\frac 23\right )^n.$ The first term is $a = 1$ , and the common ratio is $r = -\frac 23$ . The sum of the series is $\sum _{n=0}^\infty \left (-\frac 23\right )^n = \frac {1}{1-\left (-\frac 23\right )} = \frac {1}{1 + \frac 23} = \frac {1}{\left (\frac 53\right )} = \frac 35.$

(problem 7a) Find the sum of the geometric series $\sum _{n=1}^\infty \left (-\frac 45\right )^n$ The first term is $a = \answer {-4/5}$
The common ratio is $r = \answer {-4/5}$
The sum is $S= \answer {-4/9}$

(problem 7b) Find the sum of the geometric series $\sum _{n=2}^\infty \left (-\frac {2}{\pi }\right )^n$ The first term is $a = \answer {4/\pi ^2}$
The common ratio is $r = \answer {-2/\pi }$
The sum is $S= \answer {4/(\pi ^2 + 2\pi )}$

(problem 7c) Find the sum of the geometric series $\sum _{n=0}^\infty \left (\frac {2}{e}\right )^n$ The first term is $a = \answer {1}$
The common ratio is $r = \answer {2/e}$
The sum is $S= \answer {e/(e-2)}$

example 8 Write the repeating decimal $0.\overline {123} = 0.123123123...$ as a ratio of integers.
First, note that we can write this repeating decimal as an infinite series: $0.\overline {123} = 0.123 + 0.000123 + 0.000000123 + \dots$ Wring these decimals as fractions, we have $0.\overline {123} =\frac {123}{10^3} + \frac {123}{10^6}+\frac {123}{10^9} + \dots$ This is a convergent geometric series with first term, $a = \frac {123}{1000}$ and common ratio $r = \frac {1}{1000}$ . The sum is $S = \frac {a}{1-r} = \frac {\left (\frac {123}{1000}\right )}{1-\frac {1}{1000}} = \frac {\left (\frac {123}{1000}\right )}{\left (\frac {999}{1000}\right )} = \frac {123}{999} = \frac {41}{333}.$

(problem 8) Write the repeating decimal $0.\overline {45} = 0.454545...$ as a ratio of integers.

$\frac {9}{20}$ $\frac {45}{98}$ $\frac {5}{11}$

Video Lesson

Here is a detailed, lecture style video on Geometric Series: