8.2: Geometric Series

In Preview Activity $8.2.1$ we encountered the sum

for the long-term level of Warfarin in the patient's system. This sum has the form

where $a=5 \times 0.08$ and $r=0.08\text{.}$ Such a sum is called a finite geometric series with ratio $r\text{.}$

Activity $8.2.2$

Let $a$ and $r$ be real numbers (with $r \ne 1$) and let

$$S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.}$$

In this activity we will find a shortcut formula for $S_n$ that does not involve a sum of $n$ terms.

1. Multiply $S_n$ by $r\text{.}$ What does the resulting sum look like?
2. Subtract $rS_n$ from $S_n$ and explain why
   $$S_n - rS_n = a - ar^n\text{.}\tag{$$$8.2.2$}
3. Solve equation ($8.2.2$) for $S_n$ to find a simple formula for $S_n$ that does not involve adding $n$ terms.

The sum of the terms of a sequence is called a series. We summarize the result of Activity $8.2.2$ in the following way.

Note

A finite geometric series $S_n$ is a sum of the form

$$S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\tag{$$$8.2.3$}

where $a$ and $r$ are real numbers such that $r \ne 1\text{.}$ The finite geometric series $S_n$ can be written more simply as

$$S_n = a+ar+ar^2+ \cdots + ar^{n-1} = \frac{a(1-r^n)}{1-r}\text{.}\tag{$$$8.2.4$}

We now apply Equation ($8.2.4$) to the example involving Warfarin from Preview Activity $8.2.1$. Recall that

so $Q(n)$ is a geometric series with $a=5 \times 0.08 = 0.4$ and $r = 0.08\text{.}$ Thus,

Notice that as $n$ goes to infinity, the value of $0.08^n$ goes to 0. So,

Therefore, the long-term level of Warfarin in the blood under these conditions is $\frac{1}{2.3}\text{,}$ which is approximately 0.435 mg.

To determine the long-term effect of Warfarin, we considered a finite geometric series of $n$ terms, and then considered what happened as $n$ was allowed to grow without bound. In this sense, we were actually interested in an infinite geometric series (the result of letting $n$ go to infinity in the finite sum).

Definition $8.2.1$

An infinite geometric series is an infinite sum of the form

$$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\tag{$$$8.2.5$}

The value of $r$ in the geometric series ($8.2.5$) is called the common ratio of the series because the ratio of the ($n+1$)st term, $ar^n\text{,}$ to the $n$th term, $ar^{n-1}\text{,}$ is always $r\text{:}$

Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion

as a rational number. Observe that

This is an infinite geometric series with $a=\frac{12}{100}$ and $r = \frac{1}{100}\text{.}$

By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following activity.

Activity $8.2.3$

Let $r \ne 1$ and $a$ be real numbers and let

$$S = a+ar+ar^2 + \cdots ar^{n-1} + \cdots$$

be an infinite geometric series. For each positive integer $n\text{,}$ let

$$S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.}$$

Recall that

$$S_n = a\frac{1-r^n}{1-r}\text{.}$$

1. What should we allow $n$ to approach in order to have $S_n$ approach $S\text{?}$
2. What is the value of $\lim_{n \to \infty} r^n$ for $|r| \gt 1\text{?}$ for $|r| \lt 1\text{?}$ Explain.
3. If $|r| \lt 1\text{,}$ use the formula for $S_n$ and your observations in (a) and (b) to explain why $S$ is finite and find a resulting formula for $S\text{.}$

We can now find the value of the geometric series

Using $a = \frac{12}{100}$ and $r = \frac{1}{100}\text{,}$ we see that

The sum of a finite number of terms of an infinite geometric series is often called a partial sum of the series. Thus,

is called the $n$th partial sum of the series $\sum_{k=0}^{\infty} ar^k\text{.}$ We summarize our recent work with geometric series as follows.

Note

• An infinite geometric series is an infinite sum of the form
  $$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\tag{$$$8.2.6$}

  where $a$ and $r$ are real numbers such that $r \ne 0\text{.}$

• The $n$th partial sum $S_n$ of an infinite geometric series is
  $$S_n = a+ar+ar^2+ \cdots + ar^{n-1}\text{.}$$
• If $|r| \lt 1\text{,}$ then using the fact that $S_n = a\frac{1-r^n}{1-r}\text{,}$ it follows that the sum $S$ of the infinite geometric series ($8.2.6$) is
  $$S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r}$$

Activity $8.2.4$

The formulas we have derived for an infinite geometric series and its partial sum have assumed we begin indexing the sums at $n=0\text{.}$ If instead we have a sum that does not begin at $n=0\text{,}$ we can factor out common terms and use the established formulas. This process is illustrated in the examples in this activity.

1. Consider the sum
   $$\sum_{k=1}^{\infty} (2)\left(\frac{1}{3}\right)^k = (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{3}\right)^2 + (2)\left(\frac{1}{3}\right)^3 + \cdots\text{.}$$

   Remove the common factor of $(2)\left(\frac{1}{3}\right)$ from each term and hence find the sum of the series.

2. Next let $a$ and $r$ be real numbers with $-1\lt r\lt 1\text{.}$ Consider the sum
   $$\sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.}$$

   Remove the common factor of $ar^3$ from each term and find the sum of the series.

3. Finally, we consider the most general case. Let $a$ and $r$ be real numbers with $-1\lt r\lt 1\text{,}$ let $n$ be a positive integer, and consider the sum
   $$\sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.}$$

   Remove the common factor of $ar^n$ from each term to find the sum of the series.

Summary

• An infinite geometric series is an infinite sum of the form
  $$\sum_{k=0}^{\infty} ar^k$$

  where $a$ and $r$ are real numbers and $r \neq 0\text{.}$

• The $n$th partial sum of the geometric series $\sum_{k=0}^{\infty} ar^k$ is
  $$S_n = \sum_{k=0}^{n-1} ar^k\text{.}$$

  A formula for the $n$th partial sum of a geometric series is

  $$S_n = a \frac{1-r^n}{1-r}\text{.}$$

  If $|r| \lt 1\text{,}$ the infinite geometric series $\sum_{k=0}^{\infty} ar^k$ has the finite sum $\frac{a}{1-r}\text{.}$