Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.2: Geometric Series

Skills to Develop

In this section, we strive to understand the ideas generated by the following important questions:

  • What is a geometric series?
  • What is a partial sum of a geometric series?
  • What is a simplified form of the nth partial sum of a geometric series?
  • Under what conditions does a geometric series converge?
  • What is the sum of a convergent geometric series?

Many important sequences are generated through the process of addition. In Preview Activity \(\PageIndex{1}\), we see a particular example of a special type of sequence that is connected to a sum.

Preview Activity \(\PageIndex{1}\)

Warfarin is an anticoagulant that prevents blood clotting; often it is prescribed to stroke victims in order to help ensure blood flow. The level of warfarin has to reach a certain concentration in the blood in order to be effective. Suppose warfarin is taken by a particular patient in a 5 mg dose each day. The drug is absorbed by the body and some is excreted from the system between doses. Assume that at the end of a 24 hour period, 8% of the drug remains in the body. Let \(Q(n)\) be the amount (in mg) of warfarin in the body before the \((n + 1)\)st dose of the drug is administered.

  1. Explain why \(Q(1) = 5 \times 0.08\) mg.
  2. Explain why \(Q(2) = (5 + Q(1)) \times 0.08\) mg. Then show that \(Q(2) = (5 \times 0.08) (1 + 0.08)\) mg.
  3. Explain why \(Q(3) = (5 + Q(2)) \times 0.08\) mg. Then show that \(Q(3) = (5 \times 0.08) 1 + 0.08 + 0.082\) mg.
  4. Explain why \(Q(4) = (5 + Q(3)) \times 0.08\) mg. Then show that \(Q(4) = (5 \times 0.08) 1 + 0.08 + 0.082 + 0.083\) mg.
  5. There is a pattern that you should see emerging. Use this pattern to find a formula for \(Q(n)\), where \(n\) is an arbitrary positive integer.
  6. Complete Table 8.2 with values of \(Q(n)\) for the provided \(n\)-values (reporting \(Q(n)\) to 10 decimal places). What appears to be happening to the sequence \(Q(n)\) as \(n\) increases?
\(Q(1)\) 0.40
\(Q(2)\)  
\(Q(3)\)  
\(Q(4)\)  
\(Q(5)\)  
\(Q(6)\)  
\(Q(7)\)  
\(Q(8)\)  
\(Q(9)\)  
\(Q(10)\)  

Table 8.2: Values of \(Q(n)\) for selected values of \(n\)

 

Geometric Sums

In Preview Activity 8.2 we encountered the sum

\[(5 \times 0.08) 1 + 0.08 + 0.082 + 0.083 + · · · + 0.08^{n−1}. \nonumber\]

In order to evaluate the long-term level of Warfarin in the patient’s system, we will want to fully understand the sum in this expression. This sum has the form

\[a + ar + ar^2 + · · · + ar^{n−1} \tag{8.1}\label{8.1}\] 

where \(a = 5 \times 0.08\) and \(r = 0.08\). Such a sum is called a geometric sum with ratio r. We will analyze this sum in more detail in the next activity.

Activity \(\PageIndex{1}\)

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

\[S_n = a + ar + ar^2 + · · · + ar^{n−1}. \nonumber\] 

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 \(r\). What does the resulting sum look like?
  2. Subtract \(r S_n\) from \(S_n\) and explain why

\[S_n − r S_n = a − ar^n .\tag{8.2}\label{8.2}\] 

c.  Solve Equation (\( \ref{8.2}\)) for \(S_n\) to find a simple formula for \(S_n\) that does not involve adding \(n\) terms.

We can summarize the result of Activity \(\PageIndex{1}\) in the following way. A geometric sum Sn is a sum of the form

\[Sn = a + ar + ar^2 + · · · + ar^{n−1} , \tag{8.3}\label{8.3}\]

where \(a\) and \(r\) are real numbers such that \(r \neq 1\). The geometric sum \(S_n\) can be written more simply as

\[Sn = a + ar + ar^2 + · · · + ar^n−1 = a(1 − r^n ) 1 − r . \tag{8.4}\label{8.4}\]

We now apply Equation \(\ref{8.4}\) to the example involving warfarin from Preview Activity 8.2. Recall that

\(Q(n) = (5 \times 0.08) 1 + 0.08 + 0.082 + 0.083 + · · · + 0.08^{n−1}\) mg,

so \(Q(n)\) is a geometric sum with \(a = 5 \times 0.08 = 0.4 \) and \(r = 0.08\). Thus,

\[Q(n) = 0.4 1 − 0.08n 1 − 0.08 ! = 1 2.3 (1 − 0.08n ) . \nonumber\]

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

\[\lim_{n \rightarrow \infty}Q(n) = \lim_{n \rightarrow \infty}\dfrac{1}{2.3}(1-0.08^n)= \dfrac{1}{2.3}\approx 0.435.  \nonumber\]

Therefore, the long-term level of Warfarin in the blood under these conditions is \(\dfrac{1}{2.3}\), which is approximately \(0.435\) mg. To determine the long-term effect of Warfarin, we considered a geometric sum 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 sum (the result of letting n go to infinity in the finite sum). We call such an infinite geometric sum a geometric series.

Definition

A geometric series is an infinite sum of the form

\[a +ar +ar^2 + ... = \sum_{n=0}^\infty ar^n.  \tag{8.5}\label{8.5}\]

The value of \(r\) in the geometric series in Equation \(\ref{8.5}\) is called the common ratio of the series because the ratio of the \((n + 1)\)st term \(ar^n\) to the \(n\)th term \(ar^{n−1}\) is always \(r\).

Geometric series are very 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

\[N = 0.1212\overline{12} \nonumber\]

as a rational number.

Observe that

\[N = 0.12 + 0.0012 + 0.000012 + · · · \nonumber\]

\[= \dfrac{12}{100}+\dfrac{12}{100}\dfrac{1}{100}+ \dfrac{12}{100}\dfrac{1}{100}^2 + ... , \nonumber\]

which is an infinite geometric series with \(a = \dfrac{12}{100}\) and \(r = \dfrac{1}{100}\). In the same way that we were able to find a shortcut formula for the value of a (finite) geometric sum, we would like to develop a formula for the value of a (infinite) geometric series. We explore this idea in the following activity.

Activity \(\PageIndex{2}\)

Let \(r \neq 1\) and \(a\) be real numbers and let

\[S = a + ar + ar^2 + · · · ar^{n−1} + · · · \nonumber\]

be an infinite geometric series. For each positive integer \(n\), let

\[S_n = a + ar + ar^2 + · · · + ar^{n−1}.  \nonumber\]

Recall that

\[S_n = a \dfrac{1-r^n}{1-r}. \nonumber\]

a. What should we allow n to approach in order to have \(S_n\) approach \(S\)

b. What is the value of \(\lim_{n \rightarrow \infty} r^n\) for

  • |\(r\)| > 1?
  • |\(r\)| < 1?

Explain.

c. If |\(r\)| < 1, 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 \(S\).

 From our work in Activity 8.5, we can now find the value of the geometric series \(N = \dfrac{12} {100} + \dfrac{12}{100} \dfrac{1}{100} + \dfrac{12}{100} \dfrac{1}{100}^2+ · · · \). In particular, using \(a = \dfrac{12}{100}\) and \(r = \frac{1}{100}\), we see that

\(N = \dfrac{12}{100} \left(\dfrac{1}{1-\dfrac{1}{100}}\right) = \dfrac{12}{100} \left(\dfrac{100} {99} \right) = \dfrac{4}{33} \).

It is important to notice that a geometric sum is simply the sum of a finite number of terms of a geometric series. In other words, the geometric sum \(S_n\) for the geometric series

\[\sum_{k=0}^\infty ar^k \nonumber\]

is

\[S_n = a + ar + ar^2 +....ar^{n-1}= \sum_{k=0}^{n-1} ar^k \nonumber\]

We also call this sum \(S_n\) the \(n\)th partial sum of the geometric series. We summarize our recent work with geometric series as follows.

  • A geometric series is an infinite sum of the form

\[a + ar +ar^2 + ... = \sum_{n=0}^\infty ar^n, \tag{8.6}\label{8.6}\] 

where \(a\) and \(r\) are real numbers such that \(r \neq 0\).

  • The nth partial sum Sn of the geometric series is

\[S_n = a + ar + ar^2 + · · · + ar^{n−1} . \nonumber\]

  • If |\(r\)| < 1, then using the fact that \(S_n = a \dfrac{1−r^n}{1−r}\), it follows that the sum \(S\) of the geometric series (\(\ref{8.6}\)) is

\[ S = \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} a \dfrac{1-r^n}{1-r}= \frac{a}{1-r}.  \nonumber\]

Activity \(\PageIndex{3}\)

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

a. Consider the sum

\[\sum_{k=1}^\infty(2)\left(\dfrac{1}{3}\right)^k = (2)\left(\dfrac{1}{3}\right)+(2) \left(\dfrac{1}{3}\right)^2 + (2)\left(\dfrac{1}{3}\right)^3 + .... \nonumber\]

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

b. Next let \(a\) and \(r\) be real numbers with \(−1 < r < 1\). Consider the sum

\[\sum_{k=3}^\infty = ar^k = ar^3 + ar^4+ ar^5 + ... \nonumber\]

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

c. Finally, we consider the most general case. Let \(a\) and \(r\) be real numbers with \(−1 < r < 1\), let \(n\) be a positive integer, and consider the sum

 

\[\sum_{k=n}^\infty = ar^k = ar^n + ar^{n+1}+ ar^{n+2} + ... \nonumber\]  

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

Summary

In this section, we encountered the following important ideas:

  • A geometric series is an infinite sum of the form

\[\sum_{k=0}^\infty ar^k \nonumber\]

where \(a\) and \(r\) are real numbers and \(r \neq 0\) .

  • For the geometric series \(\sum_{k=0}^\infty ar^k\), its \(n\)th partial sum is

\[S_n = \sum_{k=0}^{n-1} ar^k \nonumber\]

An alternate formula for the \(n\)th partial sum is

\[S_n = a \dfrac{1-r^n}{1-r}. \nonumber\]

Whenever |\(r\)| < 1, the infinite geometric series \(\sum_{k=0}^{\infty}ar^k \) has the finite sum \(\dfrac{a}{ 1−r}\).

Contributors

Matt Boelkins (Grand Valley State University), David Austin (Grand Valley State University), Steve Schlicker (Grand Valley State University)