SubsectionGeometric Series
In Example7.12 we encountered the sum
\begin{equation*}
(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n1}\right)
\end{equation*}
for the longterm level of Warfarin in the patient's system. This sum has the form
\begin{equation}
a+ar+ar^2+ \cdots + ar^{n1}\label{eq82partsumgeometric1}\tag{7.1}
\end{equation}
where \(a=5 \times 0.08\) and \(r=0.08\text{.}\) Such a sum is called a finite geometric series with ratio \(r\text{.}\)
Example7.15
Let \(a\) and \(r\) be real numbers (with \(r \ne 1\)) and let
\begin{equation*}
S_n = a+ar+ar^2 + \cdots + ar^{n1}\text{.}
\end{equation*}
In this example we will find a shortcut formula for \(S_n\) that does not involve a sum of \(n\) terms.
Multiply \(S_n\) by \(r\text{.}\) What does the resulting sum look like?

Subtract \(rS_n\) from \(S_n\) and explain why
\begin{equation}
S_n  rS_n = a  ar^n\text{.}\label{eq821partialgeometricsum}\tag{7.2}
\end{equation}
Solve equation (7.2) for \(S_n\) to find a simple formula for \(S_n\) that does not involve adding \(n\) terms.
Hint
Distribute the factor of \(r\text{.}\)
Look for common terms in the two expressions being subtracted.
Observe that you can remove a factor of \(S_n\) from \(S_n  rS_n\text{.}\)
Answer
\(rS_n = ar+ar^2+ar^3 + \cdots + ar^n \text{.}\)
\(S_n  rS_n = a  ar^n\text{.}\)
\(S_n = a\frac{1r^n}{1r}\text{.}\)
Solution

Note that
\begin{equation*}
rS_n = ar+ar^2+ar^3 + \cdots + ar^n\text{.}
\end{equation*}

When we subtract \(rS_n\) from \(S_n\) the middle terms all cancel and we are left with
\begin{align*}
S_n  rS_n \amp = \left(a+ar+ar^2 + \cdots + ar^{n1}\right)\\
\amp\phantom{={}}{} \left(ar+ar^2+ar^3 + \cdots + ar^n\right)\\
\amp =a + (arar) + \left(ar^2ar^2\right) + \cdots + \left(ar^{n1}ar^{n1}\right)  ar^n\\
\amp = a  ar^n\text{.}
\end{align*}

Factoring \(S_n\) from left hand side and dividing gives us a formula for \(S_n\text{:}\)
\begin{align*}
S_n(1r) \amp = a  ar^n\\
S_n \amp = a\frac{1r^n}{1r}\text{.}
\end{align*}
The sum of the terms of a sequence is called a series. We summarize the result of Example7.15 in the following way.
Finite Geometric Series
A finite geometric series \(S_n\) is a sum of the form
\begin{equation}
S_n = a + ar + ar^2 + \cdots + ar^{n1}\text{,}\label{eq82geometricsum}\tag{7.3}
\end{equation}
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
\begin{equation}
S_n = a+ar+ar^2+ \cdots + ar^{n1} = \frac{a(1r^n)}{1r}\text{.}\label{eq82partsumgeometric}\tag{7.4}
\end{equation}
We now apply Equation (7.4) to the example involving Warfarin from Example7.12. Recall that
\begin{equation*}
Q(n)=(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n1}\right) \text{mg}\text{,}
\end{equation*}
so \(Q(n)\) is a geometric series with \(a=5 \times 0.08 = 0.4\) and \(r = 0.08\text{.}\) Thus,
\begin{equation*}
Q(n) = 0.4\left(\frac{10.08^n}{10.08}\right) = \frac{1}{2.3} \left(10.08^n\right)\text{.}
\end{equation*}
Notice that as \(n\) goes to infinity, the value of \(0.08^n\) goes to 0. So,
\begin{equation*}
\lim_{n \to \infty} Q(n) = \lim_{n \to \infty} \frac{1}{2.3} \left(10.08^n\right) = \frac{1}{2.3} \approx 0.435\text{.}
\end{equation*}
Therefore, the longterm level of Warfarin in the blood under these conditions is \(\frac{1}{2.3}\text{,}\) which is approximately 0.435 mg.
To determine the longterm 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).
Definition7.16
An infinite geometric series is an infinite sum of the form
\begin{equation}
a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\label{eq82geometricseries}\tag{7.5}
\end{equation}
The value of \(r\) in the geometric series (7.5) 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, \(ar^{n1}\text{,}\) is always \(r\text{:}\)
\begin{equation*}
\frac{ar^n}{ar^{n1}} = r\text{.}
\end{equation*}
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
\begin{equation*}
N=0.1212\overline{12}
\end{equation*}
as a rational number. Observe that
\begin{align*}
N \amp = 0.12 + 0.0012 + 0.000012 + \cdots\\
\amp = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.}
\end{align*}
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 example.
Example7.17
Let \(r \ne 1\) and \(a\) be real numbers and let
\begin{equation*}
S = a+ar+ar^2 + \cdots ar^{n1} + \cdots
\end{equation*}
be an infinite geometric series. For each positive integer \(n\text{,}\) let
\begin{equation*}
S_n = a+ar+ar^2 + \cdots + ar^{n1}\text{.}
\end{equation*}
Recall that
\begin{equation*}
S_n = a\frac{1r^n}{1r}\text{.}
\end{equation*}
What should we allow \(n\) to approach in order to have \(S_n\) approach \(S\text{?}\)
What is the value of \(\lim_{n \to \infty} r^n\) for \(r \gt 1\text{?}\) for \(r \lt 1\text{?}\) Explain.
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 \(S\text{.}\)
Hint
Let \(n\) increase without bound.
Think about what happens to powers of numbers that are less than or greater than 1.
Consider \(\frac{1r^n}{1r}\) and how the numerator tends to 1 as \(n \to \infty\) for certain values of \(r\text{.}\)
Answer

Observe that
\begin{equation*}
S = \lim_{n \to \infty} S_n\text{.}
\end{equation*}
If \(r \gt 1\text{,}\) then \(\lim_{n \to \infty} r^n = \infty\text{.}\) If \(0 \lt r \lt 1\text{,}\) then \(\lim_{n \to \infty} r^n = 0\text{.}\)

Since
\begin{equation*}
S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1r^n}{1r}
\end{equation*}
and
\begin{equation*}
\lim_{n \to \infty} r^n = 0
\end{equation*}
for \(0 \lt r \lt 1\text{,}\) we conclude that
\begin{equation*}
S = \lim_{n \to \infty} a\frac{1r^n}{1r} = \frac{a}{1r}
\end{equation*}
when \(0 \lt r \lt 1\text{.}\)
Solution

Observe that
\begin{equation*}
S = \lim_{n \to \infty} S_n\text{.}
\end{equation*}
If \(r \gt 1\text{,}\) then \(\lim_{n \to \infty} r^n = \infty\text{.}\) If \(0 \lt r \lt 1\text{,}\) then \(\lim_{n \to \infty} r^n = 0\text{.}\)

Since
\begin{equation*}
S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1r^n}{1r}
\end{equation*}
and
\begin{equation*}
\lim_{n \to \infty} r^n = 0
\end{equation*}
for \(0 \lt r \lt 1\text{,}\) we conclude that
\begin{equation*}
S = \lim_{n \to \infty} a\frac{1r^n}{1r} = \frac{a}{1r}
\end{equation*}
when \(0 \lt r \lt 1\text{.}\)
We can now find the value of the geometric series
\begin{equation*}
N = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.}
\end{equation*}
Using \(a = \frac{12}{100}\) and \(r = \frac{1}{100}\text{,}\) we see that
\begin{equation*}
N = \frac{12}{100} \left(\frac{1}{1\frac{1}{100}}\right) = \frac{12}{100} \left(\frac{100}{99}\right) = \frac{4}{33}\text{.}
\end{equation*}
The sum of a finite number of terms of an infinite geometric series is often called a partial sum of the series. Thus,
\begin{equation*}
S_n = a+ar+ar^2 + \cdots + ar^{n1} = \sum_{k=0}^{n1} ar^k\text{.}
\end{equation*}
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.
Infinite Geometric Series

An infinite geometric series is an infinite sum of the form
\begin{equation}
a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\label{eq82geometricseries2}\tag{7.6}
\end{equation}
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
\begin{equation*}
S_n = a+ar+ar^2+ \cdots + ar^{n1}\text{.}
\end{equation*}

If \(r \lt 1\text{,}\) then using the fact that \(S_n = a\frac{1r^n}{1r}\text{,}\) it follows that the sum \(S\) of the infinite geometric series (7.6) is
\begin{equation*}
S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1r^n}{1r} = \frac{a}{1r}.
\end{equation*}
In this case we say that the geometric series converges.
If \(r\geq1\text{,}\) the sequence \(S_n\) does not converge, so we can't assign a numerical value to the expression \(\sum_{n=0}^{\infty} ar^n\text{.}\) In this case we say that the geometric series does not converge.
Example7.18
The formulas we have derived for an infinite geometric series and its partial sum have assumed that 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 then use the established formulas. This process is illustrated in this example.

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

Next let \(a\) and \(r\) be real numbers with \(1\lt r\lt 1\text{.}\) Consider the sum
\begin{equation*}
\sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.}
\end{equation*}
Remove the common factor of \(ar^3\) from each term and find the sum of the series.

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
\begin{equation*}
\sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.}
\end{equation*}
Remove the common factor of \(ar^n\) from each term to find the sum of the series.
Hint
Think about how \(r = \frac{1}{3}\text{.}\)
Note that \(ar^3+ar^4+ar^5 + \cdots = ar^3(1 + r + r^2 + \cdots)\text{.}\)
Compare your work in (b).
Answer
\((2)\left(\frac{1}{3}\right) \left[1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \cdots \right] \text{.}\)
\(ar^3+ar^4+ar^5 + \cdots = \frac{ar^3}{1r}\text{.}\)
\(r^n+ar^{n+1}+ar^{n+2} + \cdots = \frac{ar^n}{1r}\text{.}\)
Solution

Factoring out \((2)\left(\frac{1}{3}\right)\) gives us
\begin{align*}
(2)\left(\frac{1}{3}\right) \left[1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \cdots \right] \amp = (2)\left(\frac{1}{3}\right)\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k\\
\amp = (2)\left(\frac{1}{3}\right) \left(\frac{1}{1\frac{1}{3}}\right)\\
\amp = (2)\left(\frac{1}{3}\right) \left(\frac{3}{2}\right)\\
\amp = 4\text{.}
\end{align*}

Factoring out \(ar^3\) gives us
\begin{align*}
ar^3+ar^4+ar^5 + \cdots \amp = ar^3\left(1+r+r^2+ \cdots \right)\\
\amp = ar^3\sum_{k=0}^{\infty} r^k\\
\amp = ar^3 \left(\frac{1}{1r}\right)\\
\amp = \frac{ar^3}{1r}\text{.}
\end{align*}

Factoring out \(ar^n\) gives us
\begin{align*}
ar^n+ar^{n+1}+ar^{n+2} + \cdots \amp = ar^n\left(1+r+r^2+ \cdots \right)\\
\amp = ar^n\sum_{k=0}^{\infty} r^k\\
\amp = ar^n \left(\frac{1}{1r}\right)\\
\amp = \frac{ar^n}{1r}\text{.}
\end{align*}