Section 7.3

Sequence

Definition 74 (Geometric Sequence)

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed nonzero real number, called the common ratio.

We find the common ratio by choosing any term after the first and dividing it by the preceding term.

In a geometric sequence with first term \(a_{1}\) and common ratio \(r\), the \(n\)th term \(a_{n}\) is given by the following

\[ a_{n}=a_{1}r^{n-1}. \]

Something to keep in mind when dealing with geometric sequences. The first value, \(a_1\), is your starting number and by definition is \(a_1=a_1\cdot r^0=a_1\). Then the next number in the sequence is \(a_2\) which is \(a_1\) times \(r\). That is,

\[a_2=a_1\cdot r\]

The next number is \(a_3\) which is \(a_2\) times \(r\). That is,

\[\begin{align*} a_3 & = a_2\cdot r\\ & = \left(a_1 \cdot r \right)\cdot r\\ & = a_1\cdot r^2 \end{align*}\]

The next number is \(a_4\) which is \(a_3\) times \(r\). That is,

\[\begin{align*} a_4 & = a_3\cdot r\\ & = \left(a_2\cdot r\right)\cdot r\\ & = \left(\left(a_1\cdot r\right)\cdot r\right)\cdot r\\ & = a_1\cdot r^3 \end{align*}\]

And so on. This commonly associated with compound interest.

Example 137

Determine \(a_{5}\) and \(a_{n}\)for the geometric sequence

\[ 6400,\,1600,\,400,\,100,\,... \]

Solution:

First,

\[ \frac{1600}{6400}=\frac{1}{4}=r \]

and \(a_{1}=6400\). Therefore, \(a_{n}=6400\cdot4^{n-1}\). Thus,

\[\begin{align*} a_{5} & =6400\cdot(\frac{1}{4})^{5-1}\\ & =6400\cdot\left(\frac{1}{4}\right)^{4}\\ & =25 \end{align*}\]

or since \(a_{4}=100\) we know that \(a_{5}=100\div4=25\).

Example 138

Determine \(r\) and \(a_{1}\) for the geometric sequence with \(a_{2}=-18\) and \(a_{5}=486\). Then define \(a_{n}\).

Solution:

Consider a table of values

\[\begin{align*} a_{1} & =a_{1}r^{0}=a_1\\ a_{2} & =a_{1}r=-18\\ a_{3} & =a_{2}\cdot r=-18\cdot r\\ a_{4} & =a_{3}\cdot r=\left(a_{2}\cdot r\right)\cdot r=a_{2}r^{2}=-18r^{2}\\ a_{5} & =a_{4}\cdot r=\left(-18r^{2}\right)\cdot r=-18r^{3}=486 \end{align*}\]

Thus, we can solve for \(r\)

\[\begin{align*} -18r^{3} & =486\\ r^{3} & =\frac{486}{-18}\\ & =-27\\ r & =-3 \end{align*}\]

Since \(a_{2}=a_{1}r^{2-1}\) we have

\[\begin{align*} -18 & =a_{1}(-3)^{1}\\ -18 & =a_{1}(-3)\\ a_{1} & =\frac{-18}{-3}\\ & =6 \end{align*}\]

Therefore, \(a_{1}=6\), \(r=-3\) and \(a_{n}=6(-3)^{n-1}\).

Example 139

A person receives a gift on the first day of each month for a year, starting with $50 on January 1, with the amount doubling each month. How much is received on December 1?

Solution:

We are given \(a_{1}=50\) and the common ratio is \(r=2\) (because the amount is doubled). Therefore,

\[\begin{align*} a_{12} & =50\cdot(2)^{12-1}\\ & =50\cdot2^{11}\\ & =102400 \end{align*}\]

which means the person will have $102400 at the end of the year.

Series

Consider,

\[\begin{align*} S_{n} & =\sum_{i=1}^{n}a_{1}r^{i-1}\\ & =a_{1}+a_{1}r+a_{1}r^{2}+\cdots+a_{1}r^{n-1}\\ r\cdot\left(S_{n}\right) & =r\left(a_{1}+a_{1}r+a_{1}r^{2}+\cdots+a_{1}r^{n-1}\right)\\ rS_{n} & =a_{1}r+a_{1}r^{2}+a_{1}r^{3}+\cdots+a_{1}r^{n}\\ S_{n}-rS_{n} & =\left(a_{1}+\cancel{a_{1}r}+\cdots+\cancel{a_{1}r^{n-1}}\right)-\left(\cancel{a_{1}r}+\cdots+\cancel{a_{1}r^{n-1}}+a_{1}r^{n}\right)\\ S_{n}(1-r) & =a_{1}-a_{1}r^{n}\\ & =a_{1}(1-r^{n})\\ S_{n} & =\dfrac{a_{1}(1-r^{n})}{1-r} \end{align*}\]

Definition 75

A geometric Series is the sum of the terms of a geometric sequence.

If a geometric sequence has first term \(a_{1}\) and a common ratio \(r\), then the sum \(S_{n}\) of the first \(n\) terms is given by the following

\[ S_{n}=\dfrac{a_{1}(1-r^{n})}{1-r}, \]

where \(r\ne1\).

\[ \lim_{n\to\infty}r^{n}=0 \]

when \(0<r<1\).

Definition 76

The sum \(S_{\infty}\) of the terms of an infinite geometric sequence with first term \(a_{1}\) and common ratio \(r\), where \(|r|<1\), is given by the following

\[ \lim_{n\to\infty}S_{n}=S_{\infty}=\dfrac{a_{1}}{1-r} \]

If \(|r|>1\), then the terms increase without bound in the absolute value, so there is no limit as \(n\to\infty.\) Therefore, if \(|r|>1\), then the terms of the sequence will not have a sum.

Example 140

A person receives a gift on the first day of each month for a year, starting with $50 on January 1, with the amount doubling each month. What is the total amount received throughout the year?

Solution:

We know that \(a_{1}=50\) and \(r=2\) and since we had 12 gifts we have

\[\begin{align*} S_{12} & =\dfrac{50(1-2^{12})}{1-2}\\ & =\dfrac{50(-4095)}{-1}\\ & =204750 \end{align*}\]

That is, the total received in the year is $204750.

Example 141

Evaluate

\[ \sum_{i=1}^{8}4\cdot5^{i} \]

Solution:

We have \(a_{i}=4\cdot5^{i}\), \(a_{1}=4\cdot5=20\), \(r=5\), and in this case \(n=8\).

\[\begin{align*} \sum_{i=1}^{8}4\cdot5^{i} & =\dfrac{20(1-5^{8})}{1-5}\\ & =\dfrac{20(-390624)}{-4}\\ & =1953120 \end{align*}\]

Example 142

Consider the geometric sequence

\[ 2,1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\dots \]

We have \(a_{1}=2\) and \(r=\frac{1}{2}\). Which means

\[\begin{align*} S_{n} & =\sum_{i=1}^{n}2\cdot\left(\frac{1}{2}\right)^{i}=\frac{2\cdot\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}\\ & =\frac{2\cdot\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}\\ & =\dfrac{2\cdot\left(1-\left(\frac{1}{2}\right)^{n}\right)}{\frac{1}{2}}\\ & =4\cdot\left(1-\left(\frac{1}{2}\right)^{n}\right)\\ & =4-4\cdot\left(\frac{1}{2}\right)^{n} \end{align*}\]

With this generalization we see

\[\begin{align*} S_{1} & =4-4\cdot\frac{1}{2}=2\\ S_{2} & =4-4\cdot\left(\frac{1}{2}\right)^{2}=3\\ S_{3} & =4-4\cdot\left(\frac{1}{2}\right)^{3}=3.5\\ S_{4} & =4-4\cdot\left(\frac{1}{2}\right)^{4}=3.75\\ \vdots\\ S_{10} & =4-4\cdot\left(\frac{1}{2}\right)^{10}=3.99609375\\ \vdots\\ \lim_{n\to\infty}S_{n} & =\lim_{n\to\infty}(4-4\cdot(\frac{1}{2})^{n})\\ & =4-4\lim_{n\to\infty}(\frac{1}{2})^{n}\\ & =4-4\cdot0\\ & =4 \end{align*}\]

We say the limit \(S_{n}\) as \(n\) increases without bounds is \(4\).

{prf:example}

label:

geoSeriesExam4

If a geometric series has a first term \(a_{1}\) and a common ratio \(r\) such that \(0<r<1\), then the finite sum is

\[\begin{align*} S_{n} & =\sum_{i=1}^{n}a_{1}\cdot r^{i}\\ & =\dfrac{a_{1}(1-r^{n})}{1-r} \end{align*}\]

The infinity sum is \(\lim_{n\to\infty}S_{n}\). Using the fact that \(\lim_{n\to\infty}r^{n}=0\) when \(0<r<1\) we have

\[\begin{align*} \lim_{n\to\infty}S_{n} & =\lim_{n\to\infty}\dfrac{a_{1}(1-r^{n})}{1-r}\\ & =\dfrac{a_{1}(1-0)}{1-r}\\ & =\dfrac{a_{1}(1)}{1-r}\\ & =\dfrac{a_{1}}{1-r} \end{align*}\]

Example 143

Evaluate

\[ \sum_{i=1}^{\infty}\left(\frac{2}{5}\right)\cdot\left(-\frac{1}{3}\right)^{i-1} \]

Solution:

First, we see that \(a_{1}=\frac{2}{5}\) and \(r=-\frac{1}{3}\). We notice that \(|r|<1\) which means the sum exists.

Next, we use the identity

\[\begin{align*} S_{\infty} & =\dfrac{a_{1}}{1-r}\\ & =\dfrac{\frac{2}{5}}{1-(-\frac{1}{3})}\\ & =\frac{2}{5}\cdot\frac{1}{\frac{4}{3}}\\ & =\frac{2}{5}\cdot\frac{3}{4}\\ & =\frac{3}{10} \end{align*}\]

Example 144

Evaluate

\[ \sum_{i=1}^{\infty}(0.9)^{i} \]

Solution:

First, \(\sum_{i=1}^{\infty}(0.9)^{i}=\sum_{i=1}^{\infty}(0.9)\cdot(0.9)^{i-1}\). Thus, \(a_{1}=0.9\) and \(r=0.9\). Since \(|r|<1\) we know the sum exists.

\[\begin{align*} S_{\infty} & =\dfrac{0.9}{1-0.9}\\ & =\dfrac{0.9}{0.1}\\ & =9 \end{align*}\]

Future Value of an Annuity

Theorem 15 (Future Value of an Annuity)

The formula for the future value of an annuity is given by the following

\[ S=R\left(\dfrac{(1+i)^{n}-1}{i}\right). \]

Here \(S\) is the future value, \(R\) is payment at the end of each period, \(i\) is interest rate per period, and \(n\) is number of periods.

“Proof:”

The formula for interest compounded annually is [ A=P(1+r)^{t} ] To avoid confusion we will redefine it as [ A=P(1+i)^{t} ] where \(i\) is the interest rate per period.

If \(A_{n}\) is a sequence of possible amounts we have [ A_{n}=A_{0}(1+i)^{n} ] where the common ratio is \(r=1+i\). To connect with the identity we will say \(A_{0}=R\). Now we have [ A_{n}=R(1+i)^{n} ]

Next, the sum of money after \(n\) years is

\[\begin{align*} S_{n} & =\sum_{i=1}^{n}A_{n}\\ & =\sum_{i=1}^{n}R(1+i)^{n}\\ & =\dfrac{R(1-(1+i)^{n})}{1-(1+i)}\\ & =\dfrac{R(1-(1+i)^{n})}{-i}\\ & =\dfrac{R(-1)\left((1+i)^{n}-1\right)}{(-1)i}\\ & =\dfrac{R\left((1+i)^{n}-1\right)}{i} \end{align*}\]