Using the Formula for Geometric Series

Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, $r$ . We can write the sum of the first $n$ terms of a geometric series as

S_{n} = a_{1} + r a_{1} + r^{2} a_{1} + . . . + r^{n - 1} a_{1}

${S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}$ .

Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first $n$ terms of a geometric series. We will begin by multiplying both sides of the equation by $r$ .

r S_{n} = r a_{1} + r^{2} a_{1} + r^{3} a_{1} + . . . + r^{n} a_{1}

$r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}$

Next, we subtract this equation from the original equation.

\begin{matrix} \frac{\begin{matrix} S_{n} = a_{1} + r a_{1} + r^{2} a_{1} + . . . + r^{n - 1} a_{1} - r S_{n} = - (r a_{1} + r^{2} a_{1} + r^{3} a_{1} + . . . + r^{n} a_{1}) \end{matrix}}{(1 - r) S_{n} = a_{1} - r^{n} a_{1}} \end{matrix}

$\begin{array}{l}\\ \frac{\begin{array}{l}\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\hfill \\ -r{S}_{n}=-\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\right)\hfill \end{array}}{\left(1-r\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\end{array}$

Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for ${S}_{n}$ , divide both sides by $\left(1-r\right)$ .

S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} r \neq 1

${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}$

A General Note: Formula for the Sum of the First n Terms of a Geometric Series

A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first $n$ terms of a geometric sequence is represented as

S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} r \neq 1

${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\text{ r}\ne \text{1}$

How To: Given a geometric series, find the sum of the first n terms.

Identify ${a}_{1},r,\text{and}n$ .
Substitute values for ${a}_{1},r$ , and $n$ into the formula ${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}$ .
Simplify to find ${S}_{n}$ .

Example 4: Finding the First n Terms of a Geometric Series

Use the formula to find the indicated partial sum of each geometric series.

${S}_{11}$ for the series $\text{ 8 + -4 + 2 + }\dots$
$\underset{6}{\overset{k=1}{{\sum }^{\text{ }}}}3\cdot {2}^{k}$

Solution

${a}_{1}=8$ , and we are given that $n=11$ .We can find $r$ by dividing the second term of the series by the first.
$r=\frac{-4}{8}=-\frac{1}{2}$

Substitute values for ${a}_{1}, r, \text{and} n$ into the formula and simplify.

$\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{11}=\frac{8\left(1-{\left(-\frac{1}{2}\right)}^{11}\right)}{1-\left(-\frac{1}{2}\right)}\approx 5.336\hfill \end{array}$
Find ${a}_{1}$ by substituting $k=1$ into the given explicit formula.
${a}_{1}=3\cdot {2}^{1}=6$

We can see from the given explicit formula that $r=2$ . The upper limit of summation is 6, so $n=6$ .

Substitute values for ${a}_{1},r$ , and $n$ into the formula, and simplify.

$\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{6}=\frac{6\left(1-{2}^{6}\right)}{1 - 2}=378\hfill \end{array}$

Use the formula to find the indicated partial sum of each geometric series.

Try It 6

${S}_{20}$ for the series $\text{ 1,000 + 500 + 250 + }\dots$

Solution

Try It 7

$\sum _{k=1}^{8}{3}^{k}$

Solution

Example 5: Solving an Application Problem with a Geometric Series

At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.

Solution

The problem can be represented by a geometric series with ${a}_{1}=26,750$ ; $n=5$ ; and $r=1.016$ . Substitute values for ${a}_{1}$ , $r$ , and $n$ into the formula and simplify to find the total amount earned at the end of 5 years.

\begin{matrix} S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r} S_{5} = \frac{26, 750 (1 - {1.016}^{5})}{1 - 1.016} \approx 138, 099.03 \end{matrix}

$\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\hfill \\ {S}_{5}=\frac{26\text{,}750\left(1-{1.016}^{5}\right)}{1 - 1.016}\approx 138\text{,}099.03\hfill \end{array}$

He will have earned a total of $138,099.03 by the end of 5 years.

Try It 8

At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?

Solution

Using the Formula for the Sum of an Infinite Geometric Series

Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first $n$ terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is $2+4+6+8+..$ .

This series can also be written in summation notation as $\sum _{k=1}^{\infty }2k$ , where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges.

Determining Whether the Sum of an Infinite Geometric Series is Defined

If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:

1 + 0.2 + 0.04 + 0.008 + 0.0016 + . .

$1+0.2+0.04+0.008+0.0016+..$ .

The common ratio $r\text{ = 0}\text{.2}$ . As $n$ gets very large, the values of ${r}^{n}$ get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with [latex]-1

A General Note: Determining Whether the Sum of an Infinite Geometric Series is Defined

The sum of an infinite series is defined if the series is geometric and [latex]-1

How To: Given the first several terms of an infinite series, determine if the sum of the series exists.

Find the ratio of the second term to the first term.
Find the ratio of the third term to the second term.
Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
If a common ratio, $r$ , was found in step 3, check to see if [latex]-1

Example 6: Determining Whether the Sum of an Infinite Series is Defined

Determine whether the sum of each infinite series is defined.

$\text{12 + 8 + 4 + }\dots$
$\frac{3}{4}+\frac{1}{2}+\frac{1}{3}+..$ .
$\sum _{k=1}^{\infty }27\cdot {\left(\frac{1}{3}\right)}^{k}$
$\sum _{k=1}^{\infty }5k$

Solution

The ratio of the second term to the first is $\frac{\text{2}}{\text{3}}$ ,
which is not the same as the ratio of the third term to the second, $\frac{1}{2}$ . The series is not geometric.
The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of $\frac{2}{3}\text{.}$ The sum of the infinite series is defined.
The given formula is exponential with a base of $\frac{1}{3}$ ; the series is geometric with a common ratio of $\frac{1}{3}\text{.}$ The sum of the infinite series is defined.
The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.

Determine whether the sum of the infinite series is defined.

Try It 9

$\frac{1}{3}+\frac{1}{2}+\frac{3}{4}+\frac{9}{8}+..$ .

Solution

Try It 10

$24+\left(-12\right)+6+\left(-3\right)+..$ .

Solution

Try It 11

$\sum _{k=1}^{\infty }15\cdot {\left(-0.3\right)}^{k}$

Solution

Series and Their Notations