Use the Formula for a 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

$S_{n}=a_{1}+r a_{1}+r^{2} a_{1}+\ldots+r^{n-1} a_{1}$

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 $r$.

$r S_{n}=r a_{1}+r^{2} a_{1}+r^{3} a_{1}+\ldots+r^{n} a_{1}$

Next, we subtract this equation from the original equation.

$\frac{-r S_{R}=-\left(r a_{1}+r^{2} a_{1}+r^{3} a_{1}+\ldots+r^{R} a_{1}\right)}{(1-r) S_{R}=a_{1}-r^{2} a_{1}}$

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

$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \quad \Gamma \neq 1$

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

$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r} \quad \Gamma \neq 1$

How To

Given a geometric series, find the sum of the first $n$ terms.

1. Identify $a_1,r$, and $n$.
   
2. Substitute values for $a_1,r$, and $n$ into the formula $S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$.
   
3. Simplify to find $S_n$.
   

Example 4

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

ⓐ $S_{11}$ for the series $8 + -4 + 2 + …$

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 $a_{1}, r$, and $n$ into the formula and simplify.

$\begin{aligned} &S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\ &S_{11}=\frac{8\left(1-\left(-\frac{1}{2}\right)^{11}\right)}{1-\left(-\frac{1}{2}\right)} \approx 5.336 \end{aligned}$

ⓑ 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 $a_{1}, \quad r$, and $n$ into the formula, and simplify.

$\begin{aligned} &S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\ &S_{6}=\frac{6\left(1-2^{6}\right)}{1-2}=378 \end{aligned}$

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

Try It #6

$S_{20}$ for the series $1,000 + 500 + 250 + …$

Try It #7

$\displaystyle\sum_{k=1}^{8} 3^k$

Example 5

Solving an Application Problem with a Geometric Series

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.

$\begin{aligned}&S_{n}=\frac{a_{1}\left(1-r^{R}\right)}{1-r} \\&S_{5}=\frac{26,750\left(1-1.016^{6}\right)}{1-1.016} \approx 138,099.03\end{aligned}$

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?

Source: Rice University, https://openstax.org/books/college-algebra/pages/9-4-series-and-their-notations
This work is licensed under a Creative Commons Attribution 4.0 License.

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 $2+4+6+8+\ldots$

This series can also be written in summation notation as $\displaystyle\sum_{k=1}^{\infty} 2 k$, 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 sequence approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:

$1+0.2+0.04+0.008+0.0016+\ldots$

The common ratio $r=0.2$. 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 $-1 < r < 1$ approach 0; the sum of a geometric series is defined when $-1 < r < 1$.

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 $−1 < r < 1$.

How To

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

1. Find the ratio of the second term to the first term.
   
2. Find the ratio of the third term to the second term.
   
3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.
   
4. If a common ratio, $r$, was found in step 3, check to see if $−1 < r < 1$. If so, the sum is defined. If not, the sum is not defined.
   

Example 6

Determining Whether the Sum of an Infinite Series is Defined

Determine whether the sum of each infinite series is defined.

ⓐ $12 + 8 + 4 + …$

ⓑ $\frac{3}{4}+\frac{1}{2}+\frac{1}{3}+...$

ⓒ $\sum_{k=1}^{\infty} 27\cdot\left(\frac{1}{3}\right)^k$

ⓓ $\displaystyle\sum_{k=1}^{\infty} 5k$

Solution

ⓐ The ratio of the second term to the first is $\frac{2}{3}$, 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 $\frac{2}{3}$. 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 $\frac{1}{3}$. 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

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

Try It #10

$24+(−12)+6+(−3)+...$

Try It #11

$\sum_{k=1}^{\infty} 15\cdot(-0.3)^k$

Finding Sums of Infinite Series

When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first $n$ terms of a geometric series.

$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$

We will examine an infinite series with $r=\frac{1}{2}$. What happens to $r^{n}$ as $n$ increases?

$\begin{aligned} &\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \\ &\left(\frac{1}{2}\right)^{3}=\frac{1}{8} \\ &\left(\frac{1}{2}\right)^{4}=\frac{1}{16} \end{aligned}$

The value of $r^{n}$ decreases rapidly. What happens for greater values of $n$?

$\begin{aligned} &\left(\frac{1}{2}\right)^{10}=\frac{1}{1,024} \\ &\left(\frac{1}{2}\right)^{20}=\frac{1}{1,048,576} \\ &\left(\frac{1}{2}\right)^{30}=\frac{1}{1,073,741,824} \end{aligned}$

As $n$ gets very large,$r^{n}$ gets very small. We say that, as $n$ increases without bound, $r^{n}$ approaches 0. As $r^{n}$ approaches $0,1-r^{n}$ approaches 1. When this happens, the numerator approaches $a_{1}$. This give us a formula for the sum of an infinite geometric series.

Formula for the Sum of an Infinite Geometric Series

The formula for the sum of an infinite geometric series with $−1 < r < 1$ is

$S=\frac{a_1}{1−r}$

How To

Given an infinite geometric series, find its sum.

1. Identify $a_1$ and $r$.
   
2. Confirm that $–1 < r < 1$.
   
3. Substitute values for $a_1$ and $r$ into the formula, $S=\frac{a_1}{1−r}$.
   
4. Simplify to find $S$.
   

Example 7

Finding the Sum of an Infinite Geometric Series

Find the sum, if it exists, for the following:

ⓐ $10+9+8+7+…$

ⓑ $248.6+99.44+39.776+…$

ⓒ $\sum_{k=1}^{\infty} 4,374 \cdot (-\frac{1}{3})^{k-1}$

ⓓ $\sum_{k=1}^{\infty} \frac{1}{9}\cdot \left(\frac{4}{3}\right)^k$

Solution

ⓐ There is not a constant ratio; the series is not geometric.

ⓑ There is a constant ratio; the series is geometric. $a_{1}=248.6$ and $r=\frac{99.44}{248.6}=0.4$, so the sum exists. Substitute $a_{1}=248.6$ and $r=0.4$ into the formula and simplify to find the sum:

$\begin{aligned} &S=\frac{a_{1}}{1-r} \\ &S=\frac{248.6}{1-0.4}=414 . \overline{3} \end{aligned}$

ⓒ The formula is exponential, so the series is geometric with $r=-\frac{1}{3}$. Find $a_{1}$ by substituting $k=1$ into the given explicit formula:

$a_{1}=4,374 \cdot\left(-\frac{1}{3}\right)^{1-1}=4,374$

Substitute $a_{1}=4,374$ and $r=-\frac{1}{3}$ into the formula, and simplify to find the sum:

$\begin{aligned} &S=\frac{a_{1}}{1-r} \\ &S=\frac{4,374}{1-\left(-\frac{1}{3}\right)}=3,280.5 \end{aligned}$

ⓓ The formula is exponential, so the series is geometric, but $r > 1$. The sum does not exist.

Example 8

Finding an Equivalent Fraction for a Repeating Decimal

Find an equivalent fraction for the repeating decimal $0. \overline{3}$

Solution

We notice the repeating decimal $0. \overline{3}=0.333 \ldots$ so we can rewrite the repeating decimal as a sum of terms.

$0 . \overline{3}=0.3+0.03+0.003+\ldots$

Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to $0.1$ in the second term, and the second term multiplied to $0.1$ in the third term.

$0 . \overline{3}=0.3+(0.1) \underbrace{(0.3)}+(0.1) \underbrace{(0.1)(0.3)}$

First Term Second Term

Notice the pattern; we multiply each consecutive term by a common ratio of $0.1$ starting with the first term of $0.3$. So, substituting into our formula for an infinite geometric sum, we have

$S_{n}=\frac{a_{1}}{1-r}=\frac{0.3}{1-0.1}=\frac{0.3}{0.9}=\frac{1}{3}$.

Find the sum, if it exists.

Try It #12

$2+\frac{2}{3}+\frac{2}{9}+...$

Try It #13

$\displaystyle\sum_{k=1}^{\infty} 0.76 k+1$

Try It #14

$\displaystyle\sum_{k=1}^{\infty}\left(-\frac{3}{8}\right)^{k}$