<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Sums of Infinite Geometric Series

## Identify series with calculable sums

0%
Progress
Practice Sums of Infinite Geometric Series
Progress
0%
Finding the Sum of an Infinite Geometric Series

Your task as Agent Infinite Geometric Series, should you choose to accept it, is to find the sum of the geometric series \begin{align*}\sum \limits_{n=1}^{\infty}3 \left(\frac{1}{3}\right)^{n-1}\end{align*}.

### Guidance

In the previous concept we explored partial sums of various infinite series and observed their behavior as \begin{align*}n\end{align*} became large to see if the sum of the infinite series was finite. Now we will focus our attention on geometric series. Look at the partial sums of the infinite geometric series below:

Series \begin{align*}\sum \limits_{n=1}^{\infty}3(1)^{n-1}\end{align*} \begin{align*}\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}\end{align*} \begin{align*}\sum \limits_{n=1}^{\infty}5 \left(\frac{6}{5}\right)^{n-1}\end{align*} \begin{align*}\sum \limits_{n=1}^{\infty}(-2)^{n-1}\end{align*} \begin{align*}\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}
\begin{align*}S_5\end{align*} 15 30.508 37.208 11 1.506
\begin{align*}S_{10}\end{align*} 30 37.747 129.793 \begin{align*}-341\end{align*} 1.5
\begin{align*}S_{50}\end{align*} 150 40 227485.954 \begin{align*}-3.753 \times 10^{14}\end{align*} 1.5
\begin{align*}S_{100}\end{align*} 300 40 2070449338 \begin{align*}-4.226 \times 10^{29}\end{align*} 1.5

From the table above, we can see that the two infinite geometric series which have a finite sum are \begin{align*}\sum \limits_{n=1}^{\infty}10 \left(\frac{3}{4}\right)^{n-1}\end{align*} and \begin{align*}\sum \limits_{n=1}^{\infty}2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}. The two series both have a common ratio, \begin{align*}r\end{align*}, such that \begin{align*}\left | r \right \vert < 1\end{align*} or \begin{align*}-1 < r < 1\end{align*}.

Take a look at the formula for the sum of a finite geometric series: \begin{align*}S_n=\frac{a_1\left(1-r^n\right)}{1-r}\end{align*}. What happens to \begin{align*}r^n\end{align*} if we let \begin{align*}n\end{align*} get very large for an \begin{align*}r\end{align*} such that \begin{align*}\left |r \right \vert <1\end{align*}? Let’s take a look at some examples.

\begin{align*}r\end{align*} values \begin{align*}r^5\end{align*} \begin{align*}r^{25}\end{align*} \begin{align*}r^{50}\end{align*} \begin{align*}\ldots\end{align*} \begin{align*}r^n\end{align*} or \begin{align*}r^{\infty}\end{align*}
\begin{align*}\frac{5}{6}\end{align*} \begin{align*}0.40188\end{align*} \begin{align*}0.01048\end{align*} \begin{align*}0.00011\end{align*} \begin{align*}0\end{align*}
\begin{align*}- \frac{4}{5}\end{align*} \begin{align*}-0.32768\end{align*} \begin{align*}-0.00378\end{align*} \begin{align*}0.00001\end{align*} \begin{align*}0\end{align*}
\begin{align*}1.1\end{align*} \begin{align*}1.61051\end{align*} \begin{align*}10.83471\end{align*} \begin{align*}117.39085\end{align*} keeps growing
\begin{align*}- \frac{1}{3}\end{align*} \begin{align*}-0.00412\end{align*} \begin{align*}-1.18024 \times 10^{-12}\end{align*} \begin{align*}1.39296 \times 10^{-24}\end{align*} \begin{align*}0\end{align*}

This table shows that when \begin{align*}\left | r \right \vert <1\end{align*}, \begin{align*}r^n=0\end{align*}, for large values of \begin{align*}n\end{align*}. Therefore, for the sum of an infinite geometric series in which \begin{align*}\left | r \right \vert <1, S_{\infty}=\frac{a_1 \left(1-r^n\right)}{1-r}=\frac{a_1\left(1-0\right)}{1-r}=\frac{a_1}{1-r}\end{align*}.

#### Example A

Find the sum of the geometric series if possible. \begin{align*}\sum \limits_{n=1}^{\infty}100 \left(\frac{8}{9}\right)^{n-1}\end{align*}.

Solution: Using the formula with \begin{align*}a_1=100\end{align*}, \begin{align*}r=\frac{8}{9}\end{align*}, we get \begin{align*}S_{\infty}=\frac{100}{1- \frac{8}{9}}=\frac{100}{\frac{1}{9}}=900\end{align*}.

#### Example B

Find the sum of the geometric series if possible. \begin{align*}\sum \limits_{n=1}^{\infty}9 \left(\frac{4}{3}\right)^{n-1}\end{align*}.

Solution: In this case, \begin{align*}\left | r \right \vert=\frac{4}{3}>1\end{align*}, therefore the sum is infinite and cannot be determined.

#### Example C

Find the sum of the geometric series if possible. \begin{align*}\sum \limits_{n=1}^{\infty}5(0.99)^{n-1}\end{align*}

Solution: In this case \begin{align*}a_1=5\end{align*} and \begin{align*}r=0.99\end{align*}, so \begin{align*}S_{\infty}=\frac{5}{1-0.99}=\frac{5}{0.01}=500\end{align*}.

Intro Problem Revisit In this case \begin{align*}a_1=3\end{align*} and \begin{align*}r=\frac{1}{3}\end{align*}, so \begin{align*}S_{\infty}=\frac{3}{1-\frac{1}{3}}=\frac{3}{\frac{2}{3}}=\frac{9}{2}=4.5\end{align*}.

### Guided Practice

Find the sums of the following infinite geometric series, if possible.

1. \begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{9} \left(- \frac{3}{2}\right)^{n-1}\end{align*}

2. \begin{align*}\sum \limits_{n=1}^{\infty}4 \left(\frac{7}{8}\right)^{n-1}\end{align*}

3. \begin{align*}\sum \limits_{n=1}^{\infty}3(-1)^{n-1}\end{align*}

1. \begin{align*}\left | r \right \vert=\left |- \frac{3}{2} \right \vert=\frac{3}{2}>1\end{align*} so the infinite sum does not exist.

2. \begin{align*}a_1=4\end{align*} and \begin{align*}r=\frac{7}{8}\end{align*} so \begin{align*}S_{\infty}=\frac{4}{1- \frac{7}{8}}=\frac{4}{\frac{1}{8}}=32\end{align*}.

3. \begin{align*}\left | r \right \vert=\left |-1 \right \vert=1 \ge 1\end{align*}, therefore the infinite sum does not converge. If we observe the behavior of the first few partial sums we can see that they oscillate between 0 and 3.

This pattern will continue so there is no determinable sum for the infinite series.

### Explore More

Find the sums of the infinite geometric series, if possible.

1. \begin{align*}\sum \limits_{n=1}^{\infty} 5 \left(\frac{2}{3}\right)^{n-1}\end{align*}
2. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{10} \left(- \frac{4}{3}\right)^{n-1}\end{align*}
3. \begin{align*}\sum \limits_{n=1}^{\infty} 2 \left(- \frac{1}{3}\right)^{n-1}\end{align*}
4. \begin{align*}\sum \limits_{n=1}^{\infty} 8(1.1)^{n-1}\end{align*}
5. \begin{align*}\sum \limits_{n=1}^{\infty} 6(0.4)^{n-1}\end{align*}
6. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{3}{7}\right)^{n-1}\end{align*}
7. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{5}{3} \left(\frac{1}{6}\right)^{n-1}\end{align*}
8. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{5} (1.05)^{n-1}\end{align*}
9. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{4}{7} \left(\frac{6}{7}\right)^{n-1}\end{align*}
10. \begin{align*}\sum \limits_{n=1}^{\infty} 15 \left(\frac{11}{12}\right)^{n-1}\end{align*}
11. \begin{align*}\sum \limits_{n=1}^{\infty} 0.01 \left(\frac{3}{2}\right)^{n-1}\end{align*}
12. \begin{align*}\sum \limits_{n=1}^{\infty} 100 \left(\frac{1}{5}\right)^{n-1}\end{align*}
13. \begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{2} \left(\frac{5}{4}\right)^{n-1}\end{align*}
14. \begin{align*}\sum \limits_{n=1}^{\infty} 2.5(0.85)^{n-1}\end{align*}
15. \begin{align*}\sum \limits_{n=1}^{\infty} -3 \left(\frac{9}{16}\right)^{n-1}\end{align*}

### Vocabulary Language: English

partial sums

partial sums

A partial sum is the sum of the first ''n'' terms in an infinite series, where ''n'' is some positive integer.