7.5: Geometric Series
Learning objectives
 Find the value of a finite geometric series.
 Determine if a geometric series converges or diverges.
 If an infinite series converges, find the sum.
Introduction
In lesson 2 we began working with series (sums), and in lesson 4 we proved formulas for certain sums, such as the sum of terms in an arithmetic series. In this lesson we will focus on geometric series, the sum of terms from a geometric sequence. We will prove a formula that allows us to find the sum of a finite number of terms, and we will return to the issue of the sum of an infinite number of terms.
Finite Geometric Series
A finite geometric series is simply a geometric series with a specific number of terms. For example, consider again the series introduced in lesson 2: 50+25+12.5+....The series is geometric: the first term is 50, and the common ration is (1/2).
 The sum of the first two terms is 50 + 25 = 75. We can write this as S_{2}=75
 The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S_{3}=87.5
To find the value of S_{n} in general, we could simply add together the first n terms in a series. However, this would obviously be tedious for a large value of n. Given the regular pattern in a geometric series  every term is (1/r) of the previous term, and the n^{th} term is a_{n} = a_{1}r^{n  1} , we can use induction to prove a formula for S_{n} .
Claim: The sum of the first n terms in a geometric series is \begin{align*}S_n = \frac{a_1(1  r^n)} {1  r}\end{align*}
Proof by induction:

1. If n = 1, the n^{th} sum is the first sum, or a_{1} . Using the hypothesized equation, we get \begin{align*}S_1 = \frac{a_1(1  r^1)} {1  r} = \frac{a_1(1  r)} {1  r} = a_1\end{align*}
S1=a1(1−r1)1−r=a1(1−r)1−r=a1 . This establishes the base case. 
2. Assume that the sum of the first k terms in a geometric series is \begin{align*}S_k = \frac{a_1(1  r^k)} {1  r}\end{align*}
Sk=a1(1−rk)1−r . 
3. Show that the sum of the first k+1 terms in a geometric series is \begin{align*}S_{k + 1} = \frac{a_1(1  r^{k + 1})} {1  r}\end{align*}
Sk+1=a1(1−rk+1)1−r . 

\begin{align*}S_{k + 1} = S_k + a_{k + 1}\end{align*} Sk+1=Sk+ak+1 The \begin{align*}k + 1\end{align*} k+1 sum is the \begin{align*}k^{th}\end{align*}kth sum, plus the \begin{align*}k + 1\end{align*}k+1 term\begin{align*} = \frac{a_1(1  r^k)} {1  r} + a_1r^{k + 1  1}\end{align*} =a1(1−rk)1−r+a1rk+1−1 Substitute from step 2, and substitute the \begin{align*}k + 1\end{align*} k+1 term\begin{align*}= \frac{a_1(1  r^k)} {1  r} + \frac{a_1r^k(1  r)} {1 r}\end{align*} =a1(1−rk)1−r+a1rk(1−r)1−r The common denominator is \begin{align*}1  r\end{align*} 1−r \begin{align*}= \frac{a_1(1  r^k) + a_1r^k(1  r)} {1  r}\end{align*} =a1(1−rk)+a1rk(1−r)1−r Simplify the fraction \begin{align*}= \frac{a_1 \left[1  r^k + r^k(1  r) \right]} {1  r}\end{align*} =a1[1−rk+rk(1−r)]1−r \begin{align*}= \frac{a_1 \left[1  r^k + r^k  r^{k + 1} \right]} {1  r}\end{align*} =a1[1−rk+rk−rk+1]1−r \begin{align*}= \frac{a_1 \left[1  r^{k + 1} \right]} {1  r}\end{align*} =a1[1−rk+1]1−r It is proven.

Therefore we have shown that \begin{align*}S_{n}=\frac{a_1(1  r^n)} {1  r}\end{align*}
\begin{align*}S_n = \frac{a_1(1  r^n)} {1  r} = \frac{50 \left(1  (\frac{1} {2})^6 \right)} {1  \frac{1} {2}} = \frac{50 \left(1  \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}\end{align*}
The figure below shows the same calculation on a TI83/4 calculator:
We can use this formula, as long as the series in question is geometric.
Example 1: Find the sum of the first 10 terms of a geometric series with a_{1} = 3 and r = 5.
Solution: The sum is 58,593.
 \begin{align*}S_n = \frac{a_1(1  r^n)} {1  r} = \frac{3(1  5^7)} {1  5} = \frac{3(1  78,125)} {4} = \frac{3(78,124)} {4} = 58,593\end{align*}
 Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio (1/2), the terms gets smaller and smaller. This situation implies something important about the sum.
Infinite Geometric Series
Now let’s return to the situation in the introduction in lesson 2: you are walking a 100mile charity walk. You walk halfway, 50 miles. Then you walk half of that, 25. If you keep walking half of the remaining distance, how will you ever finish the walk?
At the end of lesson 2 we represented the sum of the infinitely many pieces of this walk as: \begin{align*}\sum_{n = 1}^\infty 50 \left(\frac{1} {2} \right)^{n  1}\end{align*} . We know that the pieces have to add up to 100 miles. But how is it possible for the sum of an infinite number of terms to be a finite number?
To find the sum of an infinite number of terms, we should consider some partial sums. Above we found three partial sums: \begin{align*}S_2 = 75 , S_3 = 87.5,\end{align*} and \begin{align*}S_6 = 98 \frac{7} {16}\end{align*} or \begin{align*}98.4375.\end{align*} Now let’s look at larger values of \begin{align*}n:\end{align*}
\begin{align*}S_7\end{align*}  \begin{align*}= \frac{50 \left(1  (\frac{1} {2})^7 \right)} {1  \frac{1} {2}} \approx 99.2\end{align*}  

\begin{align*}S_8\end{align*}  \begin{align*}= \frac{50 \left(1  (\frac{1} {2})^8 \right)} {1  \frac{1} {2}} \approx 99.6\end{align*}  
\begin{align*}S_{10}\end{align*}  \begin{align*}= \frac{50 \left(1  (\frac{1} {2})^{10} \right)} {1  \frac{1} {2}} \approx 99.9\end{align*} 
As n approaches infinity, the value of S_{n} seems to approach 100. In terms of the actual sums, what is happening is this: as n increases, the n^{th} term gets smaller and smaller, and so the n^{th} term contributes less and less to the value of S_{n} . We say that the series converges, and we can write this with a limit:
\begin{align*}\lim_{n \rightarrow \infty} S_n\end{align*}  \begin{align*}= \lim_{n \rightarrow \infty} \left(\frac{50 \left(1  (\frac{1} {2})^{n} \right)} {1  \frac{1} {2}} \right )\end{align*}  

\begin{align*}= \lim_{n \rightarrow \infty} \left(\frac{50 \left(1  (\frac{1} {2})^{n} \right)} {\frac{1} {2}} \right )\end{align*}  
\begin{align*}= \lim_{n \rightarrow \infty} \left(100 \left(1  \left(\frac{1} {2} \right)^{n} \right ) \right) \end{align*} 
As n approaches infinity, the value of \begin{align*}\left(\frac{1} {2} \right)^{n}\end{align*} gets smaller and smaller. That is, the value of this expression approaches 0. Therefore the value of \begin{align*}1  \left(\frac{1} {2} \right)^{n}\end{align*} approaches 1, and \begin{align*}100 \left(1  \left(\frac{1} {2} \right)^n \right)\end{align*} approaches 100(1) = 100.
We can do the same analysis for the general case of a geometric series, as long as the terms are getting smaller and smaller. This means that the common ratio must be a number between 1 and 1: r < 1.
\begin{align*}\lim_{n \rightarrow \infty} S_n\end{align*}  \begin{align*}= \lim_{n \rightarrow \infty} \left (\frac{a_1(1  r^n)} {1  r} \right)\end{align*}  

\begin{align*}=\frac{a_1} {1  r},\end{align*} as \begin{align*}(1  r^n) \rightarrow 1\end{align*} 
Therefore, we can find the sum of an infinite geometric series using the formula \begin{align*}S = \frac{a_1} {1  r}\end{align*}.
Example 2: Determine if the series converges. If it converges, find the sum.

a.\begin{align*}40 + 20 + 10 + 5 + ...\end{align*} b.\begin{align*}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27}+ ... \end{align*} c.\begin{align*}3 + 6 +12 + 24 + ... \end{align*}
Solution:
 a. \begin{align*}40 + 20 + 10 + 5 + ... \end{align*} converges

 The common ratio is \begin{align*}\frac{1} {2}\end{align*}. Therefore the sum converges to:


\begin{align*}\frac{40} {1  \left (\frac{1} {2} \right)} = \frac{40} {\frac{3} {2}} = 40 \left(\frac{2} {3} \right) = \frac{80} {3}\end{align*}

 b. \begin{align*}1 + \frac{1} {3} + \frac{1} {9} + \frac{1} {27} + ... \end{align*} converges.

 The common ratio is (1/3) . Therefore the sum converges to:


\begin{align*}\frac{1} {1  \frac{1} {3}} = \frac{1} {\frac{2} {3}} = \frac{3} {2}\end{align*}

 c. The series 3 + 6 + 12 + 24 + ... does not converge, as the common ratio is 2.
Remember that the idea of an infinite sum was introduced in the context of a realistic situation, albeit a paradoxical one. We can in fact use infinite geometric series to model other realistic situations. Here we will look at another example: the total vertical distance traveled by a bouncing ball.
The Infinitely Bouncing Ball
Consider a bouncing ball that bounces to a particular percent or fraction of its previous height with each bounce. If the ball is left alone, theoretically, it will continue this pattern infinitely.
We can think of the total distance as the distance the ball travels down plus the distance the ball travels back up. The downward bounces form a geometric series, and the upward bounces form a geometric series. So we can think of the total distance as the sum of two series. (You can also develop one series, which you will do in the review questions). The example below shows how to find the total distance. Example 3: A bouncing ball
 A ball is dropped from a height of 20 feet. Each time it bounces, it reaches 50% of its previous height. What is the total vertical distance the ball travels?
Solution:

We can think of the total distance as the distance the ball travels down + the distance the ball travels back up. The downward bounces form a geometric series:
 20 + 10 + 5 +...
 The upward bounces form the same series, except the first term is 10.
 So the total distance is: \begin{align*}\sum_{n=1}^\infty 20 \left ( \frac{1}{2} \right )^{n1}+\sum_{n=1}^\infty 10 \left ( \frac{1}{2} \right )^{n1}\end{align*}.
 Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
\begin{align*}\frac {20}{1\frac {1}{2}}+\frac {10}{1\frac {1}{2}}=\frac {20}{\frac {1}{2}}+\frac {10}{\frac {1}{2}}=40+20=60\end{align*} 

 So the ball travels a total vertical distance of 60 feet.
Lesson Summary
In this lesson we have found the sums of finite geometric series, and we have returned to the idea of an infinite series. A geometric series is either convergent or divergent. In this lesson you have learned to determine if a geometric series converges or diverges: a geometric series will converge if r < 1 , and the sum is \begin{align*}S=\frac{a_{1}}{1r}\end{align*}. It is important to note that other kinds of series converge besides geometric series. You will examine these other types of series when you study calculus.
Points to Consider
 How can infinitely many numbers add up to a finite number?
 How does induction help us prove formulas that simplify large sums?
Review Questions
 Find the sum of each series: a. The first term of a geometric series is 4, and the common ratio is 3. Find S_{8}. b. The first term of a geometric series is 80, and the common ratio is (1/4). Find S_{7}.
 Find the sum: 5+10+20+...+640 (Hint: if a_{n}=640 , what is n?)
 Use a geometric series to answer the question: In January, a company’s sales totaled $11,000. It is predicted that the company’s sales will increase 5% each month for the next year. At this rate, what will be the total sales for the year?
 For questions 4 – 6, determine if the series converges or diverges. If it converges, find the sum.
 3+6+12+24+...
 240+60+15+...
 9+6+4+(8/3)+...
 In this lesson, we proved the formula for the sum of a geometric series, \begin{align*}S_n=\frac {a_1(1r^n)}{1r}\end{align*}, using induction. Here you will prove this formula without induction: (1) Let S_{n}=a_{1}+a_{1}r+a_{1}r^{2}+...+a_{1}r^{n1} (2) Multiply S_{n} by r to obtain a second equation (3) Subtract the equations and solve for S_{n}.
 A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height. a. Find the total vertical distance the ball travels, using the method used in the lesson. b. Find the total vertical distance the ball travels using a single series. (Hint: write out several terms for each bounce. For example, the first bounce is: 40 feet down + 10 feet up = 50 feet traveled.)
 Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does either series converge? a. \begin{align*}1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+...\end{align*} b. \begin{align*}1+\frac {1}{4}+\frac {1}{9}+\frac {1}{16}+...\end{align*}
Review Answer
 a.\begin{align*}S_{8}=\frac {4(13^8)}{13}=13,120\end{align*} b.\begin{align*}S_{7}= \frac {80 \left ( {1 \left ( \frac{1}{4} \right )^7} \right )}{1\frac {1}{4}} \approx 106.66 \end{align*}
 \begin{align*}S_{8}=\frac {5(12^8)}{12}=1275 \end{align*}
 \begin{align*}S_{12}=\frac {11000(11.05^{12})}{11.05}= \$175,088.39 \end{align*}
 The sum does not converge because r = 2.
 The sum converges. S = 320.
 The sum converges. S = 27.
 Follow the steps below:

(1) \begin{align*} S_n=a_1+a_1{r}+a_1{r^2}+...+a_1{r^{n1}}\end{align*} (2) \begin{align*}r{S_n}=a_1{r}+a_1{r^2}+a_1{r^3}+...+a_1{r^n}\end{align*} (3) \begin{align*} S_nrS_n=a_1a_1{r^n}\end{align*} \begin{align*}\Rightarrow S_n(1r)=a(1r^n)\end{align*} \begin{align*}\Rightarrow S_n=\frac {a(1r^n)}{(1r)}\end{align*}  a.\begin{align*}\sum_{n1}^\infty 40 \left ( \frac{1}{4} \right )^{n1}+\sum_{n1}^\infty 20 \left ( \frac{1}{4} \right )^{n1}=66\frac {2}{3}\end{align*} b.\begin{align*}\sum_{n1}^\infty 50 \left ( \frac{1}{4} \right )^{n1}=66\frac {2}{3}\end{align*}
 a.This series does not converge. b.This series converges around 1.65. (The actual sum is \begin{align*}\frac {\pi^2}{6}\end{align*}.)
Vocabulary
 Converge
 If the limit of the partial sums of series exists and is finite, the series converges.
 Diverge
 If the limit of the partial sums of series does not exist or is infinite, then the series diverges.
 Geometric series
 A geometric series is the sum of terms from a geometric sequence.
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