7.4.2: Sums of Infinite Geometric Series
- Page ID
- 14796
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Sayber's mom told him to clean his room on Saturday morning.
"But, MOM! It's gonna take forever!" said Sayber.
"Oh, don't be overly dramatic," said mom.
"I am NOT being dramatic!" Sayber said.
"If I start right now, it is going to take me at LEAST an hour to clean this half alone, then it will take another half hour to clean half of the remainder, and 15 mins to clean half of THAT remainder... since I will always have half left, I will never be done!"
Do you agree with Sayber? Will Sayber be stuck with a vacuum in his hand forever? Tune in next week...
Sums of Infinite Geometric Series
Let’s return to the situation in the introduction: Poor Sayber is stuck cleaning his room. He cleans half of the room in 60 mins. Then he cleans half of what is left, 30 more minutes, half again for 15 more. If he keeps cleaning half of the remaining area, how will he ever finish the room?
We know that the pieces have to add up to some finite time period (no matter what it feels like, Sayber CAN get the room clean), 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. Three partial sums, relatively early in the series, could be: \(\ S_{2}=90\), \(\ S_{3}=105\), and \(\ S_{6}=118.125\) or \(\ 118 \frac{1}{8}\)
Now let’s look at larger values of \(\ n\):
| \(\ S_{7}\) | \(\ =\frac{60\left(1-\left(\frac{1}{2}\right)^{7}\right)}{1-\frac{1}{2}} \approx 119.06 \text { minutes }\) |
|---|---|
| \(\ S_{8}\) | \(\ =\frac{60\left(1-\left(\frac{1}{2}\right)^{8}\right)}{1-\frac{1}{2}} \approx 119.5 \text { minutes }\) |
| \(\ S_{10}\) | \(\ =\frac{60\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\frac{1}{2}} \approx 119.9 \text { minutes }\) |
As n approaches infinity, the value of Sn seems to approach 120 minutes. In terms of the actual sums, what is happening is this: as n increases, the nth term gets smaller and smaller, and so the nth term contributes less and less to the value of Sn. We say that the series converges, and we can write this with a limit:
| \(\ \lim _{n \rightarrow \infty} S_{n}\) | \(\ =\lim _{n \rightarrow \infty}\left(\frac{60\left(1-\left(\frac{1}{2}\right)^{n}\right)}{1-\frac{1}{2}}\right)\) |
|---|---|
| \(\ =\lim _{n \rightarrow \infty}\left(\frac{60\left(1-\left(\frac{1}{2}\right)^{n}\right)}{\frac{1}{2}}\right)\) | |
| \(\ =\lim _{n \rightarrow \infty}\left(120\left(1-\left(\frac{1}{2}\right)^{n}\right)\right)\) |
As n approaches infinity, the value of \(\ \left(\frac{1}{2}\right)^{n}\) gets smaller and smaller. That is, the value of this expression approaches 0. Therefore the value of \(\ 1-\left(\frac{1}{2}\right)^{n}\) approaches 1, and \(\ 120\left(1-\left(\frac{1}{2}\right)^{n}\right)\) approaches \(\ 120(1)=120\).
Therefore, no matter how long the process continues, Sayber will not spend more than 2hrs cleaning the room. Of course, it may SEEM like a lot more!
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.
| \(\ \lim _{n \rightarrow \infty} S_{n}\) | \(\ =\lim _{n \rightarrow \infty}\left(\frac{a_{1}\left(1-r^{n}\right)}{1-r}\right)\) |
|---|---|
| \(\ =\frac{a_{1}}{1-r}, \text { as }\left(1-r^{n}\right) \rightarrow 1\) |
Therefore, we can find the sum of an infinite geometric series using the formula \(\ S=\frac{a_{1}}{1-r}\).
When an infinite sum has a finite value, we say the sum converges. Otherwise, the sum diverges. A sum converges only when the terms get closer to 0 after each step, but that alone is not a sufficient criterion for convergence. For example, the sum \(\ \sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\) does not converge.
Examples
Find the sum of the convergent series: \(\ 40+-20+10+-5+\ldots\)
Solution
The common ratio is \(\ \frac{-1}{2}\). Therefore the sum converges to:
| \(\ \frac{40}{1-\left(\frac{-1}{2}\right)}=\frac{40}{\frac{3}{2}}=40\left(\frac{2}{3}\right)=\frac{80}{3}\) |
|---|
Determine if the series converges. If it converges, find the sum.
| a. \(\ 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\) | b. \(\ 3+-6+12+-24+\ldots\) |
|---|
Solution
- \(\ 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\) converges. The common ratio is (1/3) . Therefore the sum converges to:
\(\ \frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\) - 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.
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: \(\ \sum_{n=1}^{\infty} 20\left(\frac{1}{2}\right)^{n-1}+\sum_{n=1}^{\infty} 10\left(\frac{1}{2}\right)^{n-1}\).
Each sum converges, as the common ratio is (1/2). Therefore the total distance is:
| \(\ \frac{20}{1-\frac{1}{2}}+\frac{10}{1-\frac{1}{2}}=\frac{20}{\frac{1}{2}}+\frac{10}{\frac{1}{2}}=40+20=60\) |
|---|
So the ball travels a total vertical distance of 60 feet.
Determine if the following series converges or diverges. If it converges, find the sum.
240 + 60 + 15 + ...
Solution
The sum converges. S = 320.
In this lesson, we proved the formula for the sum of a geometric series, \(\ S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}\) using induction.
Prove this formula without induction:
Solution
Step 1: Let \(\ S_{n}=a_{1}+a_{1} r+a_{1} r^{2}+\ldots+a_{1} r^{n-1}\)
\(\ S_{n}=a_{1}+a_{1} r+a_{1} r^{2}+\ldots+a_{1} r^{n-1}\)
Step 2: Multiply \(\ S_{n}\) by \(\ r\) to obtain a second equation
\(\ r S_{n}=a_{1} r+a_{1} r^{2}+a_{1} r^{3}+\ldots+a_{1} r^{n}\)
Step 3: Subtract the equations and solve for \(\ S_{n}\).
\(\ \begin{array}{l}
S_{n}-r S_{n}=a_{1}-a_{1} r^{n} \\
\Rightarrow S_{n}(1-r)=a\left(1-r^{n}\right) \\
\Rightarrow S_{n}=\frac{a\left(1-r^{n}\right)}{(1-r)}
\end{array}\)
A ball is dropped from a height of 40 feet, and each time it bounces, it reaches 25% of its previous height.
- Find the total vertical distance the ball travels, using the method used in the lesson.
- Find the total vertical distance the ball travels using a single series.
Solution
- \(\ \sum_{n=1}^{\infty} 40\left(\frac{1}{4}\right)^{n-1}+\sum_{n=1}^{\infty} 20\left(\frac{1}{4}\right)^{n-1}=66 \frac{2}{3}\)
- \(\ \sum_{n=1}^{\infty} 50\left(\frac{1}{4}\right)^{n-1}=66 \frac{2}{3}\)
(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?
- \(\ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\)
- \(\ 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots\)
Solution
- This series does not converge.
- This series converges around 1.65. (The actual sum is \(\ \frac{\pi^{2}}{6}\))
Review
- Find the sum of the first 10 terms of \(\ \sum_{n=1}^{\infty}\left(\frac{1}{5}\right)^{n}\) using a graphing calculator.
- Find the sum of the first 20 terms of \(\ \sum_{n=1}^{\infty}\left(\frac{1}{5}\right)^{n}\) using a graphing calculator.
- Conjecture on the possible convergence of the series in questions 1 and 2.
Evaluate the infinite sum of each of the following geometric series:
- \(\ -2+1-\frac{1}{2}+\ldots\)
- \(\ -6+\frac{24}{5}-\frac{96}{25}+\ldots\)
- \(\ 3+\frac{3}{2}-\frac{3}{4}+\ldots\)
- \(\ -6+4-\frac{8}{3}+\ldots\)
- \(\ 1+\frac{1}{2}+\frac{1}{4}+\ldots\)
Evaluate the infinite sum of each of the following geometric series:
- \(\ \sum_{n=1}^{\infty}-3\left(\frac{1}{2}\right)^{(n-1)}\)
- \(\ \sum_{n=1}^{\infty}-2\left(\frac{4}{7}\right)^{(n-1)}\)
- \(\ \sum_{n=1}^{\infty} 7\left(\frac{-4}{5}\right)^{(n-1)}\)
- \(\ \sum_{n=1}^{\infty}-9\left(\frac{-1}{5}\right)^{(n-1)}\)
- \(\ \sum_{n=1}^{\infty} 5\left(\frac{-5}{7}\right)^{(n-1)}\)
- \(\ \sum_{n=1}^{\infty} 6\left(\frac{1}{5}\right)^{(n-1)}\)
Vocabulary
| Term | Definition |
|---|---|
| converge | If a series has a limit, and the limit exists, the series converges. |
| convergent | If a series has a limit, and the limit exists, the series is convergent. |
| divergent | If a series does not have a limit, or the limit is infinity, then the series is divergent. |
| diverges | If a series does not have a limit, or the limit is infinity, then the series diverges. |
| geometric sequence | A geometric sequence is a sequence with a constant ratio between successive terms. Geometric sequences are also known as geometric progressions. |
| geometric series | A geometric series is a geometric sequence written as an uncalculated sum of terms. |
| partial sums | A partial sum is the sum of the first ''n'' terms in an infinite series, where ''n'' is some positive integer. |