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...
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James Sousa: Infinite Geometric Series
Guidance
Let’s return to the situation in the introduction: Poor Sayber is stuck cleaning his room. He cleans half of the room in 60mins. 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:
Now let’s look at larger values of

S7 =60(1−(12)7)1−12≈119.06 minutesS8 =60(1−(12)8)1−12≈119.5 minutesS10 =60(1−(12)10)1−12≈119.9 minutes
As n approaches infinity, the value of S_{n} seems to approach 120 minutes. 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:

limn→∞Sn =limn→∞⎛⎝⎜60(1−(12)n)1−12⎞⎠⎟ =limn→∞⎛⎝⎜60(1−(12)n)12⎞⎠⎟ =limn→∞(120(1−(12)n))
As n approaches infinity, the value of
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.

limn→∞Sn =limn→∞(a1(1−rn)1−r) =a11−r, as(1−rn)→1
Therefore, we can find the sum of an infinite geometric series using the formula
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
Example A
Find the sum of the convergent series:
Solution:
The common ratio is

401−(−12)=4032=40(23)=803
Example B
Determine if the series converges. If it converges, find the sum.

a) 1+13+19+127+... b) 3+−6+12+−24+...
Solution:

a)
1+13+19+127+... converges. 
 The common ratio is (1/3) . Therefore the sum converges to:


11−13=123=32

 b) 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.
Example C
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:
∑n=1∞20(12)n−1+∑n=1∞10(12)n−1 .  Each sum converges, as the common ratio is (1/2). Therefore the total distance is:


 So the ball travels a total vertical distance of 60 feet.
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Guided Practice
For questions 1 – 3, determine if the series converges or diverges. If it converges, find the sum.
1) 3 + 6 + 12 + 24 + ...
2) 240 + 60 + 15 + ...
3) 9 + 6 + 4 + (8/3) + ...
4) In this lesson, we proved the formula for the sum of a geometric series,
Prove this formula without induction:
 a) Let S_{n} = a_{1} + a_{1}r + a_{1}r^{2} + ... + a_{1}r^{n1}
 b) Multiply S_{n} by r to obtain a second equation
 c) Subtract the equations and solve for S_{n}.
5) 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.)
6) Below are two infinite series that are not geometric. Use a graphing calculator to examine partial sums. Does either series converge?

a)
1+12+13+14+... 
b)
1+14+19+116+...
Answers
1) The sum does not converge because r = 2.
2) The sum converges. S = 320.
3) The sum converges. S = 27.
4) Follow the steps below:

(1)
Sn=a1+a1r+a1r2+...+a1rn−1 
(2)
rSn=a1r+a1r2+a1r3+...+a1rn 
(3)
Sn−rSn=a1−a1rn 
⇒Sn(1−r)=a(1−rn) 
⇒Sn=a(1−rn)(1−r)
5) a)

b)
∑n=1∞50(14)n−1=6623
6) a) This series does not converge.

b) This series converges around 1.65. (The actual sum is
π26 )
Explore More
 Find the sum of the first 10 terms of
∑n=1∞(15)n using a graphing calculator.  Find the sum of the first 20 terms of
∑n=1∞(15)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+1−12+... 
−6+245−9625+... 
3+32−34+... 
−6+4−83+... 
1+12+14+...
Evaluate the infinite sum of each of the following geometric series:
 \begin{align*}\sum_{n=1}^{\infty} 3(\frac{1}{2})^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^{\infty} 2(\frac{4}{7})^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^{\infty} 7(\frac{4}{5})^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^{\infty} 9(\frac{1}{5})^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^{\infty} 5(\frac{5}{7})^{(n1)}\end{align*}
 \begin{align*}\sum_{n=1}^{\infty} 6(\frac{1}{5})^{(n1)}\end{align*}