What would happen if the absolute value of r wasn't between 0 and 1? Would the sum just be infinite?

Yes, the limit would be either positive or negative infinity. Or, as is usually stated in these cases we would say that "the series does not converge".

So this equation only works when 'r' is between zero and one?

Actually |r|<1 i.e -1<r<1. This the required condition for the equation to work. Hope it helped

What does Zeno's paradox has to do with this?

One of Zeno's paradoxes says that an arrow cannot reach its target because it first travels half the distance, then half the remaining distance, then half the remaining distance... The total distance the arrow goes can be represented by a geometric series: 1/2 + 1/4 + 1/8 + 1/16 + ... = ∑ (1/2)^n from n=1 to oo (infinity) As the geometric series approaches an infinite number of terms, the sum approaches 1. What does this mean? The arrow of the paradox ultimately reaches its target. It takes an infinite number of steps to do it, but each step is also shorter. Thus you can piece the arrow's path into an infinite number of sections, but it is still going to complete its path in a finite time.

2:10, if `|r| > 1`, the number will become massively huge. How can we come up with the formula `Σ = a / (1 - r) if |r| > 1`? Please notice that I fully understand `Σ = a / (1 - r) if 0 < |r| < 1`. Did I miss something there?

You're right--the formula Σ = a / (1 - r) does not apply if |r| > 1. So our formula for the sum of a geometric series only applies if |r| < 1, which Sal begins to address around 2:30.

How can you find a common 'r' if the first number in the series is zero?

A geometric series cannot have it's first term be 0, since all other numbers of the series are created by multiplying the first term by the common ratio, and anything multiplied by 0 would continue to be 0, so the series would be something like: 0+0+0+0+0+...+0 If for some reason you discover that a geometric series has a first term equal to 0, then you can easily say that the total sum will also be 0.

How do you find the ratio if it isn't given?

See what you have to multiply to go from one term to the next and that's the ratio. If it's not obvious to you, you can find the first term and the second term. Divide the second term by the first term and that's your ratio.

At 2:10 he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?

No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.

Hello, i need help with finding the sum of a geometric series i would appreciate it very much if you help, since i am not able to find the ratio i cant find the sum. (5/2^n - 1/3^n) with Sigma that has an index of n=0 and tends to infinity..

The sum of this series is equal to the difference of the series 5/2^n and 1/3^n, the first of which is divergent and goes to infinity.

Is there any difference between value and absolute value?

A value can be positive or negative. An absolute value is always positive.

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Algebra (all content)

Course: Algebra (all content) > Unit 18

Lesson 7: Infinite geometric series

Proof of infinite geometric series as a limit

Name: Proof of infinite geometric series as a limit
Uploaded: 2013-11-08T19:42:24Z
Description: Sal applies limits to the formula for the sum of a finite geometric series to get the sum of an infinite geometric series.

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Sal applies limits to the formula for the sum of a finite geometric series to get the sum of an infinite geometric series. Created by Sal Khan.

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Michael
Posted 10 years ago. Direct link to Michael's post “What would happen if the ...”
What would happen if the absolute value of r wasn't between 0 and 1? Would the sum just be infinite?
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(21 votes)
Answer
- Just Keith
  Posted 10 years ago. Direct link to Just Keith's post “Yes, the limit would be e...”
  Yes, the limit would be either positive or negative infinity. Or, as is usually stated in these cases we would say that "the series does not converge".
  Comment on Just Keith's post “Yes, the limit would be e...”
  (51 votes)
Mark
Posted 10 years ago. Direct link to Mark's post “So this equation only wor...”
So this equation only works when 'r' is between zero and one?
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(11 votes)
Answer
- Anirban Mandal
  Posted 10 years ago. Direct link to Anirban Mandal's post “Actually |r|<1 i.e -1<r<1...”
  Actually |r|<1 i.e -1<r<1. This the required condition for the equation to work. Hope it helped
  Button navigates to signup page
  (22 votes)
jojishani
Posted 9 years ago. Direct link to jojishani's post “What does Zeno's paradox ...”
What does Zeno's paradox has to do with this?
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(7 votes)
Answer
- Ethan Dlugie
  Posted 9 years ago. Direct link to Ethan Dlugie's post “One of Zeno's paradoxes s...”
  One of Zeno's paradoxes says that an arrow cannot reach its target because it first travels half the distance, then half the remaining distance, then half the remaining distance...
  The total distance the arrow goes can be represented by a geometric series:
  1/2 + 1/4 + 1/8 + 1/16 + ... = ∑ (1/2)^n from n=1 to oo (infinity)
  As the geometric series approaches an infinite number of terms, the sum approaches 1.
  What does this mean? The arrow of the paradox ultimately reaches its target. It takes an infinite number of steps to do it, but each step is also shorter. Thus you can piece the arrow's path into an infinite number of sections, but it is still going to complete its path in a finite time.
  Button navigates to signup page
  (32 votes)
Austin Markwell
Posted 9 years ago. Direct link to Austin Markwell's post “How can you find a common...”
How can you find a common 'r' if the first number in the series is zero?
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(4 votes)
Answer
- Yamanqui García Rosales
  Posted 9 years ago. Direct link to Yamanqui García Rosales's post “A geometric series cannot...”
  A geometric series cannot have it's first term be 0, since all other numbers of the series are created by multiplying the first term by the common ratio, and anything multiplied by 0 would continue to be 0, so the series would be something like: 0+0+0+0+0+...+0
  If for some reason you discover that a geometric series has a first term equal to 0, then you can easily say that the total sum will also be 0.
  Comment on Yamanqui García Rosales's post “A geometric series cannot...”
  (7 votes)
rooffs
Posted 8 years ago. Direct link to rooffs's post “Hello, i need help with f...”
Hello, i need help with finding the sum of a geometric series i would appreciate it very much if you help, since i am not able to find the ratio i cant find the sum. (5/2^n - 1/3^n) with Sigma that has an index of n=0 and tends to infinity..
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Thomas B
  Posted 8 years ago. Direct link to Thomas B's post “The sum of this series is...”
  The sum of this series is equal to the difference of the series 5/2^n and 1/3^n, the first of which is divergent and goes to infinity.
  Button navigates to signup page
  (3 votes)
Cyan Wind
Posted 9 years ago. Direct link to Cyan Wind's post “2:10, if `|r| > 1`, the n...”
2:10
, if |r| > 1, the number will become massively huge. How can we come up with the formula Σ = a / (1 - r) if |r| > 1? Please notice that I fully understand Σ = a / (1 - r) if 0 < |r| < 1. Did I miss something there?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Daniel Schneider
  Posted 9 years ago. Direct link to Daniel Schneider's post “You're right--the formula...”
  You're right--the formula Σ = a / (1 - r) does not apply if |r| > 1. So our formula for the sum of a geometric series only applies if |r| < 1, which Sal begins to address around
  2:30
  .
  Comment on Daniel Schneider's post “You're right--the formula...”
  (3 votes)
Ghada Muhammad
Posted 6 years ago. Direct link to Ghada Muhammad's post “I thought the formula is ...”
I thought the formula is a(1-r^n)/1-r
why did it expands to n+1 ? (a-ar^n+1)/1-r
?
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(5 votes)
Answer
Olivia Schell
Posted 9 years ago. Direct link to Olivia Schell's post “How do you find the ratio...”
How do you find the ratio if it isn't given?
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(2 votes)
Answer
- abassan
  Posted 9 years ago. Direct link to abassan's post “See what you have to mult...”
  See what you have to multiply to go from one term to the next and that's the ratio. If it's not obvious to you, you can find the first term and the second term. Divide the second term by the first term and that's your ratio.
  Button navigates to signup page
  (4 votes)
Ashwin
Posted 9 years ago. Direct link to Ashwin's post “Is there any difference b...”
Is there any difference between value and absolute value?
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(1 vote)
Answer
- rivkabubbie
  Posted 9 years ago. Direct link to rivkabubbie's post “A value can be positive o...”
  A value can be positive or negative. An absolute value is always positive.
  Comment on rivkabubbie's post “A value can be positive o...”
  (6 votes)
Aleksandar Dragutinovic
Posted 9 years ago. Direct link to Aleksandar Dragutinovic's post “At 2:10 he says that if r...”
At
2:10
he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?
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(2 votes)
Answer
- Yamanqui García Rosales
  Posted 9 years ago. Direct link to Yamanqui García Rosales's post “No, the limit of 0/0 is u...”
  No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.
  Comment on Yamanqui García Rosales's post “No, the limit of 0/0 is u...”
  (3 votes)

Video transcript

In a previous video, we derived the formula for the sum of a finite geometric series where a is the first term and r is our common ratio. What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind blowing because, or actually more than mildly mind blowing, because you're taking the sum of an infinite things but as we see, you can actually get a finite value depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just a limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals zero to n of a times r to the k. Which would be the same thing as taking the limit as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't have to keep switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than one, for r is equal to one, and actually let me make it clear-- let's think about it for the absolute values of r is greater than one, the absolute values of r equal to one, and then the absolute value of r less than one. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than one, as this exponent explodes, as it approaches infinity, this number is just going to become massively, massively huge. And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to one, then the denominator is going to become zero. And we're going to be dividing by that denominator, and this formula just breaks down. But where this formula can be helpful, and where we can get this to actually give us a sensical result, is when the absolute value of r is between zero and one. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to zero. So let's think about the case where the absolute value of r is greater than zero, and it is less than one. What's going to happen in that case? Well, the denominator is going to make sense, right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than one, and you take it to higher and higher and higher exponents, every time you multiply it by itself, you're going to get a number with a smaller absolute value. So this term right over here, this entire term, is going to go to zero as n approaches infinity. Imagine if r was 1/2. You're talking about 1/2 to the hundredth power, 1/2 to the thousandth power, 1/2 to the millionth power, 1/2 to the billionth power. That quickly approaches zero. So this goes to zero if the absolute value of r is less than one. So this, we could argue, would be equal to a over one minus r. So for example, if I had the geometric series, if I had the infinite geometric series-- let's just have a simple one. Let's say that my first term is one, and then each successive term I'm going to multiply by 1/3. So it's one plus 1/3 plus 1/3 squared plus 1/3 to the third plus, and I were to just keep on going forever. This is telling us that that sum, this infinite sum-- I have an infinite number of terms here-- this is a pretty fascinating concept here-- will come out to this. It's going to be my first term, one, over one minus my common ratio. My common ratio in this case is 1/3. One minus 1/3, which is the same thing as one over 2/3, which is equal to 3/2, or you could view it as one and 1/2. That's a mildly amazing thing.