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Proof of infinite geometric series as a limit

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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  • duskpin ultimate style avatar for user Michael
    What would happen if the absolute value of r wasn't between 0 and 1? Would the sum just be infinite?
    (21 votes)
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  • piceratops ultimate style avatar for user Mark
    So this equation only works when 'r' is between zero and one?
    (11 votes)
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  • piceratops tree style avatar for user jojishani
    What does Zeno's paradox has to do with this?
    (7 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      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.
      (32 votes)
  • aqualine ultimate style avatar for user Austin Markwell
    How can you find a common 'r' if the first number in the series is zero?
    (4 votes)
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    • male robot hal style avatar for user Yamanqui García Rosales
      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.
      (7 votes)
  • mr pink red style avatar for user rooffs
    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..
    (1 vote)
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  • ohnoes default style avatar for user Cyan Wind
    , 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?
    (5 votes)
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  • aqualine ultimate style avatar for user Ghada Muhammad
    I thought the formula is a(1-r^n)/1-r
    why did it expands to n+1 ? (a-ar^n+1)/1-r
    ?
    (5 votes)
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  • blobby green style avatar for user Olivia Schell
    How do you find the ratio if it isn't given?
    (2 votes)
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    • piceratops ultimate style avatar for user abassan
      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.
      (4 votes)
  • leaf green style avatar for user Ashwin
    Is there any difference between value and absolute value?
    (1 vote)
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  • leaf green style avatar for user Aleksandar Dragutinovic
    At 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?
    (2 votes)
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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.