If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Proof of infinite geometric series as a limit

The video provides a proof for the sum of an infinite geometric series using limits. When the absolute value of the common ratio (r) is between 0 and 1, the limit of the series converges to a finite sum. The formula for the sum is a / (1 - r), where a is the first term. Created by Sal Khan.

Want to join the conversation?

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.